The number of solutions of the system of equa | vedclass

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(D) To find the number of solutions,we first calculate the determinant $\Delta$ of the coefficient matrix:$\Delta = \begin{vmatrix} 2 & 1 & -1 \ 1 &

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$0$

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(D) To find the number of solutions,we first calculate the determinant $\Delta$ of the coefficient matrix:$\Delta = \begin{vmatrix} 2 & 1 & -1 \ 1 & -3 & 2 \ 1 & 4 & -3 \end{vmatrix}$Expanding along the first row:$\Delta = 2((-3)(-3) - (2)(4)) - 1((1)(-3) - (2)(1)) - 1((1)(4) - (-3)(1))$$\Delta = 2(9 - 8) - 1(-3 - 2) - 1(4 + 3)$$\Delta = 2(1) - 1(-5) - 1(7)$$\Delta = 2 + 5 - 7 = 0$Since $\Delta = 0$,the system either has no solution or infinitely many solutions.Now,calculate $\Delta_x$ by replacing the first column with the constants $(7, 1, 5)$:$\Delta_x = \begin{vmatrix} 7 & 1 & -1 \ 1 & -3 & 2 \ 5 & 4 & -3 \end{vmatrix} = 7(9 - 8) - 1(-3 - 10) - 1(4 + 15) = 7(1) - 1(-13) - 1(19) = 7 + 13 - 19 = 1 
eq 0$.Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero,the system is inconsistent.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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