The number of solutions of the system of equa | vedclass

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(A) Given the system of equations: $2x + y - z = 7$ $(i)$ $x - 3y + 2z = 1$ $(ii)$ $x + 4y - 3z = 5$ $(iii)$ To find the number of solutions,we check

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(A) Given the system of equations: $2x + y - z = 7$ $(i)$ $x - 3y + 2z = 1$ $(ii)$ $x + 4y - 3z = 5$ $(iii)$ To find the number of solutions,we check the consistency of the system. Let $D$ be the determinant of the coefficient matrix: $D = \begin{vmatrix} 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & -3 \end{vmatrix} = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3) = 2(1) - 1(-5) - 1(7) = 2 + 5 - 7 = 0$. Since $D = 0$,the system is either inconsistent or has infinitely many solutions. Let us check the consistency using the augmented matrix or by eliminating variables. From $(i) + (ii) 	imes 2$: $(2x + y - z) + 2(x - 3y + 2z) = 7 + 2(1) \implies 4x - 5y + 3z = 9$. Alternatively,consider $(ii) + (iii)$: $(x - 3y + 2z) + (x + 4y - 3z) = 1 + 5 \implies 2x + y - z = 6$. Comparing this with equation $(i)$,which is $2x + y - z = 7$,we see a contradiction $(6 = 7)$. Therefore,the system is inconsistent and has $0$ solutions.

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is

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