Let A = begin{bmatrix} 2 & -1 & 3 1 & 1 & -1 | vedclass

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(B) Given the matrix equation $AX = D$,where $A = \begin{bmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \ 1

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$A$ unique solution

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(B) Given the matrix equation $AX = D$,where $A = \begin{bmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$.To determine the nature of the solution,we calculate the determinant of matrix $A$,denoted by $\Delta = |A|$.$\Delta = \begin{vmatrix} 2 & -1 & 3 \ 1 & 1 & -1 \ 0 & 0 & 1 \end{vmatrix}$Expanding along the third row:$\Delta = 0 \cdot \begin{vmatrix} -1 & 3 \ 1 & -1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 3 \ 1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & -1 \ 1 & 1 \end{vmatrix}$$\Delta = 1 \cdot (2(1) - (-1)(1)) = 1 \cdot (2 + 1) = 3$.Since $\Delta = 3 
eq 0$,the matrix $A$ is non-singular and invertible.Therefore,the system $AX = D$ has a unique solution given by $X = A^{-1}D$.

Let $A = \begin{bmatrix} 2 & -1 & 3 \\ 1 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. The system $AX = D$ has

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