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# 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
Linear Algebra
Medium
##### Suggested Book
Schaums Outline of Linear Algebra

Do you really need a hint? Try it first!

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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