Question

Find the inverse of each of the matrices,if it exists $\begin{bmatrix} 2 & 1 \\\ 7 & 4 \end{bmatrix}$

Answer 1

Let A=$\begin{bmatrix} 2 & 1 \\\ 7 & 4 \end{bmatrix}$

We know that$A = IA$

$\therefore$ $\begin{bmatrix} 2 & 1 \\\ 7 & 4 \end{bmatrix}$= $\begin{bmatrix} 1 & 0 \\\ 0 & 1 \end{bmatrix}$ A

⇒ $\begin{bmatrix} 1 & \frac12 \\\ 7 & 4 \end{bmatrix}$=$\begin{bmatrix} \frac12 & 0 \\\ 0 & 1 \end{bmatrix}$A (R1$→$ $\frac{1}{2}R_1$)

⇒$\begin{bmatrix} 1 & \frac12 \\\ 0 & \frac12 \end{bmatrix}$=$\begin{bmatrix} \frac12 &0\\\ -\frac72 & 1 \end{bmatrix}$A (R2→R2-7R1)

⇒ $\begin{bmatrix} 1 & 0 \\\ 0 & \frac12 \end{bmatrix}$=$\begin{bmatrix} 4 & -1 \\\ -\frac72 & 1 \end{bmatrix}$A (R1->R1-R2)

⇒ $\begin{bmatrix} 1 & 0 \\\ 0 & 2 \end{bmatrix}$ = $\begin{bmatrix} 4 & -1 \\\ -7 & 2 \end{bmatrix}$A (R2->2R2)

$\therefore$ A-1=$\begin{bmatrix} 4 & -1 \\\ -7 & 2 \end{bmatrix}$