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Mathematics Question on Matrices

Find the inverse of each of the matrices, if it exists. [3−1 −42]\begin{bmatrix} 3 & -1 \\\ -4 & 2 \end{bmatrix}

Answer

Let A=[3−1 −42]\begin{bmatrix} 3 & -1 \\\ -4 & 2 \end{bmatrix}

We know that A=IAA = IA

\begin{bmatrix} 3 & -1 \\\ -4 & 2 \end{bmatrix}$$=A$$\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}

⇒ \begin{bmatrix} 1 & -1 \\\ 0 & 2 \end{bmatrix}$$= A$$\begin{bmatrix} 1 & 0 \\\ 2 & 1 \end{bmatrix} (C1→C1+2C2)(C_1\rightarrow C_1+2C_2)

⇒ \begin{bmatrix} 1 & 0 \\\ 0 & 2 \end{bmatrix}$$=A$$\begin{bmatrix} 1 & 1 \\\ 2 & 3 \end{bmatrix} (C2→C2+C1)(C_2\rightarrow C_2+C_1)

⇒ \begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}$$=A [112 232]\begin{bmatrix} 1 & \frac12 \\\ 2 & \frac32 \end{bmatrix} (C2→12C2)(C_2\rightarrow \frac{1}{2}C_2)

∴A−1=\therefore A^{-1}= [112 232]\begin{bmatrix} 1 & \frac12 \\\ 2 & \frac32 \end{bmatrix}