Respuesta :
Given the system of equations as shown below
[tex]\begin{gathered} -7x\text{ -2y+z = -2 ------ equation 1} \\ 6x-y-z=-18\text{ ------ equation 2} \\ 5x+4y-z\text{ = 18 ------- equation 3} \end{gathered}[/tex]The above system of equations can be expressed in the matrix form
[tex]\begin{gathered} A\times X\text{ = B} \\ \Rightarrow X=A^{-1}\times\text{ B} \\ \text{where} \\ X\text{ is the matrix of unkown variables (x, y, z)} \\ A^{-1}\text{ is the inverse matrix of matrix A} \\ B\text{ is the matrix B} \end{gathered}[/tex]Thus, we have
[tex]\begin{gathered} \begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix}\times\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix}=\begin{bmatrix}{-2} & {} & {} \\ {-18} & {} & {} \\ {18} & {} & {}\end{bmatrix} \\ A\times X\text{ = B} \end{gathered}[/tex]This implies that
[tex]\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix}=\begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix}^{-1}\times\begin{bmatrix}{-2} & {} & {} \\ {-18} & {} & {} \\ {18} & {} & {}\end{bmatrix}[/tex]Where
[tex]\begin{gathered} X=\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix} \\ A^{-1}=\text{ }\begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix}^{-1} \\ B=\text{ }\begin{bmatrix}{-2} & {} & {} \\ {-18} & {} & {} \\ {18} & {} & {}\end{bmatrix} \end{gathered}[/tex]Solving for X Using crammer's rule, which states that
[tex]\begin{gathered} x=\frac{D_x}{D} \\ y=\frac{D_y}{D} \\ z=\frac{D_z}{D} \\ ^{}where_{} \\ D,D_x,D_y,D_z\text{ are }\det er\min ants\text{ } \end{gathered}[/tex]To find the deteminant D,
From the matrix A,
[tex]\begin{gathered} A\text{ = }\begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix} \\ \\ D=-7\cdot\det \begin{pmatrix}-1 & -1 \\ 4 & -1\end{pmatrix}-\mleft(-2\mright)\det \begin{pmatrix}6 & -1 \\ 5 & -1\end{pmatrix}+1\cdot\det \begin{pmatrix}6 & -1 \\ 5 & 4\end{pmatrix} \\ \Rightarrow-7(1+4)+2(-6+5)+1(24+5) \\ =-7(5)+2(-1)+1(29) \\ =-8 \end{gathered}[/tex]Thus, D = -8
To find x,
[tex]\begin{gathered} x=\frac{D_x}{D} \\ D_x\text{ = }\det er\min ant{\text{ of }\begin{bmatrix}{-2} & {-2} & {1} \\ {-18} & {-1} & {-1} \\ {18} & {4} & {-1}\end{bmatrix}} \\ \Rightarrow D_x=\text{ -2(1-(-4))-(-2)(18}-(-18))+1(-72-(-18)) \\ =-2(5)+2(36)+1(-54) \\ =-10+54-72 \\ =8 \\ Thus, \\ x\text{ = }\frac{D_x}{D}\text{ =}\frac{8}{-8}=-1 \end{gathered}[/tex]To find y,
[tex]\begin{gathered} y=\text{ }\frac{D_y}{D} \\ D_y\text{ = }\det er\min ant\text{ of }\begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-18} & {-1} \\ {5} & {18} & {-1}\end{bmatrix} \\ D_y\text{ = -7(18-(-18))-(-2)(-6-(-5))+1(108-(-90))} \\ =-7(36)+2(-1)+1(198) \\ =-56 \\ \text{Thus,} \\ y=\text{ }\frac{D_y}{D}=\frac{-56}{-8}=7 \end{gathered}[/tex]To find z,
[tex]\begin{gathered} z=\text{ }\frac{D_z}{D} \\ D_z\text{ = }\det er\min ant\text{ of }\begin{bmatrix}{-7} & {-2} & {-2} \\ {6} & {-1} & {-18} \\ {5} & {4} & {18}\end{bmatrix} \\ =-7(-18-(-72))-(-2)(108-(-90)+(-2)(24-(-5)) \\ =-7(54)+2(198)-2(29) \\ =-40 \\ \text{Thus,} \\ z=\text{ }\frac{D_z}{D}=-\frac{-40}{-8}=5 \end{gathered}[/tex]Hence, the solution to the system of equations is
(-1, 7, 5).
The correct option is B.
