Let A = $\begin{bmatrix} 2 & -1 & 1 \\[0.3em] 3 & -1 & 1 \\[0.3em] 4 & -1 & 2 \\[0.3em] -1 &1 &-1 \end{bmatrix}$and B = $\begin{bmatrix} 8 \\[0.3em] 6 \\[0.3em] 7 \\[0.3em] 4 \end{bmatrix}$

• Equation in matrix form is:

$\begin{bmatrix} 2 & -1 & 1 \\[0.3em] 3 & -1 & 1 \\[0.3em] 4 & -1 & 2 \\[0.3em] -1 &1 &-1 \end{bmatrix}$$ \begin{bmatrix} x \[0.3em] y \[0.3em] z \end{bmatrix}$= $\begin{bmatrix} 8 \[0.3em] 6 \[0.3em] 7 \[0.3em] 4 \end{bmatrix}$   * By using$R_2 - R_1; R_3 - 2 R_1; R_4+R_1;$ $\begin{bmatrix} 2 & -1 & 1 \[0.3em] 1 & 0 & 0 \[0.3em] 0 & 1 & 0 \[0.3em] 1 &0 &0 \end{bmatrix}$$ \begin{bmatrix} x \[0.3em] y \[0.3em] z \end{bmatrix}$= $\begin{bmatrix} 8 \[0.3em] -2 \[0.3em] -9 \[0.3em] 12 \end{bmatrix}$   * By using$R_4 - R_2;$ $\begin{bmatrix} 2 & -1 & 1 \[0.3em] 1 & 0 & 0 \[0.3em] 0 & 1 & 0 \[0.3em] 0 &0 &0 \end{bmatrix}$$ \begin{bmatrix} x \[0.3em] y \[0.3em] z \end{bmatrix}$= $\begin{bmatrix} 8 \[0.3em] -2 \[0.3em] -9 \[0.3em] 14 \end{bmatrix}$   * Augmented matrix [A,B] =$\begin{bmatrix} 2 & -1 & 1 , &8 \[0.3em] 1 & 0 & 0, &-2 \[0.3em] 0 & 1 & 0, & -9 \[0.3em] 0 &0 &0, &14 \end{bmatrix}$

• Rank of A = number of non-zero rows = 3
• Rank of [A,B] = number of non-zero rows = 4
• Since Rank of A is less than Rank of [A,B], hence the given set of equation is inconsistent and there is no solution.