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Find X and Y, if: (i) X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] and X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] (ii) 2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] and 3 X+2 Y=\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]
CBSE | Class 12 | Exercise 3.2 | Maths | Matrices | NCERT
Find X and Y, if: (i) X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] and X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right] (ii) 2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] and 3 X+2 Y=\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]

(i) Adding both the expressions we get,
(X+Y)+(X-Y)=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]+\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]
2 X=\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]
\mathrm{X}=\frac{1}{2}\left[\begin{array}{ll}10 & 0 \\ 2 & 8\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right]

Subtracting both the expressions we get,

(X+Y)-(X-Y)=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]-\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]
2 \mathrm{Y}=\left[\begin{array}{ll}4 & 0 \\ 2 & 2\end{array}\right]

\mathrm{Y}=\frac{1}{2}\left[\begin{array}{ll}4 & 0 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]

(ii) 2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]_{\ldots . .(1)} and

3 X+2 Y=\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right] \ldots…..(2)

Multiplying equation (1) by 2 we get,

4 X+6 Y=2\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]_{\ldots \ldots .}…(3)

Multiplying equation (2) by 3 we get,

9 X+6 Y=3\left[\begin{array}{rr}2 & -2 \\ -1 & 5\end{array}\right]=\left[\begin{array}{rc}6 & -6 \\ -3 & 15\end{array}\right]……. (4)

Subtracting equation (4) from (3) we get,

-5 X=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right]-\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right]=\left[\begin{array}{cc}-2 & 12 \\ 11 & -15\end{array}\right]

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] B BYJU’S

since We know that

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

Substituting the value of X we get,

2\left[\begin{array}{ll}\frac{2}{5} & \frac{-12}{5} \\ \frac{-1}{5} & 3\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]

\left[\begin{array}{lr}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]

3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{ll}\frac{4}{5} & -\frac{24}{5} \\ -\frac{22}{5} & 6\end{array}\right]

therefore,

3 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}{ll}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right]

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