First slide

Transpose of matrix

Question

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

Moderate

A

$- 1$
B

$- 2$
C

$3$
D

$2$

Solution

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

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