Using elementary tansormations, find the inverse of each of the matrices, if it exists in

Engineering

Mathematics

Multiplication of Matrices

Symmetric and Skew Symmetric Matrices

Question

Using elementary tansormations, find the inverse of each of the matrices, if it exists in
$[\begin{array}{cc} 2 & 1 \\ 7 & 4 \end{array}]$

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$A = [\begin{array}{cc} 2 & 1 \\ 7 & 4 \end{array}]$

|A| = 8 – 7 = 1 ≠ 0

Hence, A^–1 exists.

Now, A = IA

[\begin{array}{cc} 2 & 1 \\ 7 & 4 \end{array}] = [\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] A

[\begin{array}{cc} 1 & \frac{1}{2} \\ 7 & 4 \end{array}] = [\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & 1 \end{array}] A (R_{1} \to \frac{1}{2} R_{1})

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

(R₂ → R₂ – 7R₁)

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

(R₂ → 2R₂)

[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{cc} 4 & - 1 \\ - 7 & 2 \end{array}] A (R_{1} \to R_{1} - \frac{1}{2} R_{2})

Hence,

A^{- 1} = [\begin{array}{cc} 4 & - 1 \\ - 7 & 2 \end{array}]

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