Question

$The roots of \left|\begin{array}{ccc}1+x&1-x&1-x\\1-x &1+x&1-x\\1-x&1-x& 1+x\end{array}\right|= 0 are......,Select Proper option from the given options.(a) 0,1(b) 0,-1(c) 0,-3(d) 0,3$

Answer 1

Answer:

Roots are (0,3).

Hence option d is correct.

Step-by-step explanation:

To find the roots of given Determinant,first simplify the determinant by applying some properties of Determinant

$\left|\begin{array}{ccc}1+x&1-x&1-x\\1-x &1+x&1-x\\1-x&1-x& 1+x\end{array}\right|= 0$

R1 -> R1+R2+R3

$\left|\begin{array}{ccc}3-x&3-x&3-x\\1-x &1+x&1-x\\1-x&1-x& 1+x\end{array}\right|= 0$

take (3-x) common from R1

$(3-x) \left|\begin{array}{ccc}1&1&1\\1-x &1+x&1-x\\1-x&1-x& 1+x\end{array}\right|= 0$

Now apply C1 -> C1-C2

C2 -> C2-C3

$(3-x) \left|\begin{array}{ccc}1-1&1-1&1\\1-x-1-x &1+x-1+x&1-x\\1-x-1+x&1-x-1-x& 1+x\end{array}\right|= 0$

$(3-x) \left|\begin{array}{ccc}0&0&1\\-2x &2x&1-x\\0&-2x& 1+x\end{array}\right|= 0$

Now expand the determinant along R1

$(3 - x)( - 4 {x}^{2} ) = 0 \\ \\ 3 - x = 0 \\ \\ - x = - 3 \\ \\ x = 3 \\ \\ or \\ \\ - 4 {x}^{2} = 0 \\ \\ {x}^{2} = 0 \\ \\ x = 0$

Roots are (0,3).

Hence option d is correct.

Hope it helps you.