Question

For how many values of x, is the matrix \(\left[ {\begin{array}{*{20}{c}} x-1&{\rm{1}}&1\\ { {\rm{1}}}&x-1&{ 1}\\ { {\rm{1}}}&{\rm{1}}&{ x-1} \end{array}} \right]\) singular?
1. Infinite
2. 2
3. 2, -1
4. -1, 1

Answer 1

Correct Answer - Option 2 : 2

Concept: 

Consider A is the square matrix;

If the matrix A is singular then |A| = 0

Calculations:

Given the matrix \(\left[ {\begin{array}{*{20}{c}} x-1&{\rm{1}}&1\\ { {\rm{1}}}&x-1&{ 1}\\ { {\rm{1}}}&{\rm{1}}&{ x-1} \end{array}} \right]\) singular

We know that the matrix is singular if |A| = 0

⇒ \(\rm\begin{vmatrix} x-1 &1 &1 \\ 1 &x-1 &1 \\ 1 &1 &x-1 \end{vmatrix} = 0\)

⇒ (x - 1)[(x -1)(x - 1) - 1 × 1] -1[(x - 1) × 1 - 1 × 1] + 1[1 × 1 - (x - 1) × 1] = 0

⇒ (x - 1)(x² - 2x) -1(x - 2) + 1(2 - x) = 0

⇒ x(x - 1)(x - 2)-1(x - 2) - 1(x - 2) = 0

⇒ (x - 2)(x(x - 1) - 1 - 1 ) = 0

⇒ (x - 2)(x² - x - 2 ) = 0

⇒ (x - 2)(x2 - 2x + x - 2 ) = 0

⇒ (x - 2)((x - 2)(x + 1) = 0

So, x = 2, -1

∴ The total value of x is two.