Question 25 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 by Teachoo

If A = [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

OR

If A = [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 ].

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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)]

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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