Question 25
Last updated at Dec. 21, 2017 by Teachoo

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Question 25
If A = [■8(3&2&1@4&−1&2@7&3&−3)], then find A-1 and hence solve the following system of equations : 3x + 4y + 7z = 14, 2x − y + 3z = 4, x + 2y − 3z = 0
Given
3x + 4y + 7z = 14,
2x − y + 3z = 4,
x + 2y − 3z = 0
Write equation as AX = B
[■8(3&4&7@2&−1&3@1&2&−3)][■8(𝑥@𝑦@𝑧)] = [■8(14@4@0)]
i.e. A’X = B
Hence A =[■8(3&2&1@4&−1&2@7&3&−3)], A’= [■8(3&4&7@2&−1&3@1&2&−3)] , X = [■8(𝑥@𝑦@𝑧)] & B =[■8(14@4@0)]
Now,
A’X = B
X = (A’)-1 B
X = (A-1)’ B
Finding A-1
Calculating |A|
|A|= |■8(3&2&1@4&−1&2@7&3&−3)|
= 3(−1(-3) – 3(2)) − 2 (4(–3) − 7(2)) + 1 (4(3) − 7(–1))
= 3(–3) – 2(–26) + 1(19) = –9 + 52 + 19
= 62
So, |A|≠ 0
∴ The system of equation is consistent & has a unique solutions
Now, A-1 = 1/(|A|) adj (A)
adj A = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)]
A = [■8(3&2&1@4&−1&2@7&3&−3)]
𝐴11 = (–1)1+1 [–1(–3) – 3(2) ] = 3 – 6 = –3
𝐴12 = (–1)1+2 [4(–3) – 7(2)]= −[−12−14] = − (–26) = 26
𝐴13 = (–1)1+3 [4(3) – 7(–1)] = 12 + 7 = 19
𝐴21 = (–1)2+1 [2(–3) – 3(1)] = –(–6 – 3) = 9
𝐴22 = (–1)2+2 [3(–3) – 7(1)] = −9 − 7 = −16
𝐴23 = (–1)2+3 [3(3) – 7(2)] = – [9 – 14] = 5
𝐴31 = (–1)3+1 [2(2) – (–1)(1)] = 4 + 1 = 5
𝐴32 = (–1)3+2 [3(2) – 4(1)] = – [6 – 4] = –2
𝐴33 = (–1)3+3 [3(–1) – 4(2)] = –3 – 8 = –11
Thus adj A = [■8(−3&9&5@26&−16&−2@19&5&−11)]
& |A| = 62
Now,
A-1 = 1/(|A|) adj A
A-1 = 1/62 [■8(−3&9&5@26&−16&−2@19&5&−11)]
Solving
X = (A-1)’ B
[■8(𝑥@𝑦@𝑧)] = 1/62 [■8(−3&9&5@26&−16&−2@19&5&−11)]^′ [■8(14@4@0)]
[■8(𝑥@𝑦@𝑧)] = 1/62 [■8(−3&26&19@9&−16&5@5&−2&−11)][■8(14@4@0)]
" " [■8(𝑥@𝑦@𝑧)]" =" 1/62 [█(−3(14)+26(4)+19(0)@9(14)+(−16)(4)+5(0)@5(14)+(−2)(4)+(−11)(0))]
" " [■8(𝑥@𝑦@𝑧)]" =" 1/62 [■8(62@62@62)]
" " [■8(𝑥@𝑦@𝑧)]" =" [■8(1@1@1)]
"∴ x = 1, y = 1 and z = 1 "
Question 25
If A = [■8(2&1&1@1&0&1@0&2&−1)] find the inverse of A using elementary row transformations and hence solve the following matrix equation XA = [1 0 1 ].
Given A = [■8(2&1&1@1&0&1@0&2&−1)]
We know that A = IA
[■8(2&1&1@1&0&1@0&2&−1)] = [■8(1&0&0@0&1&0@0&0&1)] A
R1 ↔ R2
[■8(𝟏&0&1@2&1&1@0&2&−1)] = [■8(0&1&0@1&0&0@0&0&1)] A
R2 → R2 – 2R1
[■8(1&0&1@𝟐−𝟐(𝟏)&1−2(0)&1−2(1)@0&2&−1)] = [■8(0&1&0@1−2(0)&0−2(1)&0−2(0)@0&0&1)] A
[■8(1&0&1@𝟎&1&−1@0&2&−1)] = [■8(0&1&0@1&−2&0@0&0&1)] A
R3 → R3 – 2R2
[■8(1&0&1@0&1&−1@0−2(0)&𝟐−𝟐(𝟏)&−1−2(−1))] = [■8(0&1&0@1&−2&0@0−2(1)&0−2(−2)&1−2(0))] A
[■8(1&0&1@0&1&−1@0&𝟎&1)] = [■8(0&1&0@1&−2&0@−2&4&1)] A
R1 → R1 – R3
[■8(1−0&0−0&𝟏−𝟏@0&1&−1@0&0&1)] = [■8(0−(−2)&1−4&0−1@1&−2&0@−2&4&1)] A
[■8(1&0&𝟎@0&1&−1@0&0&1)] = [■8(2&−3&−1@1&−2&0@−2&4&1)] A
R2 → R2 + R3
[■8(1&0&0@0+0&1+0&−𝟏+𝟏@0&0&1)] = [■8(2&−3&−1@1+(−2)&−2+4&0+1@−2&4&1)] A
[■8(1&0&0@0&1&𝟎@0&0&1)] = [■8(2&−3&−1@−1&2&1@−2&4&1)] A
𝐼 = [■8(2&−3&−1@−1&2&1@−2&4&1)] A
R1 → R1 – R3
[■8(1−0&0−0&𝟏−𝟏@0&1&−1@0&0&1)] = [■8(0−(−2)&1−4&0−1@1&−2&0@−2&4&1)] A
[■8(1&0&𝟎@0&1&−1@0&0&1)] = [■8(2&−3&−1@1&−2&0@−2&4&1)] A
R2 → R2 + R3
[■8(1&0&0@0+0&1+0&−𝟏+𝟏@0&0&1)] = [■8(2&−3&−1@1+(−2)&−2+4&0+1@−2&4&1)] A
[■8(1&0&0@0&1&𝟎@0&0&1)] = [■8(2&−3&−1@−1&2&1@−2&4&1)] A
𝐼 = [■8(2&−3&−1@−1&2&1@−2&4&1)] A
This is similar to
I = A-1 A
Hence A-1 = [■8(2&−3&−1@−1&2&1@−2&4&1)]
Now,
We need to solve
XA = [1 0 1 ]
X = [1 0 1 ]A–1
Putting value of A–1
X = [1 0 1 ][■8(2&−3&−1@−1&2&1@−2&4&1)]
X = [1 0 1 ]_(1 × 3) [■8(2&−3&−1@−1&2&1@−2&4&1)]_(3 × 3)
X = [■8(1(2)+0(−1)+1(−2)&1(−3)+0(2)+1(4)&1(−1)+0(1)+1(1))]_(1 × 3)
X = [■8(2−2&−3+4&−1+1)]
X = [■8(0&1&0)]

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