Question

Show that $\left|\begin{array}{ccc}x&a&a\\a&x&a\\a&a&x\end{array}\right|$ = (x + 2a)(x - a)²

Answer 1

Explanation:

It is a question of Determinant Evaluation.

I am going to solve it using The Properties of Determinant.

The benefit of using "The Properties of Determinant" is that the Expansion of the determinant becomes easier and shorter.

I have solved the question in detail.

Kindly see the Attachment for the detailed answer.

I hope it will help you.

Answer 2

Solution:
$\left|\begin{array}{ccc}x&a&a\\a&x&a\\a&a&x\end{array}\right|$

R3 -> R3- R2

$\left|\begin{array}{ccc}x&a&a\\a&x&a\\0&a-x&x-a\end{array}\right|$

C3 -> C3- C2

$\left|\begin{array}{ccc}x&a&0\\a&x&a-x\\0&a-x&2x-2a\end{array}\right|$

take common (x-a) from R3

$(x-a)\left|\begin{array}{ccc}x&a&0\\a&x&a-x\\0&-1&2\end{array}\right|$

C2 -> C2+ C3

$(x-a)\left|\begin{array}{ccc}x&a&0\\a&a&a-x\\0&1&2\end{array}\right|$

R2 -> R2- R1

$(x-a)\left|\begin{array}{ccc}x&a&0\\a-x&0&a-x\\0&1&2\end{array}\right|$

taking (a-x) common from R2

$(x-a)^{2}\left|\begin{array}{ccc}x&a&0\\-1&0&-1\\0&1&2\end{array}\right|$

now expand the determinant

$( {x - a})^{2} (x + 2a)$
=R.H.S

hence proved