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"

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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