$
\begin{aligned}
& X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \ldots \text {....(i) } \\
& X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \ldots \text { (ii) }
\end{aligned}
$
Adding (i) and (ii) we get,
$
\begin{aligned}
& 2 x=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
7+3 & 0+0 \\
2+0 & 5+3
\end{array}\right] \\
& =\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \\
& \therefore x=\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]
\end{aligned}
$
Subtracting (ii) from (i),
$
\begin{aligned}
& 2 y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& \Rightarrow 2 y=\left[\begin{array}{ll}
7-3 & 0-0 \\
2-0 & 5-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
4 & 0 \\
2 & 2
\end{array}\right] \\
& \therefore 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]
\end{aligned}
$
Hence $x=\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right], y=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$.

Matrices — Mathematics STD 10 — Question

Answer

Need a full question paper?

Explore more

Similar questions