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Question

(B) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 2 & 1 & 1 \ 1 & -2 & -1 \ 0 & 3 & -5 \end{bmatrix},

Vedclass · Accepted Answer

$x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$

Vedclass · Answer

(B) The given system of equations can be written in the form $AX = B$,where$A = \begin{bmatrix} 2 & 1 & 1 \ 1 & -2 & -1 \ 0 & 3 & -5 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, B = \begin{bmatrix} 1 \ \frac{3}{2} \ 9 \end{bmatrix}$.First,we calculate the determinant of $A$:$|A| = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) = 2(13) + 5 + 3 = 26 + 8 = 34 
eq 0$.Since $|A| 
eq 0$,the system has a unique solution $X = A^{-1}B$.The matrix of cofactors is:$C_{11} = 13, C_{12} = 5, C_{13} = 3$$C_{21} = 8, C_{22} = -10, C_{23} = -6$$C_{31} = 1, C_{32} = 3, C_{33} = -5$Thus,$adj(A) = \begin{bmatrix} 13 & 8 & 1 \ 5 & -10 & 3 \ 3 & -6 & -5 \end{bmatrix}$.$A^{-1} = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1 \ 5 & -10 & 3 \ 3 & -6 & -5 \end{bmatrix}$.$X = A^{-1}B = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1 \ 5 & -10 & 3 \ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \ \frac{3}{2} \ 9 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13 + 12 + 9 \ 5 - 15 + 27 \ 3 - 9 - 45 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 34 \ 17 \ -51 \end{bmatrix} = \begin{bmatrix} 1 \ \frac{1}{2} \ -\frac{3}{2} \end{bmatrix}$.Therefore,$x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$.

Solve the system of linear equations using the matrix method: $2x + y + z = 1$,$x - 2y - z = \frac{3}{2}$,and $3y - 5z = 9$.

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