To find the inverse of the matrix by elementary transformation method.

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

To find the inverse of the matrix $A = [\begin{array}{lll} 1 2 1 \\ 0 1 1 \\ 3 1 1 \end{array}]$ by elementary transformation method.

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

A = AI

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

R₃ → R₃ – 3R₁

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

R₃ → R₃ – 5R₂

R₁ → R₁ – 2R₂

$[\begin{array}{l} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & - 7 \end{array}] = A [\begin{array}{l} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ - 8 & 0 & 1 \end{array}]$

$R_{3} \to \frac{- 1}{7} R_{3}$

$[\begin{array}{l} 1 & 0 & - 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array}] = A [\begin{array}{l} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ \frac{8}{7} & 0 & \frac{- 1}{7} \end{array}]$

R₂ → R₂ – R₃

R₁ → R₁ + 2R₃

$[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = A [\begin{array}{l} 1 + \frac{16}{7} & - 2 & \frac{- 2}{7} \\ \frac{8}{7} & 1 & \frac{- 1}{7} \\ \frac{8}{7} & 0 & \frac{- 1}{7} \end{array}]$

$A^{- 1} = [\begin{array}{l} \frac{25}{7} & - 2 & \frac{- 2}{7} \\ \frac{8}{7} & 1 & \frac{- 1}{7} \\ \frac{8}{7} & 0 & \frac{- 1}{7} \end{array}]$

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