Question

If A’=\(\begin{bmatrix}8&2\\6&4\end{bmatrix}\) and B’=\(\begin{bmatrix}9&5\\7&3\end{bmatrix}\). Find (A+2B)’.

(a) \(\begin{bmatrix}26&20\\10&12\end{bmatrix}\)

(b) \(\begin{bmatrix}26&12\\20&10\end{bmatrix}\)

(c) \(\begin{bmatrix}26&10\\20&12\end{bmatrix}\)

(d) \(\begin{bmatrix}26&20\\12&10\end{bmatrix}\)

This question was posed to me in an interview for job.

My doubt stems from Transpose of a Matrix topic in division Matrices of Mathematics – Class 12

Answer 1

The correct option is (b) \(\begin{bmatrix}26&12\\20&10\end{bmatrix}\)

The best explanation: Given that A’=\(\begin{bmatrix}8&2\\6&4\end{bmatrix}\) and B’=\(\begin{bmatrix}9&5\\7&3\end{bmatrix}\)

Calculating the transpose of A’ and B’, we get

A=\(\begin{bmatrix}8&6\\2&4\end{bmatrix}\) and B=\(\begin{bmatrix}9&7\\5&3\end{bmatrix}\)

⇒(A+2B)=\(\begin{bmatrix}8&6\\2&4\end{bmatrix}\)+2\(\begin{bmatrix}9&7\\5&3\end{bmatrix}\)

=\(\begin{bmatrix}8+18&6+14\\2+10&4+6\end{bmatrix}\)=\(\begin{bmatrix}26&20\\12&10\end{bmatrix}\)

Hence, (A+2B)’=\(\begin{bmatrix}26&12\\20&10\end{bmatrix}\).