Question

Let \(A = \begin{bmatrix} 0 & -2 \\[0.3em] 2 & 0 \end{bmatrix}\). If \(M\) and \(N\) are two matrices given by \(M = \displaystyle\sum_{k=1}^{10} A^{2k}\) and \(N = \displaystyle\sum_{k=1}^{10} A^{2k-1}\)
then \(MN^2 \) is :

• A. a non-identity symmetric matrix
• B. a skew-symmetric matrix
• C. neither symmetric nor skew-symmetric matrix
• D. an identity matrix

Answer 1

The Correct Option is A

\(A = \begin{bmatrix} 0 & -2 \\[0.3em] 2 & 0 \end{bmatrix}\)
\(A^2 = \begin{bmatrix} 0 & -2 \\[0.3em]  2 & 0 \end{bmatrix} \begin{bmatrix} 0 & -2 \\[0.3em] 2 & 0 \end{bmatrix}\)=\( \begin{bmatrix} -4 & 0 \\[0.3em] 0 & -4 \end{bmatrix}\)=\(−4I\)
\(M = A_2 + A_4 + A_6 + … + A^{20}\)
\(= –4I + 16l – 64I + …\) upto \(10\) terms
\(= –I [4 – 16 + 64 … + \)upto \(10\) terms\(]\)
\(=−I⋅4[\frac {(−4)^{10}−1}{−4−1}]\)
\(=\frac 45(2^{20}−1)I\)
\(= A – 4A + 16A + …\) upto \(10\) terms
\(=A[\frac {(−4)^{10}−1}{−4−1}]\)

\(=−(\frac {2^{20}−1}{5})A\)

\(N^2=\frac {(2^{20}−1)^2}{2^5}\)

\(A^2=−\frac {4}{25}(2^{20}−1)^2t\)

\(MN^{2}=−\frac {16}{125}(2^{20}−1)^3 \)
\(I=KI\  (K≠±1)\)
\((MN^2)^T = (KI)^T = KI\)
∴ \(A\) is correct

So, the correct option is (A): a non-identity symmetric matrix.