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

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

A = IA

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

R₂ → 2R₁ – R₂

R₃ → R₃ – R₁

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

R₂ → 3R₂ – R₃

[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 6 & 0 \\ 0 & - 3 & 3 \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 7 & - 3 & - 1 \\ - 1 & 0 & 1 \end{array}] A

R₁ → 3R₁ – R₂

[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & - 3 & 3 \end{array}] = [\begin{array}{ccc} - 4 & 3 & 1 \\ 7 & - 3 & - 1 \\ - 1 & 0 & 1 \end{array}] A

R_{1} \to \frac{R_{1}}{3}, R_{2} \to \frac{1}{6} R_{2}, R_{3} \to \frac{1}{6} R_{3}

∴ A^{- 1} = [\begin{array}{ccc} \frac{- 4}{3} & 1 & \frac{1}{3} \\ \frac{7}{6} & \frac{- 1}{2} & \frac{- 1}{6} \\ \frac{5}{6} & \frac{- 1}{2} & \frac{1}{6} \end{array}]