Question

Which of the following is the inverse of the matrix A=$\begin{bmatrix}8&1\\1&2\end{bmatrix}$?

(a) $\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\\frac{1}{15}&\frac{8}{15}\end{bmatrix}$

(b) $\begin{bmatrix}\frac{1}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{1}{15}\end{bmatrix}$

(c) $\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$

(d) $\begin{bmatrix}\frac{2}{15}&\frac{1}{15}\\\frac{1}{15}&\frac{4}{15}\end{bmatrix}$

This question was posed to me in an online quiz.

The origin of the question is Invertible Matrices topic in chapter Matrices of Mathematics – Class 12

Answer 1

The correct option is (c) $\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$

Easy explanation: Consider the matrix A=$\begin{bmatrix}8&1\\1&2\end{bmatrix}$

Using the elementary row operation, we write A=IA

Applying R2→8R2-R1  and R2→R2/15, we get

$\begin{bmatrix}8&1\\0&1\end{bmatrix}$=$\begin{bmatrix}1&0\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$A

Applying R1→R1-R2  and R1→R1/8, we get

$\begin{bmatrix}1&0\\0&1\end{bmatrix}$=$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$A

A^-1=$\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}$.