Understand the problem

The value of  \(\lambda\)  for which the system of linear equations 2x – y – z = 12 , x -2y +z = -4 , x +y + \(\lambda\)z = 4 has no solution is 
Source of the problem
Sample Questions (MMA) :2019
Topic
Linear Algebra
Difficulty Level
Medium
Suggested Book
Schaums Outline of Linear Algebra 

Start with hints

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Write down the case in the matrix form & try to use the knowledge of system of linear cases 2x  -y -z =12, x -2y+ z =-4,x + y + \(\lambda\) z  =4
The corresponding matrix format that you will get is \begin{pmatrix}  2&-1&-1\\  1&-2&1\\  1&1&\lambda\\  \end{pmatrix} \begin{pmatrix} x \\ y\\ z \end{pmatrix}=\begin{pmatrix} 12\\-4\\4\\ \end{pmatrix}   The  system has no solution if  \begin{pmatrix} 12\\-4\\4\\ \end{pmatrix}  does not belong to the Sp { \begin{pmatrix} 2 \\ 1\\ 1 \end{pmatrix},\begin{pmatrix} -1 \\ -2\\1 \end{pmatrix},\begin{pmatrix} -1 \\ 1\\\lambda \end{pmatrix}} ———–(1) So, what is the possible value of \(\lambda\)? Can you think the condition (1) in some other ways say in terms of determinants?

If the determinant of matrix  A=\begin{pmatrix}  2&-1&-1\\  1&-2&1\\  1&1&\lambda\\  \end{pmatrix}

is zero then it has no solution

det A = 0 => \(\lambda\) = -2

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