1.

For what value of x matrix A is singular?A = \(\begin{bmatrix}x-1 & 1 & 1 \\[0.3em]1 & x-1 & 1 \\[0.3em]1 &1 & x-1\end{bmatrix}\)

Answer»

|A| = \(\begin{vmatrix}x-1 & 1 & 1 \\[0.3em]1 & x-1 & 1 \\[0.3em]1 &1 & x-1\end{vmatrix}\) 

Expanding along the first row,

|A| = (x - 1)\(\begin{vmatrix}x-1 & 1 \\[0.3em]1 & x-1 \\[0.3em]\end{vmatrix}\) - 1\(\begin{vmatrix}1 & 1 \\[0.3em]1 & x-1 \\[0.3em]\end{vmatrix}\) + \(\begin{vmatrix}1 & x-1 \\[0.3em]1 & 1 \\[0.3em]\end{vmatrix}\) 

= (x–1) ((x–1) (x–1)– 1×1) – 1((x–1) – 1×1) + 1(1×1 – 1×(x–1)) 

= (x–1) (x2 – 2x + 1 – 1) – 1(x–1 – 1) + 1(1 – x+1) 

= x(x–1) (x– 2) – 1(x–2) – (x–2) 

= (x– 2) {x(x–1) – 1 – 1} 

= (x– 2) (x2 – x – 2)

For singular |A| = 0,

(x– 2) (x2 – x – 2) = 0 

(x– 2) (x2 – 2x + x – 2) = 0 

(x–2)(x–2)(x+1) = 0 

∴ x = –1 or 2 

Also, 

|A| = 28 

⇒ 7x2 + 3x – 6 = 28

⇒ 7x2 + 3x – 34 = 0 

⇒ 7x2 + 17x – 14x – 34 = 0 

⇒ x(7x+ 17) – 2(7x +17) = 0 

⇒ (x–2)(7x +17) = 0


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