Skip to content
Question of 146

Q.Solve the system of linear equations by matrix method:
2x + 3y + 3z = 5
x - 2y + z = -4
3x - y - 2z = 3

Karnataka PUCKarnataka II PUC Board 2020Subjective· 5mImportance★★★★★
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 →

x=1, y=2, z=−1x=1,\ y=2,\ z=-1.

Concept. For AX=BAX=B with ∣A∣≠0|A|\neq 0, the unique solution is X=A−1B=1∣A∣ (adj A) BX=A^{-1}B=\dfrac{1}{|A|}\,(\text{adj }A)\,B.

Setup.

A=[2331−213−1−2],X=[xyz],B=[5−43].A=\begin{bmatrix}2&3&3\\1&-2&1\\3&-1&-2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}5\\-4\\3\end{bmatrix}.

Determinant.

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

=2(4+1)−3(−2−3)+3(−1+6)=10+15+15=40≠0.=2(4+1)-3(-2-3)+3(-1+6)=10+15+15=40\neq 0.

Cofactors and adjoint. The cofactor matrix is

[5553−131191−7],adj A=[5395−131511−7] (transpose).\begin{bmatrix}5&5&5\\3&-13&11\\9&1&-7\end{bmatrix},\qquad \text{adj }A=\begin{bmatrix}5&3&9\\5&-13&1\\5&11&-7\end{bmatrix}\ (\text{transpose}).

Solve X=140(adj A)BX=\tfrac{1}{40}(\text{adj }A)B: …

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.