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Matrix & Determinants
Problem Solved: 61
(a) Given,
x y
𝑦
𝑥+ 𝑦
𝑥
6
3[
]=[
]+[
]
z w
𝑧+ 𝑤
3
−1 2𝑤
𝑦
𝑥+ 𝑦
3𝑥 3𝑦
𝑥
6
0 0
 [
]−[
]−[
]=[
]
𝑧+ 𝑤
3
−1 2𝑤
0 0
3𝑧 3𝑤
(3𝑥 − 𝑥 − 4)
(3𝑦 − 6 − 𝑥 − 𝑦)
0 0
 [
]=[
]
(3𝑧 + 1 − 𝑧 − 𝑤) (3𝑤 − 2𝑤 − 3)
0 0
 [

2𝑥 − 4
2𝑦 − 𝑥 − 6
0 0
]=[
] then,
0 0
2𝑧 − 𝑤 + 1
𝑤−3

∴ 2𝑥 − 4 = 𝑜
 2𝑥 = 4
 𝑥=2

And

2y – 2– 6 =0
 2y – 8 = 0
 2y = 8
 Y=4
∴ x=2; y=4; z=1; w=3 (𝑎𝑛𝑠𝑤𝑒𝑟)
(b) Given,
1 −1
A+B = [
] … … … … …
...
(𝑖𝑖)
1 4
By adding equation (i) & (ii) we get,
1 −1
3 1
A+B+A−B = [
]+[
]
3 0
1 4
1 + 3 −1 + 1
(a) 2A = [
]
3+1 0+4
4 0
(b) 2A = [
]
4 4
2 0
(c) A = [
]
2 2

𝑤−3=0
=> 𝑤 = 3

2z – 3 + 1 = 0
=> 2z – 2 = 0
=> 2z = 2
=> z = 1

By subtracting equation (i) & (ii) we get,
1
A+B-A+B = [
3

−1
3 1
]−[
]
0
1 4

1 − 3 −1 − 1
]
3−1 0−4
2 −2
 2B = [
]
2 −4
−1 −1
(d) B = [
]
1 −2


2B = [

2 0
−1 −1
∴ The value of AB = [
]×[
]
2 2
1 −2
(−2 + 0) (−2 + 0)
=[
]
(−2 + 2) (−2 − 4)
−2 −2
=[
]
0 −6
−2 −2
∴ The value of AB = [
] (ans)
0 −6
(c) Given,
7
2
3
X-Y=[
0
X+Y=[

0
]…………
...
(ii)
3

By adding equation (i) & (ii) we get,
7 0
3 0
X+Y+X-Y = [
]+[
]
2 5
0 3
(7 + 3) (0 + 0)

2X = [
]
(2 + 0) (5 + 3)
10 0

2X = [
]
2 8
5 0

X=[
]
1 4
By subtracting equation (i) & (ii) we get,
7 0
3 0
X+Y-X-Y = [
]−[
]
2 5
0 3
(7 − 3) (0 − 0)
 2Y = [
]
(2 − 0) (5 − 3)
4 0
 2Y = [
]
2 2
2 0
 Y=[
]
1 1

∴ The value of (x and y): x = [

2 0
5 0
] 𝑎𝑛𝑑 Y = [
] (ans)
1 1
1 4

(d) Given,
3 4
A= [
]
6 2
7 8
B= [
]
4 3
3 4
7 8
∴ A+B = [
]+[
]
6 2
4 3
(3 + 7) (4 + 8)
=[
]
(6 + 4) (2 + 3)
10 12
= [
]
10 5
3
6
3
= [
4

∴ AT = [

4T
]
2
6
]
2

7 8T
∴ BT = [
]
4 3
7 4
= [
]
8 3
L
...
S = (𝐴 + 𝐵)T
10 12 T
= [
]
10 5
10 10
=[
]
12 5
R
...
S = A 𝑇 + B 𝑇
7 4
3 6
=[
]+ [
]
8 3
4 2
(3 + 7) (6 + 4)
=[
]
(4 + 8) (2 + 3)
10 10
=[
]
12 5
∴ L
...
S = R
...
S [SHOWED]

(e) Given,
x 𝑧
1 −1
3 5
2 [y 𝑡 ] + 3 [
] = 3[
]
0 2
4 6
2𝑥 2𝑧
3 −3
9 15

[
]+[
]=[
]
2𝑦 2𝑡
0 6
12 18
2𝑥 2𝑧
3 −3
9 15

[
]=[
] −[
]
2𝑦 2𝑡
12 −6
12 18
2𝑥 2𝑧
9 − 3 15 + 3

[
]=[
]
2𝑦 2𝑡
12 − 0 18 − 6
2𝑥 2𝑧
6 18

[
]=[
] then,
2𝑦 2𝑡
12 12
∴ 2𝑥 = 6
=> 𝑥 = 3

∴ 2𝑧 = 18
=> 𝑧 = 9

∴ 2𝑦 = 12
=> 𝑦 = 6

∴ The value of (x,y,z & t) are (3,6,9 & 6) (ans)
(f) Given,
2 −1 3
A=[ 4
2 0]
−2 7 3
2 −1 3
B=[ 4
2 0]
−2 7 3
∴ 2A + 3x = 5B
6 −2 7
2 −1 3

3x = 5 [8 0 9] − 2 [ 4
2 0]
3 −1 5
−2 7 3
4 −2 6
30 −10 35

3x = [40
4 0]
0
45] − [ 8
−4 14 6
15 −5 25
(30 − 4) (−10 + 12) (35 − 6)
(0 − 4)
(45 − 0)]

3x = [(40 − 8)
(15 + 4) (−5 − 14) (25 − 6)
26 −8 29

3x = [32 −4 45]
19 −19 19
26/3 −8/3 29/3

X = [32/3 −4/3 45/3] (𝑎𝑛𝑠)
19/3 −19/3 19/3

∴ 2𝑡 = 12
=> 𝑡 = 6



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