English

\[\Left( Ii \Right) a = \Begin{Bmatrix}X - 1 and 1 and 1 \\ 1 and X - 1 and 1 \\ 1 and 1 and X - 1\End{Bmatrix}\]

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Question

For what value of x the matrix A is singular? 

\[A = \begin{bmatrix}x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1\end{bmatrix}\]

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Solution

Matrix A will be singular if [ A ] = 0

\[\Rightarrow \left( x - 1 \right)\left[ \left( x - 1 \right)^2 - 1 \right] - 1\left( x - 1 - 1 \right) + 1\left[ 1 - \left( x - 1 \right) \right] = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( x^2 - 2x \right) - 1\left( x - 2 \right) + 1\left( 2 - x \right) = 0\]
\[ \Rightarrow x^3 - 2 x^2 - x^2 + 2x - x + 2 - x + 2 = 0\]
\[ \Rightarrow x^3 - 3 x^2 + 4 = 0\]
\[ \Rightarrow \left( x - 2 \right)^2 \left( x + 1 \right) = 0\]
\[ \Rightarrow x = 2 \text{ or }x = - 1\]

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Chapter 5: Determinants - Exercise 6.1 [Page 11]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.1 | Q 12.2 | Page 11
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