Find the inverse of the following matrix by using elementary row transformations :

Engineering

Mathematics

Introduction to Matrix

Algebra of Matrices

Inverse of a Matrix

Question

Find the inverse of the following matrix by using elementary row transformations :
$[\begin{array}{ccc} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}]$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$[\begin{array}{ccc} 1 & 1 & 2 \\ 3 & 1 & 1 \\ 2 & 3 & 1 \end{array}]$

Let A = IA

$[\begin{array}{lll} 1 1 2 \\ 3 1 1 \\ 2 3 1 \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A$

Applying R₂ → R₂ – 3R₁ and R₃ → R₃ – 2R₁

$[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 1 & - 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 2 & 0 & 1 \end{array}] A$

R₃ → 2R₃ + R₂ and R₃ → 1/11R₃

$[\begin{array}{ccc} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 7 / 11 & 1 / 11 & 2 / 11 \end{array}]$

R₁ → 2R₁ + R₂ and R₂ → R₂ – 5R₃

$[\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & - 2 & 0 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{ccc} - 1 & 1 & 0 \\ 2 / 11 & 6 / 11 & - 10 / 11 \\ - 7 / 11 & 1 / 11 & 2 / 11 \end{array}]$

$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - 1 \end{array}] = [\begin{array}{ccc} - 4 / 11 & 10 / 11 & - 2 / 11 \\ 2 / 11 & 6 / 11 & - 10 / 11 \\ - 7 / 11 & 1 / 11 & 2 / 11 \end{array}]$

R₁ → 1/2, R₂ → –1/2R₂ and R₃ → – R₃

$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = \frac{1}{11} [\begin{array}{ccc} - 2 & 5 & - 1 \\ - 1 & - 3 & - 5 \\ 7 & - 1 & - 2 \end{array}] A$

Hence $A^{- 1} = [\begin{array}{ccc} - 2 & 5 & - 1 \\ - 1 & - 3 & 5 \\ 7 & - 1 & - 2 \end{array}] A$