Question

Find the inverse of A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\).

(a) \(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)

(b) \(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}\)

(c) \(\begin{bmatrix}-\frac{1}{7}&-\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)

(d) \(\begin{bmatrix}0&\frac{3}{7}\\\frac{4}{7}&\frac{5}{7}\end{bmatrix}\)

This question was posed to me in an internship interview.

My question is taken from Invertible Matrices topic in portion Matrices of Mathematics – Class 12

Answer 1

The correct choice is (a) \(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)

For explanation I would say: Consider the matrix A=\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)

By using the elementary row operations, we write A=IA

\(\begin{bmatrix}5&3\\4&1\end{bmatrix}\)=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)A

Applying R1→R1-3R2  and R1→R1/(-7), we get

\(\begin{bmatrix}1&0\\4&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\0&1\end{bmatrix}\)

Applying R2→R2-4R1, we get

\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\)A

⇒A^-1=\(\begin{bmatrix}-\frac{1}{7}&\frac{3}{7}\\\frac{4}{7}&-\frac{5}{7}\end{bmatrix}\).