Skip to content
Question of 146

Q.Solve the following system of linear equations by matrix method: 3x + y + z = 10, 2x - y - z = 0, x - y + 2z = 1. OR Using elementary transformations find the inverse of [[3, 2, 1], [2, 4, 3], [2, -1, 2]].

Punjab PsebPSEB Punjab Class 12 Board 2017Subjective· 6mImportance★★★★★
0% · 0/146 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Writing the system as AX=B and computing X=A⁻¹B (via the adjoint of A) gives x=2, y=3, z=1.

The system: 3x+y+z=103x+y+z=10, 2x−y−z=02x-y-z=0, x−y+2z=1x-y+2z=1

In matrix form AX=BAX=B: A=[3112−1−11−12]A=\begin{bmatrix}3&1&1\\2&-1&-1\\1&-1&2\end{bmatrix}, X=[xyz]X=\begin{bmatrix}x\\y\\z\end{bmatrix}, B=[1001]B=\begin{bmatrix}10\\0\\1\end{bmatrix}

Determinant of A:

∣A∣=3[(−1)(2)−(−1)(−1)]−1[(2)(2)−(−1)(1)]+1[(2)(−1)−(−1)(1)]|A| = 3[(-1)(2)-(-1)(-1)] - 1[(2)(2)-(-1)(1)] + 1[(2)(-1)-(-1)(1)]

=3(−2−1)−1(4+1)+1(−2+1)=−9−5−1=−15= 3(-2-1) - 1(4+1) + 1(-2+1) = -9-5-1=-15 (nonzero, so a unique solution exists)

Cofactors of A:

C11=−3, C12=−5, C13=−1C_{11}=-3,\ C_{12}=-5,\ C_{13}=-1

C21=−3, C22=5, C23=4C_{21}=-3,\ C_{22}=5,\ C_{23}=4

C31=0, C32=5, C33=−5C_{31}=0,\ C_{32}=5,\ C_{33}=-5

adj(A)=[−3−30−555−14−5]\text{adj}(A) = \begin{bmatrix}-3&-3&0\\-5&5&5\\-1&4&-5\end{bmatrix} (transpose of the cofactor matrix)

A−1=1∣A∣adj(A)=1−15[−3−30−555−14−5]A^{-1} = \dfrac{1}{|A|}\text{adj}(A) = \dfrac{1}{-15}\begin{bmatrix}-3&-3&0\\-5&5&5\\-1&4&-5\end{bmatrix}

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.