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  • If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y , 2X +3Y =[2 3 4 0]

If X and Y are 2 × 2 matrices, then solve the following matrix equations for X and Y
2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]

Answers (1)

We have the given matrix equations,

2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right], 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]

By subtracting equation (i) from (ii), we get

\begin{array}{l} (3 X+2 Y)-(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]-\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow 3 X+2 Y-2 X-3 Y=\left[\begin{array}{cc} -2-2 & 2-3 \\ 1-4 & -5-0 \end{array}\right] \\ \Rightarrow 3 X-2 X+2 Y-3 Y=\left[\begin{array}{cc} -4 & -1 \\ -3 & -5 \end{array}\right] \\ \Rightarrow X-Y=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right] \end{array}

By adding equations (i) and (ii), we get

\begin{aligned} &(3 X+2 Y)+(2 X+3 Y)=\left[\begin{array}{cc} -2 & 2 \\ 1 & -5 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{Y}+2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{cc} -2+2 & 2+3 \\ 1+4 & -5+0 \end{array}\right]\\ &\Rightarrow 3 \mathrm{X}+2 \mathrm{X}+2 \mathrm{Y}+3 \mathrm{Y}=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5 X+5 Y=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow 5(X+Y)=\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\frac{1}{5}\left[\begin{array}{cc} 0 & 5 \\ 5 & -5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{ll} \frac{1}{5} \times 0 & \frac{1}{5} \times 5 \\ \frac{1}{5} \times 5 & \frac{1}{5} \times-5 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \end{aligned}

By adding equations (iii) and (iv), we get

\\ (\mathrm{X}-\mathrm{Y})+(\mathrm{X}+\mathrm{Y})=\left[\begin{array}{ll} -4 & -1 \\ -3 & -5 \end{array}\right]+\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right] \\ \Rightarrow \mathrm{X}-\mathrm{Y}+\mathrm{X}+\mathrm{Y}=\left[\begin{array}{ll} -4+0 & -1+1 \\ -3+1 & -5-1 \end{array}\right] \\ \Rightarrow \mathrm{X}+\mathrm{X}-\mathrm{Y}+\mathrm{Y}=\left[\begin{array}{ll} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow 2 \mathrm{X}=\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\frac{1}{2}\left[\begin{array}{lc} -4 & 0 \\ -2 & -6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{ll} \frac{1}{2} \times-4 & \frac{1}{2} \times 0 \\ \frac{1}{2} \times-2 & \frac{1}{2} \times-6 \end{array}\right] \\ \Rightarrow \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]

Substituting the matrix A in equation (iv), we get

\begin{array}{l} {\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]+\mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]} \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0 & 1 \\ 1 & -1 \end{array}\right]-\left[\begin{array}{cc} -2 & 0 \\ -1 & -3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 0-(-2) & 1-0 \\ 1-(-1) & -1-(-3) \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 1+1 & -1+3 \end{array}\right] \\ \Rightarrow \mathrm{Y}=\left[\begin{array}{cc} 2 & 1 \\ 2 & 2 \end{array}\right] \\ \mathrm{X}=\left[\begin{array}{lc} -2 & 0 \\ -1 & -3 \end{array}\right]_{\text {and }} \mathrm{Y}=\left[\begin{array}{ll} 2 & 1 \\ 2 & 2 \end{array}\right] \end{array}

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