# Question 2, Exercise 2.1

Solutions of Question 2 of Exercise 2.1 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Let $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bmatrix}$, $B=\begin{bmatrix}1 & -2 & -3 \\ 0 & -1 & 5\end{bmatrix}$ and $C=\begin{bmatrix}0 & 1 & -2\\0 & -1 & -1\end{bmatrix}$. Find $2A+3B-4C.$

Given: $A=\begin{bmatrix}2 & -5 & 1\\ 3 & 0 & -4\end{bmatrix}$, $B=\begin{bmatrix}1 & -2 & -3 \\ 0 & -1 & 5\end{bmatrix}$ and $C=\begin{bmatrix}0 & 1 & -2\\0 & -1 & -1\end{bmatrix}$.

Then $2A=\begin{bmatrix}4 & -10 & 2\\6 & 0 & -8\end{bmatrix}$, $3B=\begin{bmatrix}3 & -6 & -9\\0 & -3 & 15\end{bmatrix}$, $4C=\begin{bmatrix}0 & 4 & -8\\0 & -4 & -4\end{bmatrix}$. \begin{align}&2A+3B-4C\\&=\begin{bmatrix} 4 & -10 & 2\\ 6 & 0 & -8\end{bmatrix}+\begin{bmatrix}3 & -6 & -9\\0 & -3 & 15\end{bmatrix}-\begin{bmatrix}0 & 4 & -8 \\ 0 & -4 & -4\end{bmatrix}\\ &=\begin{bmatrix}4+3 & -10-6 & 2-9 \\6+0 & 0-3 & -8+15 \end{bmatrix} -\begin{bmatrix}0 & 4 & -8 \\0 & -4 & -4 \end{bmatrix} \\ &=\begin{bmatrix}7 & -16 & -7 \\6 & -3 & 7\end{bmatrix}-\begin{bmatrix}0 & 4 & -8 \\0 & -4 & -4 \end{bmatrix}\\ &=\begin{bmatrix}7-0 & -16-4 & -7+8 \\6-0 & -3+4 &7+4\end{bmatrix}\\ &=\begin{bmatrix}7 & -20 & 1\\6 & 1 & 11 \end{bmatrix}\end{align}