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

Answer: (A)

a non-identity symmetric matrix

Answer 2

$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.