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

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

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

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

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Given :

[\begin{array}{cc} 1 & 2 \\ 2 & - 1 \end{array}]

We know that A = IA

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

R₂ → R₂ – 2R₁

[\begin{array}{cc} 1 & 2 \\ 2 - 2 & - 1 - 4 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 - 2 & 1 - 0 \end{array}] A

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

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

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

R₁ → R₁ – 2R₂

[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 1 - \frac{4}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{- 1}{5} \end{array}] A

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

\Rightarrow I = \frac{1}{5} [\begin{array}{cc} 1 & 2 \\ 2 & - 1 \end{array}] A

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

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