Page 5.6 Ex. 5.1
Q1.
Answer :
We know that if a matrix is of order m×n, then it has mn elements.
The possible orders of a matrix with 8 elements are given below:
1×8, 2×4, 4×2, 8×1
Thus, there are 4 possible orders of the matrix.
The possible orders of a matrix with 5 elements are given below:
1×5, 5×1
Thus, there are 2 possible orders of the matrix.
Q2.
Answer :
i) a22+b21
Here,a22=4 and b21=-3 ⇒a22+b21=4-3=1
ii) a11b11+a22b22
Here,a11=2, b11=2, a22=4 and b22=4⇒a11b11+a22b22=2×2+4×4⇒a11b11+a22b22=4+16⇒a11b11+a22b22=20
Q3.
Answer :
The order of R1 is 1×4 and the order of C2 is 3×1.
Q4.
Answer :
i) Here,
aij=i . j, 1≤i≤2 and 1≤j≤3
a11=1×1=1, a12=1×2=2, a13=1×3=3a21=2×1=2, a22=2×2=4 and a23=2×3=6
Required matrix = A = 123246
ii) Here,
aij=2i-j
a11=21-1=2-1=1 , a12=21-2=2-2=0 , a13=21-3=2-3=-1a21=22-1=4-1=3 , a22=22-2=4-2=2 and a23=22-3=4-3=1
Required matrix = A = 10-1321
iii) Here,
aij=i+j
a11=1+1=2 , a12=1+2=3 , a13=1+3=4a21=2+1=3 , a22=2+2=4 and a23=2+3=5
Required matrix = A = 234345
iv) Here,
aij=i+j22
a11=1+122=222=42=2 , a12=1+222=322=92 , a13=1+322=422=162=8a21=2+122=322=92 , a22=2+222=422=162=8 and a23=2+322=522=252
Required matrix = A = 2928928252
Q5.
Answer :
i) i+j22
Here,
a11=1+122=222=42=2 , a12=1+222=322=92a21=2+122=322=92 , a22=2+222=422=162=8
So, the required matrix is 2 9292 8.
ii) aij=i-j22
Here,
a11=1-122=022=02=0 , a12=1-222=-122=12a21=2-122=122=12 , a22=2-222=022=02=0
So, the required matrix is 012120.
iii) aij=i-2j22
Here,
a11=1-2122=1-222=-122=12 , a12=1-2222=1-422=-322=92a21=2-2122=2-222=02=0 , a22=2-2222=2-422=-222=42=2
So, the required matrix is 129202.
iv) aij=2i+j22
Here,
a11=21+122=2+122=322=92 , a12=21+222=422=162=8a21=22+122=4+122=522=252 , a22=22+222=4+222=622=362=18
So, the required matrix is 92825218.
v) aii2i-3j2
Here,
a11=21-312=2-32=-12=12 , a12=21-322=2-62=-42=2a21=22-312=4-32=12 , a22=22-322=4-62=-22=1
So, the required matrix is 122121.
vi) aij=-3i+j2
Here,
a11=-31+12=-3+12=-22=1 , a12=-31+22=-3+22=-12=12a21=-32+12=-6+12=-52=52 , a22=-32+22=-6+22=-42=2
So, the required matrix is 112522.
Page 5.7 Ex. 5.1
Q6.
Answer :
i) aij=i+j
Here,
a11=1+1=2 , a12=1+2=3 , a13=1+3=4 , a14=1+4=5a21=2+1=3 , a22=2+2=4 , a23=2+3=5 , a24=2+4=6a31=3+1=4 , a32=3+2=5 , a33=3+3=6 and a34=3+4=7
So, the required matrix is 234534564567.
ii) aij=i-j
Here,
a11=1-1=0 , a12=1-2=-1 , a13=1-3=-2 , a14=1-4=-3a21=2-1=1 , a22=2-2=0 , a23=2-3=-1 , a24=2-4=-2a31=3-1=2 , a32=3-2=1 , a33=3-3=0 and a34=3-4=-1
So, the required matrix is 0-1-2-310-1-2210-1.
iii) aij=2i
Here,
a11=21=2 , a12=21=2 , a13=21=2 , a14=21=2a21=22=4 , a22=22=4 , a23=22=4 , a24=22=4a31=23=6 , a32=23=6 , a33=23=6 and a34=23=6
So, the required matrix is 222244446666.
iv) aij=j
Here,
a11=1 , a12=2 , a13=3 , a14=4a21=1 , a22=2 , a23=3 , a24=4a31=1 , a32=2 , a33=3 and a34=4
So, the required matrix is 123412341234.
v) aij=12-3i+j
Here,
a11=12-31+1=12-2=1 , a12=12-31+2=12-1=12 , a13=12-31+3=120=02=0 , a14=12-31+4=121=12a21=12-32+1=12-5=52 , a22=12-32+2=12-4=2 , a23=12-32+3=12-3=32 , a24=12-32+4=12-2=1a31=12-33+1=12-8=4 , a32=12-33+2=12-7=72 , a33=12-33+3=12-6=3 and a34=12-33+4=12-5=52
So, the required matrix is 112012522321472352.
Q7.
Answer :
i)aij=2i+ij
Here,
a11=21+11=2+11=31=3 , a12=21+12=4+12=52 , a13=21+13=6+13=73a21=22+21=4+21=61=6 , a22=22+22=8+22=102=5 , a23=22+23=12+23=143a31=23+31=6+31=91=9 , a32=23+32=12+32=152 , a33=23+33=18+33=213=7a41=24+41=8+41=121=12 , a42=24+42=16+42=202=10 and a43=24+43=24+43=283
So, the required matrix is 3527365143915271210283.
ii)aij=i-ji+j
Here,
a11=1-11+1=02=0 , a12=1-21+2=-1 3 , a13=1-31+3=-2 4=-1 2a21=2-12+1=13 , a22=2-22-2=00=0 , a23=2-32+3=-15a31=3-13+1=24=12 , a32=3-23+2=15 , a33=3-33+3=06=0a41=4-14+1=35 , a42=4-24+2=26=13 and a43=4-34+3=17
So, the required matrix is 0-13-12130-1512150351317.
iii)aij=i
Here,
a11=1 , a12= 1, a13=1a21=2 , a22=2 , a23=2a31=3, a32=3 , a33=3a41=4, a42=4 and a43=4
So, the required matrix is 111222333444.
Q8.
Answer :
Since the corresponding elements of two equal matrices are equal,
3x+4y2x-2ya+b2a-b-1=2245-5-1⇒3x+4y=2 …1 ⇒x-2y=4⇒x=4+2y …2 Putting the value of x in eq. 1, we get 34+2y+4y=2 ⇒12+6y+4y=2 ⇒12+10y=2 ⇒10y=2-12⇒10y=-10⇒ y=-1010=-1Putting the value of y in eq. 2, we get x=4+2-1 ⇒x=4-2=2a+b=5 ⇒a=5-b …3⇒2a-b=-5 …4 Putting the value a in eq. 4, we get 25-b-b=-5⇒10-2b-b=-5⇒10-3b=-5⇒-3b=-15 ⇒ b=-15-3 ⇒b=5Putting the value of b in eq. 3, we get a=5-5⇒a=0
∴ x=2 , y=-1 , a=0 and b=5
Q9.
Answer :
Since the corresponding elements of two equal matrices are equal,
2x-3ya-b31x+4y3a+4b=1-231629
⇒2x-3y=1 …1 x+4y=6 ⇒x=6-4y …2Putting the value of x in eq. 1, we get 26-4y-3y=1 ⇒12-8y-3y=1 ⇒12-11y=1 ⇒-11y=-11 ⇒y=-11-11=1Putting the value of y in eq. 2, we get x=6-41⇒x=6-4 ⇒x=2 Now,a-b=-2 ⇒a=-2+b …3 3a+4b=29 …4 Putting the value of a in eq. 4, we get 3-2+b+4b=29 ⇒-6+3b+4b=29 ⇒-6+7b=29 ⇒7b=29+6⇒7b=35 ⇒b=357=5 Putting the value of b in eq. 3, we get a=-2+5 ⇒a=3 ∴ a=3, b=5, x=2 and y=1
Q1o.
Answer :
Since all the corresponding elements of a matrix are equal,
2a+ba-2b5c-d4c+3d=4-31124⇒2a+b=4 ⇒b=4-2a …1a-2b=-3 …2Putting the value of b in eq. 2, we geta-24-2a=-3⇒a-8+4a=-3⇒5a-8=-3⇒5a=-3+8⇒5a=5⇒a=1Putting the value of a in eq. 1, we getb=4-21⇒b=4-2⇒b=25c-d=11 ⇒5c-11=d … 34c+3d=24 …4Putting the value of d in eq. 4, we get4c+35c-11=24⇒4c+15c-33=24⇒19c-33=24⇒19c=24+33⇒19c=57⇒c=5719=3Putting the value of c in eq. 3, we get53-11=d⇒15-11=d⇒d=4∴ a=1, b=2, c=3 and d=4
Q11.
Answer :
We have, A=2x+12y0y2-5y , B=x+3y2+20-6
The corresponding elements of two equal matrices are equal.
∴2x+1=x+3⇒2x-x=3-1⇒x=2 and 2y=y2+2⇒y2-2y+2=0⇒y=-b±b2-4ac2a⇒y=2±4-82⇒y=2±-42Also, y2-5y=-6⇒y2-5y+6=0⇒y2-3y-2y+6=0⇒yy-3-2y-3=0⇒y-3y-2=0⇒y-3=0 or y-2=0⇒y=3 or y=2
Now, 3y=y2+2 and y2-5y=-6 must have a common value.
Thus, A and B cannot be equal for any value of y.
Q12.
Answer :
Since all the corresponding elements of a matrix are equal,
A=x-232z18zy+26z , B=yz66yx2yHere,x-2=y …1 z=3 …2 18z=6y …3Putting the value of z in eq. 3, we get183=6y⇒54=6y⇒y=546=9Putting the value of y in eq. 1, we getx-2=9⇒x=9+2⇒x=11∴ x=11, y=9 and z=3
Q13.
Answer :
Since all the corresponding elements of a matrix are equal,
x3x-y2x+z3y-w=3247x=3 …13x-y=2 …2Putting the value of x in eq. 2, we get33-y=2⇒9-y=2⇒-y=-7⇒ y=72x+z=4 …3Putting the value of x in eq. 3, we get23+z=4⇒6+z=4⇒z=4-6⇒z=-23y-w=7 …4Putting the value of y in eq. 4, we get37-w=7⇒21-w=7⇒21-7=w⇒w=14∴ x=3 , y=7, z=-2 and w=14
Q14.
Answer :
Given: x-yz2x-yw=-1405Since all the corresponding elements of a matrix are equal,x-y=-1⇒x=-1+y …12x-y=0 … 2z=4 w=5Putting the value of x in eq. 2, we get2-1+y-y=0⇒-2+2y-y=0⇒-2+y=0⇒y=2Putting the value of y in eq. 1, we getx=-1+2⇒x=1
Q15.
Answer :
Here,x+10=3x+4 ∵ All the corresponding elements of the matrix are equal⇒x-3x=4-10⇒-2x=-6∴ x=3Also,y2+2y=3⇒y2+2y-3=0⇒y2+3y-y-3=0⇒yy+3-1y+3=0⇒y+3y-1=0⇒y+3=0 or y-1=0⇒y=-3 or y=1Now,-4=y2-5y⇒y2-5y+4=0⇒y2-4y-y+4=0⇒yy-4-1y-4=0⇒y-4y-1=0⇒y-4=0 or y-1=0⇒y=4 or y=1
Since y2+2y=3 and y2-5y=-4 must hold good simultaneously, we take the common solution of these
two equations.
Thus,
y = 1, x = 3 and y = 1
Q16.
Answer :
Since all the corresponding element of a matrix are equal,
x+3=0⇒x=-3 Also,2y-7=3y-2⇒2y-3y=-2+7⇒-y=5⇒y=-5 z+4=6 ⇒z=6-4⇒z=2 a-1=-3⇒a=-3+1⇒a=-2 3b=-21⇒b=-7 z+2c=0⇒2=-2c⇒c=-1
Thus,
x = -3, y = -5 , z =2, a = -2 , b = -7 and c = -1
Q17.
Answer :
i)6
This is a matrix that contains only one element.
ii)100020003
For a diagonal matrix which is not scalar, all elements except those in the leading diagonal should be zero and the
elements in the diagonal should not be equal.
iii)123054006
Here, all elements below the main diagonal in upper triangular matrix are zero.
100260345
Here, all elements above the main diagonal in lower triangular matrix are zero.
Q18.
Answer :
According to the data, dealer A sold 5 deluxe cars, 3 premium cars and 4 standard cars in January. Also, dealer B sold 7 deluxe cars, 2 premium cars and 3 standard cars in January.
The above information can be given by
534723
Total sales over the period of January-February reveal that dealer A sold 8 deluxe cars,7 premium cars and 6 standard cars, while dealer B sold 10 deluxe cars, 5 premium cars and 7 standard cars.
This information can be given by
8761057
Page 5.17 Ex. 5.2
Q1.
Answer :
i) 3-214+-2413⇒3-2-2+41+14+3⇒1227
ii) 213035-125+1-232610-31⇒ 2+11-23+3 0+23+65+1-1+02-35+1⇒3-16296-1-16
Q2.
Answer :
i) 2A-3B⇒ 2A-3B=22432-313-25⇒ 2A-3B=4864-39-615⇒ 2A-3B=4-38-96+64-15⇒ 2A-3B=1-112-11ii B-4C⇒B-4C=13-25-4-2534⇒B-4C=13-25–8201216⇒B-4C=1+83-20-2-125-16⇒B-4C=9-17-14-11iii 3A-C⇒3A-C=32432–2534⇒3A-C=61296–2534⇒3A-C=6+212-59-36-4⇒3A-C=8762iv 3A-2B+3C⇒ 3A-2B+3C=32432-213-25+3-2534⇒ 3A-2B+3C=61296-26-410+-615912⇒ 3A-2B+3C=6-2-612-6+159+4+96-10+12⇒ 3A-2B+3C=-221228
Q3.
Answer :
i) A+B=2357+-102341
It is not possible to add these matrices because the number of elements in A are not equal to the
number of elements in B. So, A + B does not exist.
⇒ B+C=-102341+-123210⇒ B+C=-1-10+22+33+24+11+0⇒ B+C= -225551
ii) 2B+3A=2-102341+32357
It is not possible to add these matrices because the number of elements in B are not equal to the
number of elements in A. So, 2B + 3A does not exist.
⇒3C-4B=3-123210-4-102341⇒3C-4B=-369630–40812164⇒3C-4B=-3+46-09-86-123-160-4⇒3C-4B=161-6-13-4
Q4.
Answer :
Here,2A-3B+4C=2-102314-30-251-31+41-5260-4⇒2A-3B+4C=-204628-0-6153-93+4-208240-16⇒2A-3B+4C=-2-0+40+6-204-15+86-3+242+9+08-3-16⇒2A-3B+4C=2-14-32711-11
Q5.
Answer :
Here,A=2000-50009 , B=10001000-4 and C=-600030004i A-2B⇒A-2B=2000-50009-210001000-4 ⇒A-2B=2000-50009-20002000-8⇒A-2B=0000-700017=diag0 -7 17iiB+C-2A⇒B+C-2A=10001000-4+-600030004-22000-50009⇒B+C-2A=10001000-4+-600030004-4000-1000018⇒B+C-2A=1-6-40+0-00+0-00+0-01+3+100+0-00+0-00+0-0-4+4-18⇒B+C-2A=-900014000-18=diag-9 14 -18iii 2A+3B-5C⇒2A+3B-5C=22000-50009+310001000-4-5-600030004⇒2A+3B-5C=4000-1000018+30003000-12–300001500020⇒2A+3B-5C=4+3+300+0-00+0-00+0-0-10+3-150+0-00+0-00+0-018-12-20⇒2A+3B-5C=37000-22000-14=diag37 -22 -14
Q6.
Answer :
Here, LHS=A+B+C =2113-10024+97-1354216+2-431-10945 =2+91+71-13+3-1+50+40+22+14+6+2-431-10945 =11806442310+2-431-10945 =11+28-40+36+14-14+02+93+410+5 =134373411715RHS=A+B+C =2113-10024+97-1354216+2-431-10945 =2113-10024+9+27-4-1+33+15-14+02+91+46+5 =2113-10024+113244411511 =2+111+31+23+4-1+40+40+112+54+11 =134373411715∴ LHS=RHS
Hence proved.
Q7.
Answer :
Given: X+Y+X-Y=5209+360-1⇒2X=5+32+60+09-1⇒2X=8808⇒X=128808⇒X=4404Now,X+Y-X-Y=5209-360-1⇒X+Y-X+Y=5-32-60-09+1⇒2Y=2-4010⇒Y=122-4010⇒Y=1-205∴ X=4404 and Y=1-205
Q8.
Answer :
Given: 2X+Y=10-32⇒2X+3214=10-32⇒2X=10-32-3214⇒2X=1-30-2-3-12-4⇒2X=-2-2-4-2⇒X=12-2-2-4-2⇒X=-1-1-2-1
Q9.
Answer :
Given: 2X-Y=6-60-421 …1 X+2Y=325-21-7 …2Multiplying eq. 1 by eq. 2, we get22X-Y=26-60-421⇒4X-2Y=12-120-842 …3From eq. 3 and eq. 4, we get 4X-2Y+X+2Y=12-120-842+325-21-7⇒5X=12+3-12+20+5-8-24+12-7⇒5X=15-105-105-5⇒X=1515-105-105-5⇒X=3-21-21-1Putting the value of X in eq. 2, we getX+2Y=325-21-7⇒3-21-21-1+2Y=325-21-7⇒2Y=325-21-7-3-21-21-1⇒2Y=3-32+25-1-2+21-1-7+1⇒Y=02200-3
Q10.
Answer :
Here,X-Y+X+Y=111110100+351-1141180⇒2X=1+31+51+11-11+10+41+110+80+0⇒2X=4620241280⇒X=124620241280⇒X=231012640Now,X-Y-X+Y=111110100-351-1141180⇒X-Y-X-Y=1-31-51-11+11-10-41-110-80-0⇒-2Y=-2-4020-4-10-80⇒Y=-12-2-4020-4-10-80⇒Y=120-102540∴ X=231012640 and Y=120-102540
Page 5.18 Ex. 5.2
Q11.
Answer :
Here,A=9-14-213-12-1049⇒A=9-1-1-24+1-2-01-43-9⇒A=8-35-2-3-6
Q12.
Answer :
Given: 5A+3B+2C=0000⇒59178+315712+2C=0000⇒4553540+3152136+2C=0000⇒45+35+1535+2140+36+2C=0000⇒48205676+2C=0000⇒2C=0000-48205676⇒C=12-48-20-56-76⇒C=-24-10-28-38
Q13.
Answer :
Given: 2A+3X=5B⇒22-242-51+3X=5804-236⇒4-484-102+3X=40020-101530⇒3X=40020-101530-4-484-102⇒3X=40-40+420-8-10-415+1030-2⇒3X=36412-142528⇒X=1336412-142528⇒X=12434-143253283
Q14.
Answer :
Given: A+B+C=000000⇒1-32202+2-1-110-1+C=000000⇒1+2-3-12-12+10+02-1+C=000000⇒3-41301+C=000000⇒C=000000-3-41301⇒C=0-30+40-10-30-00-1⇒C=-34-1-30-1
Q15.
Answer :
i) Given: x-y2-24×6+3-2210-1=60052x+y5⇒x-y+32-2-2+24+1x+06-1=60052x+y5⇒x-y+3005×5=60052x+y5⇒x-y+3=6⇒x-y=6-3⇒x-y=3 …1Also,x=2x+y⇒-x=y …2Putting the value of y in eq. 1, we getx–x=3⇒2x=3⇒x=32Putting the value of x in eq. 2, we get-32=y⇒y=-32
ii) xy+2z-3+y 45=4912⇒x+yy+2+4z-3+5=4912⇒x+yy+6z+2=4912∴ x+y=4 …1Also, y+6=9⇒y=3z+2=12⇒z=10Putting the value of y in eq. 1, we getx+3=4 ⇒x=4-3⇒ x=1 ∴ x=1, y=3 and z=10
Q16.
Answer :
Given: 2345x+1y01=70105⇒68102x+1y01=70105⇒6+18+y10+02x+1=70105⇒78+y102x+1=70105∴ 8+y=0⇒y=-8Also,2x+1=5⇒2x=4⇒x=42=2∴ x=2 and y=-8
Q17.
Answer :
Given: λ102345+2123-1-32=44104214⇒λ02λ3λ4λ5λ+246-2-64=44104214⇒λ+20+42λ+63λ-24λ-65λ+4=44104214⇒λ+2=4 ⇒λ=4-2 ∴ λ=2
Q18.
Answer :
i) 2A+B+X=0⇒2-1234+3-215+X=0000⇒-2468+3-215+X=0000⇒-2+34-26+18+5+X=0000⇒12713+X=0000∴ X=-1-2-7-13
ii) 2A+3X=5B⇒2804-236+3X=52-242-51⇒1608-4612+3X=10-102010-255⇒3X=10-102010-255-1608-4612⇒3X=10-16-10-020-810+4-25-65-12⇒3X=-6-101214-31-7⇒X=13-6-101214-31-7∴ X=-2-1034143-313-73
Q19.
Answer :
i) 3xyzt=x6-12t+4x+yz+t3⇒3x3y3z3t=x+46+x+y-1+z+t2t+3∴ 3x=x+4 ⇒3x-x=4 ⇒2x=4 ⇒x=2 Also,3y=6+x+y⇒3y-y=6+x⇒2y=6+x …1Putting the value of x in eq. 1, we get 2y=6+2⇒2y=8 ⇒y=4Now,3t=2t+3⇒3t-2t=3⇒t=33z=-1+z+t ⇒3z-z=-1+t ⇒2z=-1+t …2 Putting the value of t in eq. 2, we get2z=-1+3⇒2z=2⇒ z=1∴ x=2, y=4, z=1 and t=3
ii) 2x57y-3+3412=7141514⇒2x10142y-6+3412=7141514⇒2x+310+414+12y-6+2=7141514⇒2x+314152y-4=7141514∴ 2x+3=7 ⇒2x=4 ⇒x=2 Also,2y-4=14⇒2y=18⇒y=9
Q20.
Answer :
Number of different types of posts in any college is given by
X = 15611
Total number of posts of each kind in all the colleges = 30X
= 3015611
= 4501803030
Page 5.38 Ex. 5.3
Q1.
Answer :
i) ab-baa-bba⇒a×a+b×ba×-b+b×a-b×a+a×b -b×-b+a×a⇒a2+b2-ab+ab-ab+abb2+a2⇒a2+b200a2+b2
ii) 1-223123-32-1⇒1×1+-2×-31×2+-2×21×3+-2×-12×1+3×-32×2+3×22×3+3×-1⇒1+62-43+22-94+66-3⇒7-25-7103
iii) 2343454561-35024305⇒2×1+3×0+4×32×-3+3×2+4×02×5+3×4+4×53×1+4×0+5×33×-3+4×2+5×03×5+4×4+5×54×1+5×0+6×34×-3+5×2+6×04×5+5×4+6×5⇒2+0+12-6+6+010+12+203+0+15-9+8+015+16+254+0+18-12+10+020+20+30⇒1404218-15622-270
Q2.
Answer :
i)
AB=5-1672134⇒AB=10-35-412+216+28⇒AB=713334 …1Also,BA=21345-167⇒BA=10+6-2+715+24-3+28⇒BA=1653925 …2∴ AB ≠ BA From eqs. (1) and (2)
ii) AB= -1100-11234123010110⇒AB=-1+0+0-2+1+0-3+0+00+0+10-1+10+0+02+0+44+3+46+0+0⇒AB=-1-1-31006116 …1Also,BA=123010110-1100-11234⇒BA=-1+0+61-2+90+2+120+0+00-1+00+1+0-1+0+01-1+00+1+0⇒BA=58140-11-101 …2∴ AB ≠ BA From eqs. (1) and (2)
iii) AB= 130110410010100051⇒AB=0+3+01+0+00+0+00+1+01+0+00+0+00+1+04+0+00+0+0⇒AB=310110140 …1Also,BA=010100051130110410⇒BA=0+1+00+1+00+0+01+0+03+0+00+0+00+5+40+5+10+0+0⇒BA=110130960 …2∴ AB ≠ BA From eqs. (1) and (2)
Q3.
Answer :
i) AB=1-223123231⇒AB=1-42-63-22+64+96+3⇒AB=-3-418139
Since the number of columns in B is greater then the number of rows in A, BA does not exists.
ii) AB=32-10-11456012⇒AB=12+015+218+4-4+0-5+0-6+0-4+0-5+1-6+2⇒AB=121722-4-5-6-4-4-4Also, BA=45601232-10-11⇒BA=12-5-68+0+60-1-20+0+2⇒BA=114-32
iii) AB=1-1230132⇒AB=0+-1+6+6⇒AB=11Also,BA=01321-123⇒BA=00001-1233-3692-246
iv) abcd+abcdabcd⇒ac+bd+a2+b2+c2+d2 a2+b2+c2+d2+ac+bd
Page 5.39 Ex. 5.3
Q4.
Answer :
i AB=13-12-1-130-1-23-1-12-1-69-4⇒AB=-2-3+63+6-9-1-3+4-4+1+66-2-9-2+1+4-6-0+69+0-9-3-0+4⇒AB=1003-53001 …1Also,BA=-23-1-12-1-69-413-12-1-130-1⇒BA=-2+6-3-6-3+02-3+1-1+4-3-3-2+01-2+1-6+18-12-18-9+06-9+4⇒BA=1-900-500-271 …2∴ AB ≠ BA From eqs. (1) and (2)
Q5.
Answer :
i) 13-1-4+3-2-11135246⇒ 1+33-2-1-1-4+1135246⇒41-2-3135246⇒4+212+420+6-2-6-6-12-10-18⇒61626-8-18-28
ii) 123102201012246⇒1+4+00+0+32+2+6246⇒5310246⇒10+12+60⇒82
iii) 1-10223102201-012102⇒1-102231-00-12-22-10-01-2⇒1-102231-1010-1⇒1-1-1+00+10+20+00-22+3-2+00-3⇒0-1120-25-2-3
Q6.
Answer :
Here,A2=AA⇒A2=10011001⇒A2=1+00+00+00+1⇒A2=1001 …1B2=BB⇒B2=100-1100-1⇒B2=1+00-00-00+1⇒B2=1001 …2C2=CC⇒B2=01100110⇒B2=0+10+00+01+0⇒B2=1001 …3We know, I2=1001 …4
⇒A2=B2=C2=I2 From eqs. (1), (2), (3) and (4)
Q7.
Answer :
Given: A=2-132Now,A2=AA⇒A2=2-1322-132⇒A2=4-3-2-26+6-3+4⇒A2=1-41213A2-2B+I⇒3A2-2B+I=31-4121-204-17+1001⇒3A2-2B+I=3-12363-08-214+1001⇒3A2-2B+I=3-0+1-12-8+036+2+03-14+1⇒3A2-2B+I=4-2038-10
Q8.
Answer :
Given: A-2IA-3I⇒ A-2IA-3I=42-11-21001 42-11-31001⇒ A-2IA-3I=42-11-2002 42-11-3003⇒ A-2IA-3I=4-22-0-1-01-24-32-0-1-01-3⇒ A-2IA-3I=22-1-112-1-2⇒ A-2IA-3I=2-24-4-1+1-2+2⇒ A-2IA-3I=0000⇒ A-2IA-3I=0Hence proved.
Q9.
Answer :
Given: A=1101Now,A2=AA⇒A2=11011101⇒A2=1+01+10+00+1⇒A2=1201A3=A2A⇒A3=12011101⇒A3=1+01+20+00+1⇒A3=1301
Hence proved.
Q10.
Answer :
Given: A=abb2-a2-abNow,A2=AA⇒A2=abb2-a2-ababb2-a2-ab⇒A2=a2b2-a2b2ab3-ab3-a3b+a3b-a2b2+a2b2⇒A2=0000⇒A2=0 Hence proved.
Q11.
Answer :
Given: A=cos 2θsin 2θ-sin 2θcos 2θNow,A2=AA⇒A2=cos 2θsin 2θ-sin 2θcos 2θcos 2θsin 2θ-sin 2θcos 2θ⇒A2=cos22θ-sin22θcos2θsin2θ+cos2θsin2θ-cos2θsin2θ-sin2θcos2θ-sin22θ+cos22θ⇒A2=cos2×2θ2sin2θcos2θ-2sin2θcos2θcos2×2θ ∵ cos2θ-sin2θ=cos2θ⇒A2=cos 4θsin2×2θ-sin2×2θcos 4θ ∵ sin2θ=2sinθcosθ⇒A2=cos 4θsin 4θ-sin 4θcos 4θ
Q12.
Answer :
Here,AB=2-3-5-1451-3-4-1351-3-5-135⇒AB=-2-3+56+9-1510+15-251+4-5-3-12+15-5-20+25-1-3+43+9-125+15-20⇒AB=000000000⇒AB=03×3 …1BA=-1351-3-5-1352-3-5-1451-3-4⇒BA=-2-3+53+12-155+15-202+3-5-3-12+15-5-15+20-2-3+53+12-155+15-20⇒BA=000000000 ⇒BA=03×3 …2⇒AB=BA=03×3 From eqs. (1) and (2)
Q13.
Answer :
Here,AB=0c-b-c0ab-a0a2abacabb2bcacbcc2⇒AB=0+abc-abc0+b2c-b2c0+bc2-bc2-a2c+0+a2c-abc+0+abc-ac2+0+ac2a2b-a2b+0ab2-ab2+0abc-abc+0⇒AB=000000000⇒AB=O3×3 …1BA=a2abacabb2bcacbcc20c-b-c0ab-a0⇒BA=0-abc+abca2c+0-a2c-a2b+a2b+00-b2c+b2cabc+0-abc-ab2+ab2+00-bc2+bc2ac2+0-ac2-abc+abc+0⇒BA=000000000⇒BA=O3×3 …2 ⇒AB=BA=O3×3 From eqs. (1) and (2)
Q14.
Answer :
Given: AB=2-3-5-1451-3-42-2-4-1341-2-3⇒AB=4+3-5-4-9+10-8-12+15-2-4+52+12-104+16-152+3-4-2-9+8-4-12+12⇒AB=2-3-5-1451-3-4⇒AB=ABA=2-2-4-1341-2-32-3-5-1451-3-4⇒BA=4+2-4-6-8+12-10-10+16-2-3+43+12-125+15-162+2-3-3-8+9-5-10+12⇒BA=2-2-4-1341-2-3⇒BA=B
Page 5.40 Ex. 5.3
Q15.
Answer :
Given: A=-11-13-33555Now,A2=AA⇒A2=-11-13-33555-11-13-33555⇒A2=1+3-5-1-3-51+3-5-3-9+153+9+15-3-9+15-5+15+255-15+25-5+15+25⇒A2=-1-9-13273351535B2=BB⇒B2=0431-3-3-1440431-3-3-144⇒B2=0+4-30-12+120-12+120-3+34+9-123+9-120+4-4-4-12+16-3-12+16⇒B2=100010001A2-B2⇒A2-B2=-1-9-13273351535-100010001⇒A2-B2=-1-1-9-0-1-03-027-13-035-015-035-1⇒A2-B2=-2-9-13263351534
Q16.
Answer :
i) ABC=ABC ⇒120-10110-1203 1-1 =120-101 10-12031-1 ⇒1-2+00+4+0-1-0+00+0+3 1-1 =120-1011-0-1-20-3 ⇒-14-131-1 =120-1011-3-3 ⇒-1-4-1-3 =1-6-0-1-0-3 ⇒-5-4=-5-4∴ LHS=RHSHence proved.
ii) ABC=ABC⇒4231123011-110122-11 12-1301001 =423112301 1-110122-1112-1301001⇒4+0+6-4+2-34+4+31+0+4-1+1-21+2+23+0+2-3+0-13+0+1 12-1301001=4231123011-3+02-0+0-1-1+10+3+00+0+00+1+22-3+04-0+0-2-1+1⇒10-5115-255-4412-1301001=423112301-22-1303-14-2⇒10-15+020-0+0-10-5+115-6+010-0+0-5-2+55-12+010-0+0-5-4+4=-8+6-38+0+12-4+6-6-2+3-22+0+8-1+3-4-6+0-16+0+4-3+0-2⇒-520-4-110-2-710-5=-520-4-110-2-710-5∴ LHS=RHSHence proved.
Q17.
Answer :
i) AB+C = AB+AC⇒1-102-1021+011-1=1-102-1021+1-102011-1⇒1-102-1+00+12+11-1=-1-20-10+40+2+0-11+10+20-2⇒1-102-1130=-3-142+-122-2⇒-1-31-00+60+0=-3-1-1+24+22-2⇒-4160=-4160∴ LHS=RHSHence proved.
ii) AB+C=AB+AC⇒2-111-120111+1-101=2-111-120111+2-111-121-101⇒2-111-120+11-11+01+1=0-12-10+11+10+2-1+2+2-0-2-11+0-1+1-1+01+2⇒2-111-121012=-111221+2-310-13⇒2-10-21+10+2-1+20+4=-1+21-31+12+02-11+3⇒1-22214=1-22214∴ LHS=RHSHence proved.
Q18.
Answer :
Given: AB-C=AB-AC⇒10-23-10-21105-4-213-102-152-1100-11=10-23-10-21105-4-213-102-10-23-10-211152-1100-11⇒10-23-10-2110-15-5-4-2-2+11-13-0-1-00+12-1=0-0+25+0-0-4+0-40+2-015-1+0-12-3+00-2-1-10+1+08+3+2- 1-0-05+0+22+0-23+1+015-1-06-0+0-2-1+0-10+1-1-4+0+1⇒10-23-10-211-10-6-103-111=25-8214-15-3-913-1704146-3-10-3⇒-1-0+20+0-2-6+0-2-3+1-00-0+0-18-3+02-1-10+0+112+3+1=2-15-7-8-02-414-14-15-6-3+3-9+1013+3⇒1-2-8-20-210116=1-2-8-20-210116∴ LHS=RHSHence proved.
Q19.
Answer :
We have,
Given: A=0102020324042-1-324301-12-23-34-40⇒A=0102020324040-32+3-2-44+4-4-00+6-3-63+8-6-86+00+94-9-4+128-12-8+0⇒A=010202032404-35-68-46-911-1469-58-4-8⇒A=0+6+00-9-00+11+00-14-00+6-0-6+0+1810-0-10-12+0+1616-0-8-8+0-160+18+180-27-100+33+160-42-80+18-16-12+0+3620-0-20-24+0+3232-0-16-16+0-32⇒A=6-911-14612048-2436-3749-502240816-48∴ a43=8 and a22=0
Q20.
Answer :
Given: A=010001pqrNow,A2=AA⇒A2=010001pqr010001pqr⇒A2=0+0+00+0+00+1+00+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r2A2=001pqrrpp+rqq+r2A3=A2A⇒A3=001pqrrpp+rqq+r2010001pqr⇒A3=0+0+p0+0+q0+0+r0+0+rpp+0+rq0+q+r20+0+pq+r2prp+0+q2+r2q0+p+rq+rq+r3⇒A3=pqrrpp+rqq+r2pq+r2prp+q2+r2qp+2rq+r3 …1pI+qA+rA2⇒pI+qA+rA2=p100010001+q010001pqr+r001pqrrpp+rqq+r2⇒pI+qA+rA2=p000p000p+0q000qpqq2qr+00rrprqr2r2prp+r2qrq+r3⇒pI+qA+rA2=p+0+00+q+00+0+r0+0+rpp+0+rq0+q+r20+pq+r2p0+q2+rp+r2qp+qr+qr+r3⇒pI+qA+rA2=pqrrpp+rqq+r2pq+r2pq2+r2q+rpp+2qr+r3 …2 A3=pI+qA+rA2 From eqs. (1) and (2)
Q21.
Answer :
Here,LHS= 1ww2ww21w21w+ww21w21www211ww2 =1+ww+w2w2+1w+w2w2+11+ww2+w1+w2w+11ww2 =-w2-1-w-1-w-w2-1-w-w21ww2 ∵ 1+w+w2=0 and w3=1 =-w2-w-w3-1-w2-w4-1-w2-w4 =-w1+w+w2-1-w2-w3w-1-w2-w3w =-w×0-1-w2-w-1-w2-w ∵1+w+w2=0 and w3=1 =0-0-0 =000∴ 1ww2ww21w21w+ww21w21www211ww2=000
Q22.
Answer :
Here,A2=AA⇒A2=2-3-5-1451-3-42-3-5-1451-3-4⇒A2=4+3-5-6-12+15-10-15+20-2-4+53+16-155+20-202+3-4-3-12+12-5-15+16⇒A2=2-3-5-1451-3-4∴ A2=A
Q23.
Answer :
Here,A2=AA⇒A2=4-1-430-43-1-34-1-430-43-1-3⇒A2=16-3-12-4+0+4-16+4+1212+0-12-3+0+4-12+0+1212-3-9-3+0+3-12+4+9⇒A2=100010001∴ A2=I3
Q24.
Answer :
Given: 11×102021210111=0⇒1+0+2×0+2+x2+1+0111=0⇒1+2×2+x3111=0⇒1+2x+2+x+3=0⇒6+3x=0⇒3x=-6⇒x=-63∴ x=-2
Page 5.41 Ex. 5.3
Q25.
Answer :
Given: x4121210202-4×4-1=0⇒2x+4+0x+0+22x+8-4×4-1=0⇒2x+4x+22x+4×4-1=0⇒2×2+4x+4x+8-2x-4=0⇒2×2+6x+4=0⇒x2+3x+2=0⇒x2+2x+x+2=0⇒xx+2+1x+2=0⇒x+2x+1=0⇒x+2=0 or x+1=0⇒x=-2 or x=-1∴ x=-2,-1
Q26.
Answer :
Given: 1-1×01-1213111011=0⇒0-2+x1-1+x-1-3+x011=0⇒-2+xx-4+x011=0⇒0+x-4+x=0⇒2x-4=0⇒2x=4⇒x=42∴ x=2
Q27.
Answer :
Given: A=3-24-2Now,A2=AA⇒A2=3-24-23-24-2⇒A2=9-8-6+412-8-8+4⇒A2=1-24-4A2-A+2I=1-24-4-3-24-2+21001⇒A2-A+2I=1-3-2+24-4-4+2+2002⇒A2-A+2I=-200-2+2002⇒A2-A+2I=-2+20+00+0-2+2⇒A2-A+2I=0000⇒A2-A+2I=0Hence proved.
Q28.
Answer :
Given: A=31-12Now,A2=AA⇒A2=31-1231-12⇒A2=9-13+2-3-2-1+4⇒A2=85-53A2=5A+λI⇒85-53=531-12+λ1001⇒85-53=155-510+λ00λ⇒85-53=15+λ5+0-5+010+λ⇒85-53=15+λ5-510+λThe corresponding elements of two equal matrices are equal.∴ 8=15+λ ⇒8-15=λ ⇒-7=λ∴ λ=-7
Q29.
Answer :
Given: A=31-12Now,A2=AA⇒A2=31-1231-12⇒A2=9-13+2-3-2-1+4⇒A2=85-53A2-5A+7I2⇒A2-5A+7I2=85-53-531-12+71001⇒A2-5A+7I2=85-53-155-510+7007⇒A2-5A+7I2=8-15+75-5+0-5+5+03-10+7⇒A2-5A+7I2=0000⇒A2-5A+7I2=0Hence proved.
Q30.
Answer :
Given: A=23-10Now,A2=AA⇒A2=23-1023-10⇒A2=4-36+0-2+0-3+0⇒A2=16-2-3A2-2A+3I2⇒ A2-2A+3I2=16-2-3-223-10+31001⇒ A2-2A+3I2=16-2-3-46-20+3003⇒ A2-2A+3I2=1-4+36-6+0-2+2+0-3+0+3⇒ A2-2A+3I2=0000=0 Hence proved.
Q31.
Answer :
We have, A=2312∴A2=AA⇒A2=23122312⇒A2=4+36+62+23+4⇒A2=71247⇒A3=A2A⇒A3=712472312⇒A3=14+1221+248+712+14⇒A3=26451526Now, A3-4A2+A⇒ A3-4A2+A=26451526-471247+2312⇒ A3-4A2+A=26451526-28481628+2312⇒ A3-4A2+A=26-28+245-48+315-16+126-28+2⇒ A3-4A2+A=0000=O Hence proved.
Q32.
Answer :
Given: A=53127Now,A2=AA⇒A2=5312753127⇒A2=25+3615+2160+8436+49⇒A2=613614485A2-12A-I⇒A2-12A-I=613614485-1253127-1001⇒A2-12A-I=613614485-603614484-1001⇒A2-12A-I=61-60-136-36+0144-144+085-84-1⇒A2-12A-I=0000Since A is satisfying the equation A2-12A-I, A is the root of the equation A2-12A-I .
Q33.
Answer :
Given: A=3-5-42Now,A2=AA⇒A2=3-5-423-5-42⇒A2=9+20-15-10-12-820+4⇒A2=29-25-2024A2-5A-14I⇒A2-5A-14I=29-25-2024-53-5-42-141001⇒A2-5A-14I=29-25-2024-15-25-2010-140014⇒A2-5A-14I=29-15-14-25+25+0-20+20+024-10-14⇒A2-5A-14I=0000
Q34.
Answer :
i)Given: A=2312Now,A2=AA⇒A2=23122312⇒A2=4+36+62+23+4⇒A2=71247A3=A2A⇒A3=712472312⇒A3=14+1221+248+712+14⇒A3=26451526A3-4A2+A⇒A3-4A2+A=26451526-471247+2312⇒A3-4A2+A=26451526-28481628+2312⇒A3-4A2+A=26-28+245-48+315-16+126-28+2⇒A3-4A2+A=0000=0Hence proved.
ii) Given: A= 31-12Now,A2=AA⇒A2=31-1231-12⇒A2=9-13+2-3-2-1+4⇒A2=85-53A2-5A+7I⇒A2-5A+7I=85-53-531-12+71001⇒A2-5A+7I=85-53-155-510+7007⇒A2-5A+7I=8-15+75-5+0-5+5+03-10+7⇒A2-5A+7I=0000=0Hence proved.Given: A2-5A+7I=0⇒A2=5A-7I …1⇒A3=A5A-7I Multilpying by A on both sides⇒A3=5A2-7AI⇒A3=55A-7I-7A From eq. 1⇒A3=25A-35I-7A⇒A3=18A-35I ⇒A4=18A-35IA Multilpying by A on both sides⇒A4=18A2-35A⇒A4=185A-7I-35A From eq. 1 ⇒A4=90A-126I-35A⇒A4=55A-126I⇒A4=5531-12-1261001⇒A4=16555-55110-12600126⇒A4=165-12655-0-55-0110-126⇒A4=3955-55-16
Q35.
Answer :
Given: A=3-24-2Now,A2=AA⇒A2=3-24-23-24-2⇒A2=9-8-6+412-8-8+4⇒A2=1-24-4A2=kA-2I2⇒1-24-4=k3-24-2-21001⇒1-24-4=3k-2k4k-2k-2002⇒1-24-4=3k-2-2k-04k-0-2k-2The corresponding elements of two equal matrices are equal.∴1=3k-2 ⇒1+2=3k ⇒3=3k ⇒k=1
Q36.
Answer :
Given: A=10-17Now,A2=AA⇒A2=10-1710-17⇒A2=1-00+0-1-70+49⇒A2=10-849A2-8A+kI=0⇒10-849-810-17+k1001=0⇒10-849-80-856+k00k=0⇒1-8+k0-0+0-8+8+049-56+k=0⇒-7+k00-7+k=0000The corresponding elements of two equal matrices are equal.∴-7+k=0 ⇒k=7
Q37.
Answer :
Here,fx=x2-2x-3⇒fA=A2-2A-3I2Now, A2=AA⇒A2=12211221⇒A2=1+42+22+24+1⇒A2=5445fA=A2-2A-3I2⇒fA=5445-21221-31001⇒fA=5445-2442-3003⇒fA=5-2-34-4-04-4-05-2-3⇒fA=5-5005-5⇒fA=0
Q38.
Answer :
i) Given: A=2312Now,A2=AA⇒A2=23122312⇒A2=4+36+62+23+4⇒A2=71247A2=λA+μI⇒71247=λ2312+μ1001⇒71247=2λ3λλ2λ+μ00μ⇒71247=2λ+μ3λ+0λ+02λ+μ⇒71247=2λ+μ3λλ2λ+μThe corresponding elements of two equal matrices are equal.∴ 7=2λ+μ …1 12=3λ⇒λ=123=4Putting the value of λ in eq. 1, we get 7=24+μ⇒7-8=μ∴ μ=-1
ii) We have, A=2312⇒A2=AA⇒A2=23122312⇒A2=4+36+62+23+4⇒A2=71247Now, A3=A2A⇒A3=712472312⇒A3=14+1221+248+712+14⇒A3=26451526Now, A3-4A2+A⇒A3-4A2+A=26451526-471247+2312⇒A3-4A2+A=26451526-28481628+2312⇒A3-4A2+A=26-28+245-48+315-16+126-28+2⇒A3-4A2+A=0000=0Hence proved.
Q39.
Answer :
Here, 2070101-21-x14x7x010x-4x-2x=100010001⇒-2x+0+7x28x+0-28x14x+0-14×0+0+00+1-00+0-0-x-0+x14x-2-4x7x-0-2x=100010001⇒5x00010010x-25x=100010001The corresponding elements of two equal matrices are equal.∴ 5x=1 ⇒x=15
Q40.
Answer :
i) x110-2-3×5=0⇒x-20-3×5=0⇒x-2-3×5=0⇒x2-2x-15=0⇒x2-2x-15=0⇒x2-5x+3x-15=0⇒xx-5+3x-5=0⇒x-5x+3=0⇒x-5=0 or x+3=0⇒x=5 or x=-3
ii) 12112020110202x=0⇒1+4+12+0+00+2+202x=0⇒62402x=0⇒0+4+4x=0⇒4+4x=0⇒4x=-4∴x=-44=-1
iii) x-5-1102021203×41=0⇒x-0-20-10-02x-5-3×41=0⇒x-2-102x-8×41=0⇒x2-2x-40+2x-8=0⇒x2-48=0⇒x2=48⇒x=±48
Page 5.42 Ex. 5.3
Q41.
Answer :
Given: A=1203-450-13Now,A2=AA⇒A2=1203-450-131203-450-13⇒A2=1+6+02-8-00+10+03-12+06+16-50-20+150-3+00+4-30-5+9⇒A2=7-610-917-5-314A2-4A+3I3⇒ A2-4A+3I3=7-610-917-5-314-41203-450-13+3100010001⇒A2-4A+3I3=7-610-917-5-314-48012-16200-412+300030003⇒A2-4A+3I3=7-4+3-6-8+010-0+0-9-12+017+16+3-5-20+0-3-0+01+4+04-12+3⇒A2-4A+3I3=6-1410-2136-25-35-5
Q42.
Answer :
Given: fx=x2-2xfA=A2-2ANow, A2=AA⇒A2=012450023012450023⇒A2=0+4+00+5+40+0+60+20+04+25+08+0+00+8+00+10+60+0+9⇒A2=496202988169fA=A2-2A⇒fA=496202988169-2012450023⇒fA=496202988169-0248100046⇒fA=4-09-26-420-829-108-08-016-49-6⇒fA=472121988123
Q43.
Answer :
Given: fx=x3+4×2-xfA=A3+4A2-ANow, A2=AA⇒A2=0122-301-100122-301-10⇒A2=0+2+20-3-20+0+00-6+02+9-04-0+00-2+01+3-02-0+0⇒A2=4-50-6114-242 A3=A2A⇒A3=4-50-6114-2420122-301-10⇒A3=0-10+04+15-08-0+00+22+4-6-33-4-12+0+00+8+2-2-12-2-4+0+0⇒A3=-1019826-43-1210-16-4fA=A3+4A2-A⇒fA=-1019826-43-1210-16-4+44-50-6114-242-0122-301-10⇒fA=-1019826-43-1210-16-4+16-200-244416-8168-0122-301-10⇒fA=-10+16-019-20-18+0-226-24-2-43+44+3-12+16-010-8-1-16+16+1-4+8+0⇒fA=6-26044114
Q44.
Answer :
Given: fx=x3-6×2+7x+2fA=A3-6A2+7A+2I3Now, A2=AA⇒A2=102021203102021203⇒A2=1+0+40+0+02+0+60+0+20+4+00+2+32+0+60+0+04+0+9⇒A2=5082458013A3=A2A⇒A3=5082458013102021203⇒A3=5+0+160+0+010+0+242+0+100+8+04+4+158+0+260+0+016+0+39⇒A3=210341282334055A3-6A2+7A+2I3⇒A3-6A2+7A+2I3=210341282334055-65082458013+7102021203+2100010001⇒A3-6A2+7A+2I3=210341282334055-3004812243048078+7014014714021+200020002⇒A3-6A2+7A+2I3=21-30+7+20-0+0+034-48+14+012-12+0+08-24+14+223-30+7+034-48+14+00-0+0+055-78+21+2⇒A3-6A2+7A+2I3=000000000=0Since fA=0, A is the root of fx=x3-6×2+7x+2.
Q45.
Answer :
Given: A=122212221Now,A2=AA⇒A2=122212221122212221⇒A2=1+4+42+2+42+4+22+2+44+1+44+2+22+4+24+2+24+4+1⇒A2=988898889A2-4A-5I⇒ A2-4A-5I=988898889-4122212221-5100010001⇒ A2-4A-5I=988898889-488848884-500050005⇒ A2-4A-5I=9-4-58-8-08-8-08-8-09-4-58-8-08-8-08-8-09-4-5⇒ A2-4A-5I=000000000=0
Hence proved.
Q46.
Answer :
Given: A=320140005Here,A2=AA⇒A2=320140005320140005⇒A2=9+2+06+8+00+0+03+4+02+16+00+0+00+0+00+0+00+0+25⇒A2=1114071800025Now, A2-7A+10I3⇒A2-7A+10I3=1114071800025-7320140005+10100010001⇒A2-7A+10I3=1114071800025-2114072800035+100001000010⇒A2-7A+10I3=11-21+1014-14+00-0+07-7+018-28+100-0+00-0+00-0+025-35+10⇒A2-7A+10I3=000000000=0
Q47.
Answer :
Given: 5-7-23xyzu=-16-672⇒5x-7z5y-7u-2x+3z-2y+3u=-16-672The corresponding elements of two equal matrices are equal.∴5x-7z=-16 …1 5y-7u=-6 …2 -2y+3u=2 ⇒3u=2+2y⇒u=2+2y3 …3 -2x+3z=7 ⇒3z=7+2x ⇒z=7+2×3 …4 Putting the value of z in eq. 1, we get5x-77+2×3=-16⇒5x-49+14×3=-16⇒15x-49-14×3=-16⇒x-49=-48⇒x=-48+49∴ x=1Putting the value of x in eq. 4, we getz=7+213⇒z=93=3Putting the value of u in eq. 2, we get5y-72+2y3=-6⇒5y-14+14y3=-6⇒15y-14-14y3=-6⇒y-14=-18⇒y=-18+14∴ y=-4Putting the value of y in eq. 3, we getu=2+2-43⇒u=-2∴ xyzu=1-43-2
Q48.
Answer :
i Let A=xyzabc⇒1101xyzabc=335101⇒x+ay+bz+c0+a0+b0+c=335101⇒x+ay+bz+cabc=335101The corresponding elements of two equal matrices are equal.⇒x+a=3 …1y+b=3 …2 z+c=5 …3⇒ a=1, b=0 and c=1Putting the value of a in eq. 1, we get x+1=3⇒x=3-1∴ x=2Putting the value of b in eq. 2, we get y+b=3⇒y+0=3∴ y=3Putting the value of c in eq. 3, we get z+1=5⇒z=5-1∴ z=4∴ A=234101
ii Let A=wxyz⇒A123456=-7-8-9246⇒wxyz123456=-7-8-9246⇒w+4x2w+5x3w+6xy+4z2y+5z3y+6z=-7-8-9246The correspnding elements of two equal matrices are equal.∴ 3w+6x=-9 …1 y+4z=2 y=2-4z …2 w+4x=-7 ⇒w=-7-4x …3 2y+5z=4 …4 Putting the value of w in eq. 1, we get3-7-4x+6x=-9⇒-21-12x+6x=-9⇒-6x=12⇒x=-2Putting the value of x in eq. 3, we getw=-7-4-2 ⇒w=-7+8⇒w=1Putting the value of y in eq. 4, we get22-4z+5z=4⇒4-8z+5z=4⇒4-3z=4⇒-3z=0⇒z=0Putting the value of z in eq. 2, we get y=2-40 ⇒y=2∴ A=1-220
Q49.
Answer :
Let A = wxyz
Now,
wxyz1-214=6I2⇒w+x-2w+4xy+z-2y+4z=61001⇒w+x-2w+4xy+z-2y+4z=6006The corresponding elements of two equal matrices are equal.∴ w+x=6 ⇒w=6-x …1 -2w+4x=0 …2 Putting the value of w in eq. 2, we get-26-x+4x=0⇒-12+2x+4x=0⇒-12+6x=0⇒6x=12⇒x=2Putting the value of x in eq. 1, we getw=6-2⇒w=4Now, y+z=0⇒y=-z …3 -2y+4z=6 …4 Putting the value of y in eq. 4, we get -2-z+4z=6⇒2z+4z=6⇒6z=6⇒z=1Putting the value of z in eq. 3, we get y=-1∴ A=42-11
Q50.
Answer :
Given: A=0040Here,A2=AA⇒A2=00400040⇒A2=0+00+00+00+0⇒A2=0000A4=A2A2⇒A4=00000000⇒A4=0000A8=A4A4⇒A8=00000000⇒A8=0000A16=A8A8⇒A16=00000000∴ A16=0000Thus, A16 is a null matrix.
Q51.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral powers of matrix, we have
A1=1101=A
So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
Am=1m01 …(1)
Now, we shall show that the result is true for n=m+1.
Here,
Am+1=1m+101
By definition of integral power of matrix, we have
Am+1=AmA =1m011101 From eq. 1 =1+01+m0+00+1 =11+m01
This shows that when the result is true for n = m, it is also true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
Q52.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=a1ba1-1/a-101=ab01=A
So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
Am=ambam-1/a-101 …(1)
Now, we shall show that the result is true for n=m+1.
Here,
Am+1=am+1bam+1-1/a-101
By definition of integral power of matrix, we have
Am+1=AmA⇒Am+1=ambam-1/a-101ab01 From eq. 1⇒Am+1=ama+0amb+bam-1/a-10+00+1⇒Am+1=am+1am+1b-amb+amb-b/a-101⇒Am+1=am+1bam+1-1/a-101
This shows that when the result is true for n = m, it is also true for n = m +1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
Q53.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=cos 1θi sin1θi sin 1θcos 1θ=cosθi sinθi sinθcosθ=A
Thus, the result is true for n=1.
Step 2: Let the result be true for n = m. Then,
Am=cosmθi sinmθi sinmθcosmθ
Now we shall show that the result is true for n=m+1.
Here,
Am+1=cos m+1θi sinm+1θi sin m+1θcos m+1θ …(1)
By definition of integral power of matrix, we have
Am+1=AmA⇒Am+1=cos mθi sinmθi sin mθcos mθcosθi sinθi sinθcosθ From eq. 1⇒Am+1=cos mθ.cosθ+i sinmθ.i sinθcos mθ.i sinθ+i sinmθ.cosθi sin mθ.cosθ+cos mθ.i sinθi sin mθ.i sinθ+cos mθ.cosθ⇒Am+1=cos mθ.cosθ-sinmθ.sinθicos mθ.sinθ+sinmθ.cosθi sin mθ.cosθ+cos mθ sinθ-sin mθ.sinθ+cos mθ.cosθ⇒Am+1=cos mθ.cosθ-sinmθ.sinθicos mθ.sinθ+sinmθ.cosθi sin mθ.cosθ+cos mθ sinθcos mθ.cosθ-sin mθ.sinθ⇒Am+1=cosmθ+θi sinmθ+θi sinmθ+θcosmθ+θ⇒Am+1=cosm+1θi sinm+1θi sinm+1θcosm+1θ
This shows that when the result is true for n = m, it is true for n=m+1.
Hence, by the principle of mathematical induction, the result is valid for all n∈N.
Q54.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=cos 1α+sin 1α2sin 1α-2 sin 1αcos 1α-sin 1α=cos α+sin α 2sin α-2sin αcos α-sin α=A
So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
Am=cos mα+sin mα 2sin mα-2sin mαcos mα-sin mα …(1)
Now we shall show that the result is true for n=m+1.
Here,
Am+1=cos m+1α+sin m+1α 2sin m+1α-2sin m+1αcos m+1α-sin m+1α
By definition of integral power of matrix, we have
Am+1=Am.A⇒Am+1=cos mα+sin mα 2sin mα-2sin mαcos mα-sin mαcos α+sin α 2sin α-2sin αcos α-sin α From eq. 1⇒Am+1=cos mα+sin mαcos α+sin α -2sin mα2sin αcos mα+sin mα 2sin α+2sin mαcos α-sin α-2sin mαcos α+sin α-cos mα-sin mα2sin α-2sin mα2sin α+cos mα-sin mαcos α-sin α⇒Am+1=cos mα cosα+sin mα cosα+cos mα sinα+sin mα sinα-2sin mα sinα2sin α cos mα+2sin α sin mα+2sin mα cosα-2sin ma sinα-2sin ma cosα-2sin ma sinα-2sin α cos mα+2sin α sin mα-2sin α sin mα+cos mα cosα-sin mα cosα-cos mα sinα+sin mα sinα⇒Am+1=cosmα-α+sinmα+α-cosmα-α+cosmα+α2sinmα+α-2sinmα+αcosmα+α-sinmα+α⇒Am+1=cosmα+α+sinmα+α2sinmα+a-2sinmα+αcosmα+α-sinmα+α⇒Am+1=cosm+1α+sinm+1α2sinm+1α-2sinm+1αcosm+1α-sinm+1α
This show that when the result is true for n = m, it is also true for n = m +1.
Hence, by the principle of mathematical induction, the result is valid for all n∈N.
Page 5.43 Ex. 5.3
Q55.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=1111+1/2011001=111011001=A
Thus, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
Am=1mmm+1/201m001 …(1)
Now, we shall show that the result is true for n=m+1.
Here,
Am+1=1m+1m+1m+1+1/201m+1001⇒Am+1=1m+1m+1m+2/201m+1001
By definition of integral power of matrix, we have
Am+1=AmA⇒Am+1=1mmm+1/201m001111011001 From eq. 1⇒Am+1=1+0+01+m+01+m+mm+1/20+0+00+1+00+1+m0+0+00+0+00+0+1⇒Am+1=11+m2+2m+m2+m/2011+m001⇒Am+1=11+mm2+3m+2/2011+m001⇒Am+1=11+mm+1m+2/2011+m001
This shows that when the result is true for n = m, it is also true for n = m + 1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
Q56.
Answer :
Let Pn be the statement given by Pn : An+1=BnB+n+1C.
For n = 1, we have
P1 : A2=BB+2CHere,LHS =A2 =B+CB+C =BB+C+CB+C =B2+BC+CB+C2 =B2+2BC ∵ BC=CB and C2=O =BB+2C=RHS
Hence, the statement is true for n = 1.
If the statement is true for n = k, then
Pk : Ak+1=BkB+k+1C …(1)
For Pk+1 to be true, we must have
Pk+1 : Ak+2=Bk+1B+k+2C
Now,
Ak+2=Ak+1A =BkB+k+1CB+C From eqn. 1 =Bk+1+k+1BkCB+C =Bk+1B+C+k+1BkCB+C =Bk+2+Bk+1C+k+1BkCB+k+1BkC2 =Bk+2+Bk+1C+k+1BkBC ∵ BC=CB and C2=0 =Bk+2+Bk+1C+k+1Bk+1C =Bk+2+k+2Bk+1C =Bk+1B+k+2C
So the statement is true for n = k+1.
Hence, by the principle of mathematical induction, Pn is true for all n∈N.
Q57.
Answer :
We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
A1=a1000b1000c1=a000b000c=A
So, the result is true for n = 1.
Step 2: Let the result be true for n = m. Then,
Am=am000bm000cm …(1)
Now, we shall check if the result is true for n=m+1.
Here,
Am+1=am+1000bm+1000cm+1
By definition of integral power of matrix, we have
Am+1=AmA⇒Am+1=am000bm000cma000b000c From eq. 1⇒Am+1=aam+0+00+0+00+0+00+0+00+bbm+00+0+00+0+00+0+00+0+ccm⇒Am+1=am+1000bm+1000cm+1
This shows that when the result is true for n = m, it is also true for n=m+1.
Hence, by the principle of mathematical induction, the result is valid for any positive integer n.
Q58.
Answer :
Here,X a+b×a+2Y b+1×a+3Since XY exists, the number of columns in X is equal to the number of rows in Y.⇒ a+2=b+1 …1Similarly, since YX exists, the number of columns in Y is equal to the number of rows in X.⇒a+b=a+3⇒b=3Putting the value of b in 1, we geta+2=3+1⇒a=2
Since the order of the matrices XY and YX is not same, XY and YX are not of the same type and they are unequal.
Q59.
Answer :
i) Let A=1-232 and B=23-12AB=1-23223-12⇒AB=2+23-46-29+4⇒AB=4-1413Now,BA=23-121-232⇒BA=2+9-4+6-1+62+4⇒BA=11256
Thus, AB ≠ BA.
ii) Let A=0200 and B=1000∴ AB=0+00+00+00+0 =0000=O
Thus, AB = O while A ≠ 0 and B ≠ 0.
iii) Let A=0100 and B=1000∴ AB=O and BA=10000100 ⇒BA=0+01+00+00+0⇒BA=0100
Thus, AB = O but BA ≠ O.
iv) Let A=1000 , B=0001 and C=0002∴AB=10000001⇒AB=0+00+00+00+0=0000and AC=10000002⇒AC=0+00+00+00+0=0000
Thus,
AB = AC
But B ≠ C and A ≠ 0.
Q60.
Answer :
LHS=A+B2 =A+BA+B =AA+B+BA+B =A2+AB+BA+B2
We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
A+B2 ≠ A2+2AB+B2
Q61.
Answer :
i) LHS =A+B2 =A+BA+B =AA+B+BA+B =A2+AB+BA+B2
We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
A+B2 ≠ A2+2AB+B2
ii) LHS=A-B2 =A-BA-B =AA-B-BA-B =A2-AB-BA+B2
We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
A-B2 ≠ A2-2AB+B2
iii) LHS=A+BA-B =AA-B+BA-B =A2-AB+BA-B2
We know that a matrix does not have commutative property. So,
AB ≠ BA
Thus,
A+BA-B ≠ A2-B2
Q62.
| Shopkeepers | Notebooks In dozen | Pens In dozen | Pencils In dozen |
| A | 12 | 5 | 6 |
| B | 10 | 6 | 7 |
| C | 11 | 3 | 8 |
Here,
Cost of notebooks per dozen = 12×40 paise = Rs 4.80
Cost of pens per dozen = Rs. 12×1.25 = Rs 15
Cost ofpPencils per dozen = 12×35 paise = Rs 4.20
∴ 12561067111384.80154.20=12×4.80+5×15+6×4.2010×4.80+6×15+7×4.2011×4.80+13×15+8×4.20 =57.60+75+25.2048+90+29.4052.80+195+33.60 =157.80167.40281.40
Thus, the bills of A, B and C are Rs 157.80, Rs 167.40 and Rs 281.40, respectively.
Q63.
Answer :
Stock of various types of books in the store is given by
Physics Chemistry MathematicsX=120 96 60
Selling price of various types of books in the store is given by
Y=8.303.454.50PhysicsChemistryMathematics
Total amount received by the store from selling all the items is given by
XY=12096608.303.454.50 =1208.30+963.45+604.50 =996+331.20+270 =1597.20
Required amount = Rs 1597.20
Q64.
Answer :
The cost per contact in paise is given by
A=4010050TelephoneHousecallLetter
The number of contacts of each type made in the two cities X and Y is given by
Telephone Housecall LetterB=1000 500 50003000 1000 10000 XY
Total amount spent by the group in the two cities X and Y is given by
BA=1000500500030001000100004010050 =40000+50000+250000120000+100000+500000 =340000720000XY
Thus,
Amount spent on X = Rs 3400
Amount spent on Y = Rs 7200
Page 5.44 Ex. 5.3
Q65.
Answer :
If Rs x are invested in the first type of bond and Rs 30000-x are invested in the second type of bond, then the matrix A=x30000-x represents investment and the matrix B=51007100 represents rate of interest.
i x30000-x51007100=1800⇒5×100+730000-x100=1800⇒5x+210000-7×100=1800⇒210000-2x=180000⇒2x=30000⇒x=15000
Thus,
Amount invested in the first bond = Rs 15000
Amount invested in the second bond = Rs 30000-15000
= Rs 15000
ii x30000-x51007100=2000⇒5×100+730000-x100=2000⇒5x+210000-7×100=2000⇒210000-2x=200000⇒2x=10000⇒x=5000
Thus,
Amount invested in the first bond = Rs 5000
Amount invested in the second bond = Rs 30000-5000
= Rs 25000
Q66.
Answer :
Here, A=111333, B=3152-24 and C=42-3550Now, AB=1113333152-24⇒AB=3+5-21+2+49+15-63+6+12⇒AB=671821AC=11133342-3550⇒AC=4-3+52+5+012-9+156+15+0⇒AC=671821
So, AB = AC though B ≠ C , A ≠ O.
Page 5.49 Ex. 5.4
Q1.
Answer :
Given: A=2-3-75AT=2-7-35B=102-4 BT=120-4i 2AT=2AT⇒22-3-75T=22-7-35⇒4-6-1410T=4-14-610⇒4-14-610=4-14-610 ∴ LHS=RHSii A+BT=AT+BT⇒2-3-75+102-4T=2-7-35+120-4⇒2+1-3+0-7+25-4T=2+1-7+2-3+05-4⇒ 3-3-51T=3-5-31⇒3-5-31=3-5-31∴ LHS=RHSiii A-BT=AT-BT⇒2-3-75-102-4T=2-7-35-120-4⇒2-1-3-0-7-25+4T=2-1-7-2-3-05+4⇒1-3-99T=1-9-39⇒1-9-39=1-9-39∴ LHS=RHSiv ABT=BTAT⇒2-3-75102-4T=120-4 2-7-35⇒2-60+12-7+100-20T=2-6-7+100+120-20⇒-4123-20T=-4312-20⇒-4312-20=-4312-20∴ LHS=RHS
Q2.
Answer :
Given: A=352AT=352B=104BT=104Now, AB=352104⇒AB=30125020208⇒ABT=35200012208 …1BTAT=104352⇒BTAT=35200012208 …2⇒ ABT=BTAT From eqs. (1) and (2)
Q3.
Answer :
Given: A=1-10213121 and B=123213011AT=121-112031 and BT=120211331i A+B=1-10213121+123213011 ⇒A+B=1+1-1+20+32+21+13+31+02+11+1⇒A+B=213426132⇒A+BT=241123362 …1Now, AT+BT=121-112031+120211331⇒AT+BT=1+12+21+0-1+21+12+10+33+31+1⇒AT+BT=241123362 …2⇒A+BT=AT+BT From eqs. 1 and 2 ii AB=1-10213121 123213011 ⇒AB=1-2+02-1+03-3+02+2+04+1+36+3+31+4+02+2+13+6+1⇒AB=-11048125510⇒ABT=-14518501210 …1Now,BTAT=120211331121-112031⇒BTAT=1-2+02+2+01+4+02-1+04+1+32+2+13-3+06+3+33+6+1⇒BTAT=-14518501210 …2⇒ABT=BTAT From eqs. (1) and (2)iii 2A=21-10213121 ⇒2A=2-20426242⇒2AT=242-224062 …1Now, 2AT=2121-112031⇒2AT=242-224062 …2⇒ 2AT=2AT From eqs. (1) and (2)
Q4.
Answer :
Given: A=-245AT=-245B=13-6 BT=13-6Now, AB=-245 13-6 ⇒AB=-2-612412-24515-30⇒ABT=-245-6121512-24-30 …1 BTAT=13-6-245⇒BTAT=-245-6121512-24-30 …2∴ ABT=BTAT From eqs. (1) and (2)
Q5.
Answer :
Here,AB=24-1-10234-1221⇒AB=6-4-28+8-1-3-0+4-4+0+2⇒AB=0151-2⇒ABT=0115-2
Q6.
Answer :
i) Given: A=213410 AT=241130B=1-10250BT=105-120Now,AB=2134101-10250 ⇒AB =2+0+15-2+2+04+0+0-4+2+0⇒AB =1704-2⇒ABT=1740-2 …1BTAT=105-120241130⇒BTAT=2+0+154+0+0-2+2+0-4+2+0⇒BTAT=1740-2 …2∴ ABT=BTAT From eqs. (1) and (2)
ii) Given: A=1324AT=1234B=1425BT=1245Now,AB=1324 1425⇒AB=1+64+152+88+20⇒AB=7191028⇒ABT=7101928 …1Also, BTAT=12451234⇒BTAT=1+62+84+158+20⇒BTAT=7101928 …2∴ ABT=BTAT From eqs. (1) and (2)
Q7.
Answer :
Given: li, mi, ni are direction cosines of three mutually perpendicular vectors. ⇒l1l2+m1m2+n1n2=0 l2l3+m2m3+n2n3=0 l1l3+m1m3+n1n3=0 …1And, l12+m12+n12=1l22+m22+n22=1l32+m32+n32=1 …2 A=l1m1n1l2m2n2l3m3n3 AT=l1l2l3m1m2m3n1n2n3 Now, AAT=l1m1n1l2m2n2l3m3n3 l1l2l3m1m2m3n1n2n3 ⇒AAT=l12+m12+n12l1l2+m1m2+n1n2 l1l3+m1m3+n1n3l1l2+m1m2+n1n2l22+m22+n22l2l3+m2m3+n2n3l1l3+m1m3+n1n3l3l2+m3m2+n3n2l32+m32+n32 ⇒AAT=100010001 From eqs. 1 and 2⇒AAT=I
Q8.
Answer :
Given: A=cos αsin α-sin αcos α AT=cos α-sin αsin αcos αNow, ATA=I2Consider: LHS=ATA=cos α-sin αsin αcos αcos αsin α-sin αcos α=cos2α+sin2αcos α sin α-sin α cos αsin α cos α-cos α sin αsin2α+cos2α=1001=RHS
Hence proved.
Q9.
Answer :
Given: A=sinαcosα-cosαsinα AT=sinα-cosαcosαsinαNow, ATA=sinα-cosαcosαsinαsinαcosα-cosαsinα ⇒ATA=sinαsinα+-cosα-cosαsinαcosα+-cosαsinαcosαsinα+sinα-cosαcosαcosα+sinαsinα⇒ATA=sin2α+cos2αsinα cosα-sinα cosαsinα cosα-sinα cosαcos2α+sin2α⇒ATA=1001=I
Page 5.54 Ex. 5.5
Q1.
Answer :
Given: A=2345 AT=2435Now,A-AT=2345-2435⇒A-AT=2-23-44-35-5⇒A-AT=0-110 …1A-ATT=0-110T⇒A-ATT=01-10⇒A-ATT=-0-110⇒A-AT=-A-ATT Using eq. 1Thus, A-AT is a skew-symmetric matrix.
Q2.
Answer :
Given: A=3-41-1AT=31-4-1Now, A-AT=3-41-1-31-4-1⇒A-AT=3-3-4-11+4-1+1⇒A-AT=0-550 …1A-ATT=0-550T⇒A-ATT=05-50⇒A-ATT=-0-550⇒A-ATT=-A-AT From eq. 1Thus, A-AT is a skew-symmetric matrix.
Q3.
Answer :
Given: A=52xyz-34t-7⇒AT=5y42ztx-3-7Since A is a symmetric matrix, AT=A.⇒ 5y42ztx-3-7=52xyz-34t-7The corresponding elements of two equal matrices are equal.∴ x=4 y=2 z=z t=-3Hence, x=4, y=2, t=-3 and z can have any value.
Page 5.55 Ex. 5.5
Q4.
Answer :
Given: A=327143-258 ⇒ AT=31-2245738Let X=12A+AT=12327143-258+31-2245738=3325232445248Let Y=12A-AT=12327143-258-31-2245738=01292-120-1-9210XT=3325232445248T=3325232445248T=XYT=01292-120-1-9210T=0-12-92120192-10=-01292-120-1-9210=YThus, X is a symmetric matrix and Y is skew-symmetric matrix.Now,X+Y=3325232445248+01292-120-1-9210=327143-258=A∴ X=3325232445248 and Y=01292-120-1-9210
Q5.
Answer :
Given: A=42-13571-21AT=43125-2-171Let X=12A+AT=1242-13571-21+43125-2-171=4520525520521XT=4520525520521T=4520525520521=XLet Y=12A-AT=1242-13571-21-43125-2-171 =0-12-1120921-920YT=0-12-1120921-920T=0121-120-92-1920=-0-12-1120921-920=-YThus, X is a symmetric matrix and Y is a skew-symmetric matrix.Now, X+Y=4520525520521+0-12-1120921-920=42-13571-21=A
Q6.
Answer :
A square matrix A is called a symmetric matrix, if AT=A.Given: A=2456 AT=2546Now, A+AT=2456+2546 ⇒A+AT=49912 …1A+ATT=49912T =49912 =A+AT From eq. 1∴ A+ATT=A+ATThus, A+AT is a symmet
Q7.
Answer :
Given: A=3-41-1AT=31-4-1Let X=12A+AT=123-41-1+31-4-1=3-32-32-1XT=3-32-32-1T=3-32-32-1=XLet Y=12A-AT=123-41-1-31-4-1 =0-52520YT=0-52520T=052-520=-0-52520=Y∴ X is a symmetric matrix and Y is a skew-symmetric matrix.X+Y=3-32-32-1+0-52520=3-41-1=A
Q8.
Answer :
Here, A=3-2-43-2-5-112⇒ AT=33-1-2-21-4-52Let X=12A+AT=123-2-43-2-5-112+33-1-2-21-4-52=312-5212-2-2-52-22XT=312-5212-2-2-52-22T=312-5212-2-2-52-22=XLet Y=12A-AT=123-2-43-2-5-112-33-1-2-21-4-52=0-52-32520-33230 YT=0-52-32520-33230T=05232-5203-32-30=-0-52-32520-33230=-Y⇒ X is a symmetric matrix and Y is a skew-symmetric matrix.Now, X+Y=312-5212-2-2-52-22+0-52-32520-33230=3-2-43-2-5-112=A
Page 5.55 (Very Short Answers)
Q1.
Answer :
Given: Order of A = m×n
Order of B = n×p
Since the number of columns in A are equal to the number of rows in B, i.e. n, AB exists.
Order of AB = Number of rows in A× Number of columns in B
= m×p
Q2.
Answer :
The order of matrix A is 2×3 and the order of matrix B is 3×2.
Since the number of columns in A is equal to the number of rows in B, AB exists and it is of order 2×2.
Also, since the number of columns in B is equal to the number of rows in A, BA exists and it is of order 3×3.
Q3.
Answer :
AB=4312-43⇒AB=-16+9-4+6⇒AB=-72
Q4.
Answer :
Given: A=123 AT=123Now, AAT=123123⇒AAT=123246369
Q5.
Answer :
Let the two matrices be A=5090 and B=0012-21, such that AB=50900012-21.⇒AB=0000
Q6.
Answer :
Given: A=2357 AT=2537Now, A+AT=2357+2537⇒A+AT=2+23+55+37+7⇒A+AT =48814
Q7.
Answer :
Given: A=i00iA2=AA⇒A2=i00ii00i⇒A2=i2+00+00+00+i2⇒A2=i200i2⇒A2=-100-1 ∵ i2=-1
Q8.
Answer :
Given: A=cos xsin x-sin xcos xAT=cos x-sin xsin xcos xNow, A+AT=I⇒cos xsin x-sin xcos x+cos x-sin xsin xcos x=1001⇒cos x+cos xsin x-sin x-sin x+sin xcos x+cos x=1001⇒2cos x002cos x=1001⇒2cos x=1 ⇒cos x=12⇒x=π3
Q9.
Answer :
Given: A=cosx-sinxsinxcosx ⇒AT=cosxsinx-sinxcosxAAT=cosx-sinxsinxcosxcosxsinx-sinxcosx⇒AAT=cos2x+sin2xcosxsinx-sinxcosxcosxsinx-sinxcosxsin2x+cos2x⇒AAT=1001
Page 5.56 (Very Short Answers)
Q10.
Answer :
Given: 10y5+2×01-2=I⇒10y5+2×02-4=1001⇒1+2×0+0y+25-4=1001⇒1+2x0y+21=1001∴ 1+2x=1 and y+2=0⇒2x=1-1 and y=-2⇒2x=0⇒x=02=0
Q11.
Answer :
A2=AA⇒A2=1-1-111-1-11⇒A2=1+1-1-1-1-11+1⇒A2=2-2-22Now, A2=kA⇒2-2-22=k1-1-11⇒2-2-22=k-k-kk∴ k=2
Q12.
Answer :
A2=A.A⇒A2=11111111⇒A2=1+11+11+11+1⇒A2=2222Now, A4=A2A2⇒A4=22222222⇒A4=4+44+44+44+4⇒A4=8888Also, A4=λA⇒8888=λ1111⇒8888=λλλλ∴ λ=8
Q13.
Answer :
Here,A2=AA⇒A2=-1000-1000-1-1000-1000-1⇒A2=1+0+00+0+00+0+00+0+00+1+00+0+00+0+00+0+00+0+1⇒A2=100010001
Q14.
Answer :
Here,A2=AA⇒A2=-1000-1000-1-1000-1000-1⇒A2=1+0+00+0+00+0+00+0+00+1+00+0+00+0+00+0+00+0+1⇒A2=100010001Now, A3=A2A⇒A3=100010001-1000-1000-1⇒A3=-1+0+00+0+00+0+00+0+00-1+00+0+00+0+00+0+00+0-1⇒A3=-1000-1000-1=A
Q15.
Answer :
Here,A2=AA⇒A2=-300-3-300-3⇒A2=9+00+00+00+9⇒A2=9009Now,A4=A2A2⇒A4=90099009⇒A4=81+00+00+00+81⇒A4=810081
Q16.
Answer :
Given: x234=2⇒3x+8=2⇒3x=2-8⇒3x=-6⇒x=-63⇒x=-2
Q17.
Answer :
Here, aij=i+2j, 1≤i≤2 and 1≤j≤2∴ a11=1+21=3, a12=1+22=1+4=5⇒a21=2+21=4 and a22=2+22=2+4=6∴ A=a11a12a21a22=3546
Q18.
Answer :
Given: A+23-14=3-6-38⇒A=3-6-38-23-14⇒A=3-2-6-3-3+18-4⇒A=1-9-24
Q19.
Answer :
Here,aij=i2-j2 , 1≤i≤2 and 1≤j≤2∴ a11=12-12=1-1=0 , a12=12-22=1-4=-3a21=22-12=4-1=3 and a22=22-22=4-4=0∴ A=a11a12a21a22=0-330AT=03-30⇒AT=-0-330⇒AT=-ASince AT=-A, A is skew-symmetric.
Q20.
Answer :
Here,
AATT=ATTAT ∵ ABT=BTAT⇒ AATT=AAT ∵ ATT=A
Thus, AAT is a symmetric matrix.
Q21.
Answer :
Given: A=aij is a skew-symmetric matrix.⇒aij=-aij For all values of i, j⇒aii=-aii For all values of i⇒aij+aii=0⇒2aii=0⇒aii=0 For all values of i∑iaii=0+0+…+0 i times =0Thus,∑iaii=0
Q22.
Answer :
Given: A=aij is a skew symmetric matrix.⇒aij=-aji For all values of i, j⇒aii=-aii For all values of i⇒aij=0Now, ∑i∑jaij=a11+a12+a13+…+a21+a22+a23+…+a31+a32+a33+… =0+a12+a13+…-a12+0+a23+…-a13-a23+0+… =0Thus, ∑i∑jaij=0
Q23.
Answer :
Given: AB is symmetric.
⇒ ABT=AB⇒BTAT=AB ∵ ABT=BTAT⇒BA=AB ∵ A and B are symmetric matrices, so AT=A and BT=B
Thus, AB is also symmetric, if AB = BA.
Q24.
Answer :
If B is a skew-symmetric matrix, then BT=-B.
ABATT=ATTBTAT ∵ ABCT=CTBTAT⇒ABATT=ABTAT ∵ ATT=A⇒ABATT=A-B AT ∵ BT=-B⇒ABATT=-ABAT∴ ABAT is a skew-symmetric matrix.
Q25.
Answer :
If B is a skew-symmetric matrix, then BT=-B.
ABATT=ATTBTAT ∵ ABCT=CTBTAT⇒ABATT=ABTAT ∵ ATT=A⇒ABATT=A-B AT ∵ BT=-B⇒ABATT=-ABAT∴ ABAT is a skew-symmetric matrix.
Q26.
Answer :
Given: A is skew symmetric matrix.
⇒AT=-A
AnT=λAn⇒ATn=λAn⇒-An=λAn⇒-1n An=λAn⇒λ=-1n
Q27.
Answer :
If A is a symmetric matrix, then AT=A.Now, AnT=ATn for all n∈N⇒AnT=An ∵ AT=A
Hence, An is a symmetric matrix.
Q28.
Answer :
If A is a skew-symmetric matrix, then AT=-A.AnT=ATn For all n∈N⇒AnT=-An ∵ AT=-A ⇒AnT=-1n An⇒AnT=An, if n is even or -An, if n is odd.
Hence, An is symmetric when n is an even natural number.
Q29.
Answer :
If A is a skew-symmetric matrix, then AT=-A.AnT=ATn For all n∈N⇒AnT=-An ∵ AT=-A⇒AnT=-1n An⇒AnT=An, if n is even or -An, if n is odd.
Hence, An is skew-symmetric when n is an odd natural number.
Q30.
Answer :
Since A and B are symmetric matrices, AT=A and BT=B.
Here,
AB-BAT=ABT-BAT⇒AB-BAT=BTAT-ATBT ∵ ABT=BTAT⇒AB-BAT=BA-AB ∵ BT=B and AT=A⇒AB-BAT=-AB-BATherefore, AB-BA is skew-symmetric.
Q31.
Answer :
Let A=0000 AT=0000Since AT=A, A is a symmmetric matrix.Now, -A=-0000 ⇒-A=0000Since AT=-A, A is a skew-symmetric matrix.Thus, A=0000 is an example of a matrix that is both symmetric and skew-symmetric.
Q32.
Answer :
Given: 2130x+y012=5618⇒2602x+y012=5618⇒2+y6+00+12x+2=5618⇒2+y612x+2=5618∴ 2+y=5 and 2x+2=8⇒y=5-2 and 2x=8-2⇒y=3 and 2x=6⇒x=62=3
Q33.
Answer :
The corresponding elements of two equal matrices are equal.
Given: x+34y-4x+y=5439x+3=5 and y-4=3⇒x=5-3 and y=3+4⇒x=2 and y=7∴ x=2 and y=7
Q34.
Answer :
The corresponding elements of two equal matrices are equal.
Given: 2x-y53y=653-22x-y=6 …1y=-2Putting the value of y in eq. 12x–2=6⇒2x+2=6⇒2x=6-2⇒2x=4⇒x=42=2∴ x=2
Q35.
Answer :
Given: x-y2x5=2235The corresponding elements of two equal matrices are equal.∴x-y=2 …1 and x=3Putting the value of x in eq. 13-y=2⇒3-2=y∴ y=1
Page 5.57 (Very Short Answers)
Q36.
Answer :
The corresponding elements of two equal matrices are equal.
Given: 3x+y-y2y-x3=12-533x+y=1 …1 -y=2⇒y=-2Putting the value of y in eq. 13x+-2=1⇒3x-2=1⇒3x=1+2⇒3x=3⇒x=33=1∴ x=1
Q37.
Answer :
Given: A=123 AT=123AAT=123123⇒AAT=1+4+9⇒AAT=14
Q38.
Answer :
The corresponding elements of two equal matrices are equal.
Given: 2x+y3y04=60642x+y=6 …1 3y=0⇒y=0Putting the value of y in eq. 12x+0=6⇒2x=6⇒x=62∴ x=3
Q39.
Answer :
Given: A=1234 AT=1324A+AT=1234 +1324⇒A+AT=1+12+33+24+4⇒A+AT=2558.
Q40.
Answer :
The corresponding elements of two equal matrices are equal.
⇒a+b25b=6522⇒a+b=6 …1∴ b=2Putting the value of b in eq. 1a+2=6⇒a=6-2∴ a=4
Q41.
Answer :
If A is a matrix of order 3 × 4 and B is a matrix of order 4 × 3, then the order of matrix AB is given by the number of rows in A and number of columns in B, respectively.
Thus, the order of matrix AB is 3×3.
Q42.
Answer :
Here, A=cos α-sin αsin αcos α=I⇒cos α-sin αsin αcos α=1001The corresponding elements of equal matrices are equal.∴ cos α=1 ⇒α=0°
Q43.
Answer :
Given: 12343125=711k23⇒3+41+109+83+20=711k23⇒7111723=711k23The corresponding elements of two equal matrices are equal.∴ k=17
Q44.
Answer :
Given: A is a square matrix, such that A2=A.
Here,
I+A2-3A=I+AI+A-3A⇒I+A2-3A=I×I+I×A+A×I+A×A-3A using distributive property⇒I+A2-3A=I+A+A+A2-3A using I×I=I and IA=AI=A⇒I+A2-3A=I+2A+A-3A ∵ A2=A⇒I+A2-3A=I+3A-3A⇒I+A2-3A=I
Q45.
Answer :
Given: A=1203⇒ AT=1023Let B=12A+AT=121203+1023 =121+12+00+23+3 =122226 =1113Now, BT=1113=BTherefore, B is symmetric matrix.Let C=12A-AT=121203-1023 =121-12-00-23-3 =1202-20 =01-10 ∴ CT=01-10T=0-110=-01-10=CSo, C is a skew-symmetric matrix.Now, B+C=1113+01-10=1+01+11-13+0=1203=A∴ B=1113
Q46.
Answer :
Order of A = 2×3
Order of AT=3×2
Let order of B = m×n
Given: ATB and BAT are definedIf AT3×2Bm×n exists, then the number of columns in AT must be equal to number of rows in B.⇒m=2 If Bm×n AT3×2 exists, then the number of columns in B must be equal to number of rows in AT.⇒n=3∴ Order of B=2×3
Q47.
Answer :
In a 2×2 matrix, the total number of elements are 4 and each entry can be written in 2 ways.
Number of ways in which 4 entries can be written = 42=16 [Applying the above property]
Q48.
Answer :
We have, xx-y2x+y7=3187The corresponding elements of two equal matrices are equal.∴ x=3 x-y=1 …1 Putting the value of x in eq. 13-y=1⇒3-1=y∴ y=2
Q49.
Answer :
We know that if a matrix is of order m×n, then it has mn elements.
If the matrix has 5 elements, then the number of elements will be 1×5 or 5×1, i.e. there will be 2 possible orders of the matrix.
Q50.
Answer :
Here, aij=ij1≤ i≤2 1≤ j≤2⇒ a12=12Therefore, the value of a12 is12.
Page 5.58 (Multiple Choice Questions)
Q1.
Answer :
b a unit matrix
A2=AA⇒A2=100010ab-1100010ab-1⇒A2=1+0+00+0+00+0-00+0+00+1+00+0-0a+0-a0+b-b0+0+1⇒A2=100010001
Q2.
Answer :
(c) 1001
Here, A=i00i⇒ A2=i00ii00i=i200i2⇒ A3=A2.A=i200i2i00i=-i00-i⇒A4=A3.A=-i00-ii00i=1001
So, 1001 is repeated on multiple of 4 and 4n is a multiple of 4.
Thus,
A4n=1001
Q3.
Answer :
(a) B
Here, AB=A …1 BA=B …2⇒BAB=BB Multiplying both sides by B⇒BA=B2 From eq. 1⇒B=B2 From eq. 2
Q4.
Answer :
(a) B2 = B and A2 = A
Here, AB=A …1 BA=B …2⇒ABA=AA Multiplying both sides by A BAB=BB Multiplying both sides by A⇒AB=A2 From eq. 2 BA=B2 From eq. 1⇒A=A2 From eq. 1 B=B2 From eq. 2
Q5.
Answer :
(c) A + B
Given: AB=B and BA=A A2+B2=AA+BB⇒ A2+B2=BABA+ABAB ∵ AB=B and BA=A⇒A2+B2=BBA+AAB ∵ AB=B and BA=A⇒A2+B2=BA+AB ∵ AB=B and BA=A⇒A2+B2=A+B ∵ AB=B and BA=A
Q6.
Answer :
(d) 7
Here, A=cos 2π7-sin2π7sin2π7cos2π7⇒A2=A×A⇒A2=cos 2π7-sin2π7sin2π7cos2π7 cos 2π7-sin2π7sin2π7cos2π7⇒A2=cos22π7-sin22π7-2cos2π7sin2π72cos2π7sin2π7cos22π7-sin22π7⇒A2=cos4π7-sin4π7sin4π7cos4π7 ∵ cos2θ-sin2θ=cos2θ2sinθ cosθ=sin2θ⇒A3=A2×A⇒A3=cos4π7-sin4π7sin4π7cos4π7 cos 2π7-sin2π7sin2π7cos2π7⇒A3=cos 4π7cos2π7-sin4π7sin2π7-cos4π7sin2π7-sin4π7cos2π7sin4π7cos2π7+cos4π7sin2π7-sin2π7sin4π7+cos4π7cos2π7⇒A3=cos6π7-sin6π7sin6π7cos6π7 ∵ cosA+B=cosAcosB-sinAsinBsinA+B=sinAcosB+cosAsinB
Now we check if the pattern is same for k = 6.
Here,
A6=A3.A3⇒A6=cos 6π7-sin6π7sin6π7cos6π7 cos 6π7-sin6π7sin6π7cos6π7⇒A6=cos 12π7-sin12π7sin 12π7cos 12π7
Now, we check if the pattern is same for k = 7.
Here,
A7=A6×A⇒A7=cos 12π7-sin12π7sin 12π7cos 12π7 cos2π7-sin2π7sin2π7cos2π7⇒A7=cos 14π7-sin14π7sin 14π7cos 14π7⇒A7=cos 2π -sin2πsin 2π cos 2π ∵14π7=2π =1001
So, the least positive integral value of k is 7.
Q7.
Answer :
(a) It is not necessary that either A = O or, B = O
Let A=0200 and B=1000∴ AB=02001000=0000
Q8.
Answer :
(c) an000an000an
Here, A=a000a000a⇒ A2=a000a000aa000a000a=a2000a2000a2⇒ A3=a2000a2000a2a000a000a=a3000a3000a3This pattern is applicable on all natural numbers.∴ An=an000an000an
Q9.
Answer :
a null matrix
Since A is non-singular matrix and the determinant of a non-singular matrix is non-zero, B should be a null matrix.
Q10.
Answer :
(b) nB
Here, A=n000n000n and B=a1a2a3b1b2b3c1c2c3∴ AB=n000n000na1a2a3b1b2b3c1c2c3⇒AB=na1na2na3nb1nb2nb3nc1nc2nc3⇒AB=na1a2a3b1b2b3c1c2c3⇒AB=nB
Q11.
Answer :
(a) 1na01
Here, A=1a01⇒A2=1a011a01=1+0a+a0+00+1=12a01 A3=A2×A=12a011a01=1+0a+2a0+00+1=13a01
This pattern is applicable for all natural numbers.
∴ An=1na01
Page 5.59 (Multiple Choice Questions)
Q12.
Answer :
(a) 0
Given: AB=I3⇒ 12×0100011-2y010001=100010001⇒ 10y+x010001=100010001The corresponding elements of two equal matrices are equal.∴ y+x=0
Q13.
Answer :
(b) a = 1, b = 4
Here,A+B2=A2+B2⇒A2+AB+BA+B2=A2+B2⇒AB+BA=O⇒AB=-BA⇒1-12-1a1b-1=-a1b-11-12-1⇒a-b22a-b3=-a+2-a-1b-2-b+1⇒a-b22a-b3=-a-2a+1b+2b-1The corresponding elements of two equal matrices are equal.⇒a+1=2 and b-1=3∴ a=1 and b=4
Q14.
Answer :
(c) 1 − α2 − βγ = 0
Here, A2=I⇒αβγ-ααβγ-α=1001⇒α2+βγαβ-βαλα-αγγβ+α2=1001⇒α2+βγ00γβ+α2=1001The corresponding elements of two equal matrices are equal.⇒α2+βγ=1 ⇒1-α2-βγ=0
Q15.
Answer :
(c) kA
Here,
S=Sij⇒S=k00k ∵ Sij=k Let A=a11a12a21a22 ∵ A is square matrixNow, AS=a11a12a21a22k00k=ka11ka12ka21ka22=ka11a12a21a22=kASA=k00ka11a12a21a22=ka11ka12ka21ka22=ka11a12a21a22=kA∴ AS=SA=kA
Q16.
Answer :
(c) I
Here, A2=A …1A3=A2A =A2 From eq. 1 =A ∴ A3=A …2We know that I+A3=I3+3I2A+3IA2+A3⇒I+A3=I+3A+3A+A From eqs. 1 and 2 ⇒I+A3=I+7A⇒I+A3-7A=I
Q17.
Answer :
(b) A is a zero matrix
Let A=aij be a matrix which is both symmetric and skew-symmetric.
If A=aij is a symmetric matrix, then
aij=aji for all i, j …(1)
If A=aij is a skew-symmetric matrix, then
aij=-aji for all i, j
⇒aji=-aij for all i,j …(2)
From eqs. (1) and (2), we have
aij=-aij ⇒aij+aij=0 ⇒2aij=0 ⇒aij=0 ∴ A=aij is a zero matrix or null matrix.
Q18.
Answer :
(a) a skew-symmetric matrix
Here,
A = 05-7-50117-110
⇒AT = 0-5750-11-7110
⇒AT=-05-7-50117-110⇒AT=-A
Thus, A is a skew-symmetric matrix.
Q19.
Answer :
(d) none of these
Given: A is a square matrix.
Let A=1210⇒AA=12101210=3212
Q20.
Answer :
(a) symmetric matrix
Since A and B are symmetric matrices, we getA=A’ and B=B’ABA’=BA’ A’ =A’B’A’ =ABA ∵ A=A’ and B=B’Since ABA’=ABA, ABA is a symmetric matrix.
Q21.
Answer :
(c) x = y
Here,A=5xy0 AT=5yx0Now, A=ATThe corresponding elements of two equal matrices are equal.∴ 5xy0=5yx0⇒x=y
Q22.
Answer :
(a) 3 × 4
The order of A is 3 × 4. So, the order of A’ is 4 × 3.
Now, both A’B and BA’ are defined. So, the number of columns in A’ should be equal to the number of rows in B for A’B.
Also, the number of columns in B should be equal to number of rows in A’ for BA’.
Hence, the order of matrix B is 3 × 4.
Q23.
Answer :
(d) none of these
Given: A is a square matrix of even order.Let A=a11a12a21a22⇒A=0-330 ∵ aij=i2-j2So, it is a skew-symmetric matrix as aij=-aji.Now, A=a11a12a21a22=a11a22-a21a12=0–9=9
Q24.
Answer :
(c) θ = 2nπ + π3, n ∈ Z
Here, A=cos θ-sin θsin θcos θ ⇒AT=cos θsin θ-sin θcos θNow, AT+A=I2⇒cos θsin θ-sin θcos θ+cos θ-sin θsin θcos θ=1001⇒2cos θ002cos θ=1001⇒2cos θ=1⇒cos θ=12⇒cos θ=cosπ3⇒θ=2nπ±π3 n∈Z
Page 5.60 (Multiple Choice Questions)
Q25.
Answer :
(a) 22-4 23 4-44 2
Here, A=20-3431-572⇒ AT=24-5037-312Now, A+AT=20-3431-572+24-5037-312⇒A+AT=2+20+4-3-54+03+31+7-5-37+12+2⇒A+AT=44-8468-884A-AT=20-3431-572-24-5037-312 ⇒ A-AT=2-20-4-3+54-03-31-7-5+37-12-2 ⇒A-AT=0-4240-6-260Let P=12A+AT=1244-8468-884=22-4234-442Q=12A-AT=120-4240-6-260=0-2120-3-130Now, PT=22-4234-442T=22-4234-442=PQT=0-2120-3-130T=02-1-2031-30=-0-2120-3-130=-QThus, P is symmetric and Q is skew-symmetric. P+Q=22-4234-442+0-2120-3-130=2+02-2-4+12+23+04-3-4-14+32+0=20-3431-572=AThus, we have expressed A is the sum of a symmetric and a skew-symmetric matrix.Hence, the symmetric matrix is 22-4234-442.
Q26.
Answer :
a 0000
A diagonal matrix in which all the diagonal elements are equal is called the scalar matrix.
Q27.
Answer :
(d) 512
There are 9 elements in a 3×3 matrix and one element can be filled in two ways, either with 0 or 1.
Thus,
Total possible matrices = 29 = 512
Q28.
Answer :
(d) Not possible to find
Here,3x+75y+12-3x=0y-284We know that for two equal matrices the corresponding elements are equal.∴ 3x+7=0, 5=y-2, y+1=8 and 2-3x=4⇒3x=-7, 5+2=y, y=8-1 and-3x=4-2⇒x=-73, y=7, y=7 and x=-23
Q29.
Answer :
(c) −6, −4, −9
Given: A=023-4Here,kA=03a2b24⇒k023-4=03a2b24⇒02k3k-4k=03a2b24The corresponding elements of two equal matrices are equal.⇒2k=3a, 3k=2b and-4k=24 Now,-4k=24 ⇒k=-6Also, 2k=3a and 3k=2b⇒2-6=3a and 3-6=2b Using k=-6⇒-12=3a and -18=2b∴ a=-4 and b=-9
Q30.
Answer :
a I cos θ+J sin θHere, I cos θ+J sin θ=1001cos θ+01-10sin θ=cos θ00cos θ+0sin θ-sin θ0=cos θsin θ-sin θcos θ=B
Q31.
Answer :
(a) 17
The trace of a matrix is the sum of the diagonal elements.
∴ Tr A = 1 + 7 + 9 = 17
Q32.
Answer :
(a) nk
Here, A=aijn×nTrace of A, i.e. trA=∑aiji=1n=a11+a22+ … +ann =k+k+k+ … n times =kn
5.69 q. No. 41 solve.give the solution of this question
5.69 q. No. 41 solve