The number of solutions of the system of equa | vedclass

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(B) To find the number of solutions,we write the system in matrix form $AX = B$,where $A = \begin{bmatrix} 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & -3 \end

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(B) To find the number of solutions,we write the system in matrix form $AX = B$,where $A = \begin{bmatrix} 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & -3 \end{bmatrix}$,$X = \begin{bmatrix} x \ y \ z \end{bmatrix}$,and $B = \begin{bmatrix} 7 \ 1 \ 5 \end{bmatrix}$.First,calculate the determinant of $A$ $(|A|)$:$|A| = 2((-3)(-3) - (2)(4)) - 1((1)(-3) - (2)(1)) - 1((1)(4) - (-3)(1))$$|A| = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3)$$|A| = 2(1) - 1(-5) - 1(7) = 2 + 5 - 7 = 0$.Since $|A| = 0$,the system is either inconsistent (no solution) or has infinitely many solutions.We check the consistency using the augmented matrix $[A|B]$:$\begin{bmatrix} 2 & 1 & -1 & | & 7 \ 1 & -3 & 2 & | & 1 \ 1 & 4 & -3 & | & 5 \end{bmatrix}$Performing row operations: $R_1 \leftrightarrow R_2$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \ 2 & 1 & -1 & | & 7 \ 1 & 4 & -3 & | & 5 \end{bmatrix}$.$R_2 	o R_2 - 2R_1$ and $R_3 	o R_3 - R_1$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \ 0 & 7 & -5 & | & 5 \ 0 & 7 & -5 & | & 4 \end{bmatrix}$.$R_3 	o R_3 - R_2$ gives $\begin{bmatrix} 1 & -3 & 2 & | & 1 \ 0 & 7 & -5 & | & 5 \ 0 & 0 & 0 & | & -1 \end{bmatrix}$.The last row implies $0 = -1$,which is a contradiction.Therefore,the system has no solution.

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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