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Matrices — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMatrices2 Marks
Question
Find X and Y, if $2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right] \text { and } 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} {2} & {-2} \\ {-1} & {5} \end{array}\right]$
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Answer

2X + 3Y = $\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$ ......(1)
3X + 2Y = $\left[\begin{array}{cc} {2} & {-2} \\ {-1} & {5} \end{array}\right]$ .....(2)
Now, multiply equation (1) by 2 and equation (2) by 3, we get,
4X + 6Y = $\left[\begin{array}{ll} {4} & {6} \\ {8} & {0} \end{array}\right]$ .....(3)
9X + 6Y = $\left[\begin{array}{cc} {6} & {-6} \\ {-3} & {15} \end{array}\right]$ ....(4)
Subtracting equation (4) from (3), we get,
(4X + 6Y) – (9X + 6Y) = $\left[\begin{array}{cc} {4} & {6} \\ {8} & {0} \end{array}\right]-\left[\begin{array}{cc} {6} & {-6} \\ {-3} & {15} \end{array}\right]$
$\Rightarrow-5 \mathrm{X}=\left[\begin{array}{cc} {4-6} & {6-(-6)} \\ {8-(-3)} & {0-15} \end{array}\right]$
$=\left[\begin{array}{cc} {-2} & {12} \\ {11} & {-15} \end{array}\right]$
$\Rightarrow \mathrm{X}=-\frac{1}{5}\left[\begin{array}{cc} {-2} & {12} \\ {11} & {-15} \end{array}\right]=\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{-12}{5}} \\ {\frac{-11}{5}} & {3} \end{array}\right]$
Now, 2X + 3Y = $\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 2\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{-12}{5}} \\ {\frac{-11}{5}} & {3} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{cc} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]-\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{cc} {2-\frac{4}{5}} & {3+\frac{24}{5}} \\ {4+\frac{22}{5}} & {0-6} \end{array}\right]=\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{13}{5}} \\ {\frac{14}{5}} & {-2} \end{array}\right]$

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