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Question: Answered & Verified by Expert
Compute the following :
(i) $\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]$
(ii) $\left[\begin{array}{cc}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$
(iii) $\left[\begin{array}{ccc}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right]+\left[\begin{array}{ccc}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right]$
(iv) $\left[\begin{array}{ll}\cos ^2 x & \sin ^2 x \\ \sin ^2 x & \cos ^2 x\end{array}\right]+\left[\begin{array}{ll}\sin ^2 x & \cos ^2 x \\ \cos ^2 x & \sin ^2 x\end{array}\right]$
MathematicsMatrices
Solution:
1545 Upvotes Verified Answer
(i) $\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]=\left[\begin{array}{cc}2 a & 2 b \\ 0 & 2 a\end{array}\right]$
(ii) $\left[\begin{array}{cc}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{array}\right]+\left[\begin{array}{cc}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$
$$
=\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right]
$$
$$
=\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right]
$$
(iii) We have
$$
\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]=\left[\begin{array}{ccc}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]
$$
(iv) $\left[\begin{array}{ll}\cos ^2 x & \sin ^2 x \\ \sin ^2 x & \cos ^2 x\end{array}\right]+\left[\begin{array}{ll}\sin ^2 x & \cos ^2 x \\ \cos ^2 x & \sin ^2 x\end{array}\right]$
$$
=\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]
$$

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