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=\(\begin{bmatrix}\frac{2}{15}&-\frac{1}{15}\\-\frac{1}{15}&\frac{8}{15}\end{bmatrix}\).