Question

If $A^T$ denotes the transpose of the matrix $A = \begin{bmatrix}0&0&a\\ 0&b&c\\ d&e&f\end{bmatrix}$, where a, b, c, d, e and f are integers such that abd $\neq$ 0, then the number of such matrices for which $A^{-1} = A^T$ is

• A. $2(3!)$
• B. $3(2!)$
• C. $2^3$
• D. $3^2$

Answer 1

Answer: (C)

$A = \begin{bmatrix}0&0&a\\ 0&b&c\\ d&e&f\end{bmatrix} ,\left|A\right| = - abd \ne0$
$ c_{11} = + \left(bf - ce\right), c_{12} = - \left(- cd\right) = cd, c_{13} = + \left(- bd\right) = -bd$
$c_{21} = -\left(-ea\right) = ae, c_{22} = + \left(-ad\right) = -ad, c_{23} = -\left(0\right) = 0$
$c_{31} = + \left(-ab\right) = - ab, c_{32} = - \left(0\right) = 0, c_{33} = 0$
$ \text{Adj} \, A = \begin{bmatrix}\left(bf-ce\right)&ae&-ab\\ cd &-ad&0\\ -bd &0&0\end{bmatrix} $
$A^{-1} = \frac{1}{\left|A\right|} \left(\text{adj} \,A\right) = \frac{1}{abd} \begin{bmatrix}bf-ce&ae&-ab\\ cd &-ad&0\\ -bd &0&0\end{bmatrix} $
$A^{T} = \begin{bmatrix}0&0&d\\ 0 &b&e\\ a&c&f\end{bmatrix} $
Now $A^{-1} = A^{T}$
$ \Rightarrow \frac{1}{-abd} \begin{bmatrix}bf-ce&ae&-ab\\ cd &-ad&0\\ -bd &0&0\end{bmatrix} = \begin{bmatrix}0&0&d\\ 0 &b&e\\ a&c&f\end{bmatrix} $
$\Rightarrow \begin{bmatrix}bf-ce&ae&-ab\\ cd &-ad&0\\ -bd &0&0\end{bmatrix} = \begin{bmatrix}0&0&-abd^{2}\\ 0&-ab^{2}d&-abde\\ -a^{2}bd & -abcd&-abdf\end{bmatrix} $
$\therefore bf - ce = ae = cd = 0$ ...(i)
$abd^{2} = ab, ab^{2}d = ad, a^{2}bd = bd$ ...(ii)
$ abde = abcd = abdf = 0$ ...(iii)
From (ii),
$\left(abd^{2}\right).\left(ab^{2}d\right) .\left(a^{2}bd\right) = ab.ad.bd $
$\Rightarrow \left(abd\right)^{4} - \left(abd\right)^{2} = 0 $
$\Rightarrow \left(abd\right)^{2} \left[\left(abd\right)^{2} - 1 \right] = 0 $
$\because \ abd \neq 0 , \therefore \ abd = \pm 1$ ...(iv) From (iii) and (iv),
$e = c = f = 0$ ...(v)
From (i) and (v),
$bf = ae = cd = 0 $ ...(vi)
From (iv), (v) and (vi), it is clear that a, b, d can be any non-zero integer such that abd = $\pm$ 1
But it is only possible, if a = b = d = $\pm$ 1
Hence, there are 2 choices for each a, b and d.
there fore, there are $2 \times 2 \times 2$ choices for a, b and d.
Hence number of required matrices = $2 \times 2 \times 2=(2)^3$