The values of a for which the matrix A=aa2−1−3a+12a2+4−34a−1 is symmetric are

The values of a for which the matrix $A = (\begin{array}{ccc} a & a^{2} - 1 & - 3 \\ a + 1 & 2 & a^{2} + 4 \\ - 3 & 4 a & - 1 \end{array})$ is symmetric are

A
-1
B
-2
C
3
D
2

Solution:

$\begin{array}{l} A^{'} = A \\ \Rightarrow (\begin{array}{ccc} a & a + 1 & - 3 \\ a^{2} - 1 & 2 & 4 a \\ - 3 & a^{2} + 4 & - 1 \end{array}) = (\begin{array}{ccc} a & a^{2} - 1 & - 3 \\ a + 1 & 2 & a^{2} + 4 \\ - 3 & 4 a & - 1 \end{array}) \\ \Rightarrow a + 1 = a^{2} - 1 and 4 a = a^{2} + 4 \\ \Rightarrow a + 1 = 4 a - 4 - 1 \Rightarrow 6 - 3 a \Rightarrow a = 2 . \end{array}$

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