Question

What is the solution to the system of equations? -7x - 2y + z = -2 6x - y - z = -18 5x + 4y - z = 18 a) (7, -1, 45) b) (-1, 7, 5) c) (-1, 7, 9) d) (1, -7, -5)

Answer 1

Given the system of equations as shown below

`\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}`

The above system of equations can be expressed in the matrix form

`\begin{gathered} A* X\text{ = B} \\ \Rightarrow X=A^(-1)*\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}`

Thus, we have

`\begin{gathered} \begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix}*\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix}=\begin{bmatrix}{-2} & {} & {} \\ {-18} & {} & {} \\ {18} & {} & {}\end{bmatrix} \\ A* X\text{ = B} \end{gathered}`

This implies that

`\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {z} & {} & {}\end{bmatrix}=\begin{bmatrix}{-7} & {-2} & {1} \\ {6} & {-1} & {-1} \\ {5} & {4} & {-1}\end{bmatrix}^(-1)*\begin{bmatrix}{-2} & {} & {} \\ {-18} & {} & {} \\ {18} & {} & {}\end{bmatrix}`

Where

`\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}`

Solving for X Using crammer's rule, which states that

`\begin{gathered} x=(D_x)/(D) \\ y=(D_y)/(D) \\ z=(D_z)/(D) \\ ^{}where_{} \\ D,D_x,D_y,D_z\text{ are }\det er\min ants\text{ } \end{gathered}`

To find the deteminant D,

From the matrix A,

`\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}`

Thus, D = -8

To find x,

`\begin{gathered} x=(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{ = }(D_x)/(D)\text{ =}(8)/(-8)=-1 \end{gathered}`

To find y,

`\begin{gathered} y=\text{ }(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{ }(D_y)/(D)=(-56)/(-8)=7 \end{gathered}`

To find z,

`\begin{gathered} z=\text{ }(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{ }(D_z)/(D)=-(-40)/(-8)=5 \end{gathered}`

Hence, the solution to the system of equations is

(-1, 7, 5).

The correct option is B.