Using elementary row transformations, find the inverse of the following matrix:

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Using elementary row transformations, find the inverse of the following matrix:
$[\begin{array}{cc} 2 & 5 \\ - 3 & 1 \end{array}]$

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Given $[\begin{array}{cc} 2 & 5 \\ - 3 & 1 \end{array}]$

In order to use how elementary transformation we write A = AI

$[\begin{array}{rr} 2 5 \\ - 3 1 \end{array}] = [\begin{array}{ll} 2 5 \\ - 3 1 \end{array}] [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

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

$R_{1} \to \frac{1}{2} R_{1}$

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

R₂ → R₂ + 3R₁

$[\begin{array}{ll} 1 5 / 2 \\ 0 17 / 2 \end{array}] = A [\begin{array}{ll} 1 / 2 0 \\ 3 / 2 1 \end{array}]$

$R_{2} \to \frac{2}{17} R_{2}$

$[\begin{array}{cc} 1 & 5 / 2 \\ 0 & 1 \end{array}] = A [\begin{array}{cc} 1 / 2 & 0 \\ 3 / 17 & 2 / 17 \end{array}]$

$R_{1} \to R_{1} - \frac{5}{2} R_{2}$

${[\begin{array}{ll} 1 0 \\ 0 1 \end{array}]}_{1}^{2} = A [\begin{array}{cc} 1 / 17 & - 5 / 17 \\ 3 / 17 & 2 / 17 \end{array}]$

$∴ A^{- 1} = \frac{1}{17} [\begin{array}{cc} 1 & - 5 \\ 3 & 2 \end{array}]$

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