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Matrix A=$\begin{bmatrix}2&-1&-1\\1&-2&1\\1&1&\lambda \end{bmatrix}$

$|A|=0\implies 2(-2\lambda-1)+(\lambda-1)-3=0$

$\implies -3\lambda -6=0$

$\implies \lambda=-2$

But det(A)=0 is not a sufficient condition for no solution.

Infinitely many solutions $\implies$ |A|=0 and rank(A)=rank(A|B)

No solution $\implies$ |A|=0 and rank(A)$\neq$ rank(A|B)

For $\lambda=-2$,

Augmented matrix (A|B)

$=\begin{bmatrix} 2&-1&-1&12\\1&-2&1&-4\\1&1&-2&4\end{bmatrix}$

Apply $R1\leftarrow R1-(R2+R3)$

$\implies \begin{bmatrix} 0&0&0&12\\1&-2&1&-4\\1&1&-2&4\end{bmatrix}$

Apply $R1\leftrightarrow R3$

$\implies \begin{bmatrix}1&1&-2&4\\1&-2&1&-4\\0&0&0&12\end{bmatrix} $

So rank(A)=2 and rank(A|B)=3

Hence $\lambda=-2\implies \text{no solution}$


Alternative approach--

$2x-y-z=12$         $ \ldots (1)$

$x-2y+z=-4$          $\ldots (2)$

$x+y+\lambda z=4$           $\ldots (3)$

$equation(1)-(2)\implies  x+y+(-2)z=16$

So eqⁿ  3 will be inconsistent with this equation if $\lambda=-2$

edited by

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