The number of solutions of the system of equations 2x + y − z = 7 x − 3y + 2z = 1 x + 4y − 3z = 5 is

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(d) 0 The given system of equations can be written in matrix form as follows: \begin{bmatrix}2 & 1 & - 1 \ 1 & - 3 & 2 \ 1 & 4 & - 3\end{bmatrix} \begin{bmatrix}x \ y \ z\end{bmatrix} = \begin{bmatrix}7 \ 1 \ 5\end{bmatrix} AX = B Here, A = \begin{bmatrix}2 & 1 & - 1 \ 1 & - 3 & 2 \ 1 & 4 & - 3\end{bmatrix}, X = \begin{bmatrix}x \ y \ z\end{bmatrix}	ext{ and }B = \begin{bmatrix}7 \ 1 \ 5\end{bmatrix} Now, \left| A ight|=2 \left( 9 - 8 ight) - 1\left( - 3 - 2 ight) - 1\left( 4 + 3 ight) = 2 + 5 - 7 = 0 {	ext{ Let }C}_{ij} {	ext{ be the cofactors of the elements a }}_{ij}	ext{ in }A=\left[ a_{ij} ight]. 	ext{ Then,} C_{11} = \left( - 1 ight)^{1 + 1} \begin{vmatrix}- 3 & 2 \ 4 & - 3\end{vmatrix} = 1, C_{12} = \left( - 1 ight)^{1 + 2} \begin{vmatrix}1 & 2 \ 1 & - 3\end{vmatrix} = 5, C_{13} = \left( - 1 ight)^{1 + 3} \begin{vmatrix}1 & - 3 \ 1 & 4\end{vmatrix} = 7 C_{21} = \left( - 1 ight)^{2 + 1} \begin{vmatrix}1 & - 1 \ 4 & - 3\end{vmatrix} = - 1, C_{22} = \left( - 1 ight)^{2 + 2} \begin{vmatrix}2 & - 1 \ 1 & - 3\end{vmatrix} = - 5, C_{23} = \left( - 1 ight)^{2 + 3} \begin{vmatrix}2 & 1 \ 1 & 4\end{vmatrix} = - 7 C_{31} = \left( - 1 ight)^{3 + 1} \begin{vmatrix}1 & - 1 \ - 3 & 2\end{vmatrix} = - 1, C_{32} = \left( - 1 ight)^{3 + 2} \begin{vmatrix}2 & - 1 \ 1 & 2\end{vmatrix} = - 5, C_{33} = \left( - 1 ight)^{3 + 3} \begin{vmatrix}2 & 1 \ 1 & - 3\end{vmatrix} = - 7 adj A = \begin{bmatrix}1 & 5 & 7 \ - 1 & - 5 & - 7 \ - 1 & - 5 & - 7\end{bmatrix}^T = \begin{bmatrix}1 & - 1 & - 1 \ 5 & - 5 & - 5 \ 7 & - 7 & - 7\end{bmatrix} \Rightarrow \left( adj A ight) B = \begin{bmatrix}1 & - 1 & - 1 \ 5 & - 5 & - 5 \ 7 & - 7 & - 7\end{bmatrix}\begin{bmatrix}7 \ 1 \ 5\end{bmatrix} = \begin{bmatrix}7 - 1 - 5 \ 35 - 5 - 25 \ 49 - 7 - 35\end{bmatrix} = \begin{bmatrix}1 \ 5 \ 7\end{bmatrix} 
eq 0 So, the given system of equations has no solution.

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