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 check for consistency,we write the aug

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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 check for consistency,we write the augmented matrix or use elimination. Multiply $(ii)$ by $2$: $2x - 6y + 4z = 2$ $(iv)$ Subtract $(iv)$ from $(i)$: $(2x + y - z) - (2x - 6y + 4z) = 7 - 2 \implies 7y - 5z = 5$ $(v)$ Now,subtract $(ii)$ from $(iii)$: $(x + 4y - 3z) - (x - 3y + 2z) = 5 - 1 \implies 7y - 5z = 4$ $(vi)$ Comparing $(v)$ and $(vi)$,we have $7y - 5z = 5$ and $7y - 5z = 4$. Since $5 
eq 4$,the system is inconsistent and has no solution. Therefore,the number of solutions is $0$.

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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