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.
Question 13
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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