Find the A–1 of following matrix of by elementary transformation.(a) (b)

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Inverse of a Matrix

Question

Find the A^–1 of following matrix of by elementary transformation.

(a) $[\begin{array}{c} 1 \\ 0 \\ 1 \end{array} \begin{array}{c} 0 \\ 2 \\ 2 \end{array} \begin{array}{c} 1 \\ 3 \\ 1 \end{array}]$ (b) $[\begin{array}{c} 1 \\ 0 \\ - 1 \end{array} \begin{array}{c} 2 \\ - 2 \\ 3 \end{array} \begin{array}{c} - 2 \\ 1 \\ 0 \end{array}]$

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

A^–1 by elementary transformation

AA^–1 = I

$[\begin{array}{lll} 1 0 1 \\ 0 2 3 \\ 1 2 1 \end{array}] A^{- 1} = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}]$
R₃ → R₂ – R₁

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

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

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

R₃ → R₃ – 2R₂

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

$R_{2} \to (\frac{1}{2}) R_{3}$

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

$R_{2} \to R_{1} - R_{3} R_{2} \to R_{2} - \frac{3}{2} R_{2}$

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

$\Rightarrow A^{'} = [\begin{array}{ccc} 2 / 3 & - 1 / 3 & 1 / 3 \\ - 1 / 2 & 0 & 1 / 2 \\ 1 / 6 & 1 / 3 & - 1 / 3 \end{array}]$

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

AA^–1 = I

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

$R_{2} \to R_{1} + R_{1} R_{2} \to (\frac{1}{2}) R_{2}$

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

R₃ → R₃ – 5R₂R₁ → R₁ – 2R₁

$[\begin{array}{ccc} 1 & 0 & - 1 \\ 0 & 1 & - 1 / 2 \\ 0 & 0 & 1 / 2 \end{array}] A^{- 1} = [\begin{array}{ccc} 1 & 1 & 0 \\ 0 & - 1 / 2 & 0 \\ - 4 & 0 & 2 \end{array}]$

$R_{2} \to R_{2} + (\frac{1}{2}) R_{3} R_{2} \to R_{1} + R_{2}$

$[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] A^{- 1} = [\begin{array}{ccc} - 3 & 1 & 2 \\ - 2 & - 1 / 2 & 1 \\ - 4 & 0 & 2 \end{array}]$

$\Rightarrow A^{- 1} = [\begin{array}{ccc} - 3 & 1 & 2 \\ - 2 & - 1 / 2 & 1 \\ - 4 & 0 & 2 \end{array}]$