Find X and Y, if 2 X+3 Y= [ ll 2 & 3 4 & 0 ] and 3 X+2 Y= [ cc 2 & -2 -1 & 5 ]

2X + 3Y = [ ll 2 & 3 4 & 0 ] ......(1) 3X + 2Y = [ cc 2 & -2 -1 & 5 ] .....(2) Now, multiply equation (1) by 2 and equation (2) by 3, we get, 4X + 6Y = [ ll 4 & 6 8 & 0 ] .....(3) 9X + 6Y = [ cc 6 & -6 -3 & 15 ] ....(4) Subtracting equation (4) from (3), we get, (4X + 6Y) – (9X + 6Y) = [ cc 4 & 6 8 & 0 ]- [ cc 6 & -6 -3 & 15 ] -5 X= [ cc 4-6 & 6-(-6) 8-(-3) & 0-15 ] = [ cc -2 & 12 11 & -15 ] X=-1 5 [ cc -2 & 12 11 & -15 ]= [ cc 2 5 & -12 5 -11 5 & 3 ] Now, 2X + 3Y = [ ll 2 & 3 4 & 0 ] 2 [ cc 2 5 & -12 5 -11 5 & 3 ]+3 Y= [ cc 2 & 3 4 & 0 ] [ cc 4 5 & -24 5 -22 5 & 6 ]+3 Y= [ ll 2 & 3 4 & 0 ] 3 Y= [ ll 2 & 3 4 & 0 ]- [ cc 4 5 & -24 5 -22 5 & 6 ] 3 Y= [ cc 2-4 5 & 3+24 5 4+22 5 & 0-6 ]= [ cc 6 5 & 39 5 42 5 & -6 ] Y=1 3 [ cc 6 5 & 39 5 42 5 & -6 ] Y= [ cc 2 5 & 13 5 14 5 & -2 ]

Find X and Y, if 2 X+3 Y= [ ll 2 & 3 4 & 0 ] and 3 X+2 Y= [ cc 2 …

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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 = $\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 2\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{-12}{5}} \\ {\frac{-11}{5}} & {3} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{cc} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{ll} {2} & {3} \\ {4} & {0} \end{array}\right]-\left[\begin{array}{cc} {\frac{4}{5}} & {\frac{-24}{5}} \\ {\frac{-22}{5}} & {6} \end{array}\right]$
$\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{cc} {2-\frac{4}{5}} & {3+\frac{24}{5}} \\ {4+\frac{22}{5}} & {0-6} \end{array}\right]=\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} {\frac{6}{5}} & {\frac{39}{5}} \\ {\frac{42}{5}} & {-6} \end{array}\right]$
$\Rightarrow Y=\left[\begin{array}{cc} {\frac{2}{5}} & {\frac{13}{5}} \\ {\frac{14}{5}} & {-2} \end{array}\right]$