Question 36 (Choice 1) - CBSE Class 12 Sample Paper for 2021 Boards
Last updated at Oct. 27, 2020 by Teachoo

if A = [1 2 0 -2 1 0 0 -1 1], find A-1 . Hence Solve the system of equations;
𝑥 − 2𝑦 = 10
2𝑥 − 𝑦 − 𝑧 = 8
−2𝑦 + 𝑧 = 7

Note
: This
is similar
to
Question 33
of
CBSE Sample Paper
Maths
Class 12 - 2020

Check the answer here
https://
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Transcript

Question 36 (Choice 1) if A = [■8(1&2&0@−2&−1&−2@0&−1&1)], find A-1 . Hence Solve the system of equations; 𝑥 − 2𝑦 = 10 2𝑥 − 𝑦 − 𝑧 = 8 −2𝑦 + 𝑧 = 7
The equations are
𝑥 − 2𝑦 = 10 2𝑥 − 𝑦 − 𝑧 = 8 −2𝑦 + 𝑧 = 7
So, the equation is in the form of
[■8(1&−2&0@2&−1&−1@0&−2&1)][■8(𝑥@𝑦@𝑧)] = [■8(10@8@7)]
i.e. ATX = B
X = (AT)–1 B
X = (A−1)T B
Here, A = [■8(1&2&0@−2&−1&−2@0&−1&1)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(10@8@7)]
Finding A–1
We know that
A-1 = 1/(|A|) adj (A)
Calculating |A|
|A|= |■8(1&2&0@−2&−1&−2@0&−1&1)|
= 1(−1 − 2) − 2 (−2 – 0) + 0 (2 + 1) = –3 + 4 + 0
= 1
Since |A|≠ 0
∴ The system of equation is consistent & has a unique solution
Now finding 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(1&2&0@−2&−1&−2@0&−1&1)]
𝐴11 = −2 + 0 = –3
𝐴12 = −[2−0] = 2
𝐴13 = 1 – 0 = 2
𝐴21 = –[6−4] = –2
𝐴22 = 4 – 0 = 1
𝐴23 = –[2−0] = 1
𝐴31 = 0−(−4)= −4
𝐴32 = –[0−4] = 2
𝐴33 = −2−3 = 3
Thus adj A = [■8(−3&−2&−4@2&1&2@2&1&3)]
& |A| = 1
Now,
A-1 = 𝟏/(|𝐀|) adj A
A-1 = 1/1 [■8(−3&−2&−4@2&1&2@2&1&3)] = [■8(−3&−2&−4@2&1&2@2&1&3)]
Now,
X = (A−1)T B
[■8(𝑥@𝑦@𝑧)] = [■8(−3&−2&−4@2&1&2@2&1&3)]^𝑇 [■8(10@8@7)]
[■8(𝑥@𝑦@𝑧)] =[■8(−3&2&2@−2&1&1@−4&2&3)][■8(10@8@7)]
[■8(𝑥@𝑦@𝑧)]" =" [█(−3(10)+2(8)+2(7)@−2(10)+1(8)+1(7)@−4(10)+2(8)+3(7))]
" " [■8(𝑥@𝑦@𝑧)]" =" [■8(−30+16+14@−20+8+7@−40+16+21)]
" " [■8(𝑥@𝑦@𝑧)]" =" [■8(0@−5@−3)]
"∴ x = 0, y = –5 and z = −3"

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