Question 13, Exercise 2.2

Solutions of Question 13 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the matrices $X$ and $Y$ such that $2 X-Y=\left[\begin{array}{ccc}1 & 6 & -3 \\ 2 & 1 & 7\end{array}\right]$ and $X+3 Y=\left[\begin{array}{ccc}4 & 3 & 2 \\ 1 & -3 & 0\end{array}\right]$.

Solution.

Given: $$\begin{align*} 2X - Y = \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \cdots (i)\\ X + 3Y = \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix} \cdots (ii) \end{align*}$$ From (i) we have $$\begin{align*} Y = 2X - \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \end{align*}$$ Put value of $Y$ in (ii), we have $$\begin{align*} X + 3\left(2X - \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix}\right)&= \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix}\\ X + 6X - 3 \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} &= \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix}\\ 7X - \begin{pmatrix} 3 & 18 & -9 \\ 6 & 3 & 21 \end{pmatrix} &= \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix}\\ 7X &= \begin{pmatrix} 4 & 3 & 2 \\ 1 & -3 & 0 \end{pmatrix} + \begin{pmatrix} 3 & 18 & -9 \\ 6 & 3 & 21 \end{pmatrix}\\ 7X &= \begin{pmatrix} 7 & 21 & -7 \\ 7 & 0 & 21 \end{pmatrix}\\ X &= \dfrac{1}{7} \begin{pmatrix} 7 & 21 & -7 \\ 7 & 0 & 21 \end{pmatrix}\\ X &= \begin{pmatrix} 1 & 3 & -1 \\ 1 & 0 & 3 \end{pmatrix} \end{align*}$$ Put value of $X$ into value of $Y$. $$\begin{align*} Y &= 2X - \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \\ Y &= 2 \begin{pmatrix} 1 & 3 & -1 \\ 1 & 0 & 3 \end{pmatrix} - \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \\ Y &= \begin{pmatrix} 2 & 6 & -2 \\ 2 & 0 & 6 \end{pmatrix} - \begin{pmatrix} 1 & 6 & -3 \\ 2 & 1 & 7 \end{pmatrix} \\ Y &= \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & -1 \end{pmatrix} \end{align*}$$ Therefore, the matrices $X$ and $Y$ are: $$\begin{align*} X &= \begin{pmatrix} 1 & 3 & -1 \\ 1 & 0 & 3 \end{pmatrix} \\ Y &= \begin{pmatrix} 1 & 0 & 1 \\ 0 & -1 & -1 \end{pmatrix} \end{align*}$$

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