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

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Mathematics

Introduction to Matrix

Algebra of Matrices

Inverse of a Matrix

Question

Using elementary row transformations, find the inverse of the following matrix:
$[\begin{array}{cc} 6 & 7 \\ 8 & 9 \end{array}]$

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Given $A = [\begin{array}{ll} 6 7 \\ 8 9 \end{array}]$

In order to we elementary how transformation to find inverse, we write

A = AI

$\Rightarrow [\begin{array}{ll} 6 7 \\ 8 9 \end{array}] = A [\begin{array}{ll} 1 0 \\ 0 1 \end{array}]$

$R_{1} \to \frac{1}{6} R_{1}$

$\Rightarrow [\begin{array}{cc} 1 & 7 / 6 \\ 8 & 9 \end{array}] = A [\begin{array}{cc} 1 / 6 & 0 \\ 0 & 1 \end{array}]$

R₂ → R₂ – 8R₁

$\Rightarrow [\begin{array}{cc} 1 & 7 / 6 \\ 0 & - 1 / 3 \end{array}] = A [\begin{array}{cc} 1 / 6 & 0 \\ - 4 / 3 & 1 \end{array}]$

R₂ → – 3R₂

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

R₁ → R₁ – 7 / 6R₂

$[\begin{array}{ll} 1 0 \\ 0 1 \end{array}] = A [\begin{array}{cc} - 9 / 2 & 7 / 3 \\ 4 & - 3 \end{array}]$

Hence $A^{- 1} = [\begin{array}{cc} - 9 / 2 & 7 / 3 \\ 4 & - 3 \end{array}]$