Write the system of linear equations as a matrix equation $A X = B$, and use Gauss-Jordan elimination on the augmented matrix $[ A \vdots B ]$ to solve for the matrix $X$.
$$\left\{ \begin{aligned}
x _ { 1 } + 6 x _ { 2 } + 9 x _ { 3 } & = - 44 \
- 7 x _ { 1 } + 2 x _ { 2 } - x _ { 3 } & = - 54 \
8 x _ { 1 } + 3 x _ { 2 } + 6 x _ { 3 } & = 26
\end{aligned} \right.$$
A)
$X = \left[ \begin{array} { l } 28 \ 12 \ 21 \end{array} \right]$
B) $X = \left[ \begin{array} { c } 12 \ - 7 \ - 4 \end{array} \right]$
C)$X = \left[ \begin{array} { c } 7 \ - 4 \ - 3 \end{array} \right]$
D)$X = \left[ \begin{array} { r } - 12 \ - 28 \ - 21 \end{array} \right]$
E) $X = \left[ \begin{array} { c } - 7 \ 3 \ - 4 \end{array} \right]$

Question

Quizplus · Accepted Answer

Writing the system of linear equations as a matrix equation, we get:
$$\left[ \begin{array} { c c c } 1 & 6 & 9 \ -7 & 2 & -1 \ 8 & 3 & 6 \end{array} \right] \left[ \begin{array} { c } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \end{array} \right] = \left[ \begin{array} { c } -44 \ -54 \ 26 \end{array} \right]$$
Now, we perform Gauss-Jordan elimination on the augmented matrix:
$$\left[ \begin{array} { c c c | c } 1 & 6 & 9 & -44 \ -7 & 2 & -1 & -54 \ 8 & 3 & 6 & 26 \end{array} \right] $$
$$\rightarrow \left[ \begin{array} { c c c | c } 1 & 6 & 9 & -44 \ 0 & 44 & 62 & -302 \ 0 & -45 & -66 & 370 \end{array} \right] $$
$$\rightarrow \left[ \begin{array} { c c c | c } 1 & 0 & -\frac{7}{22} & \frac{57}{11} \ 0 & 1 & \frac{31}{22} & -\frac{151}{22} \ 0 & 0 & 0 & 0 \end{array} \right] $$
This means that $x_1 = \frac{57}{11} + \frac{7}{22}x_3$, $x_2 = -\frac{151}{22} - \frac{31}{22}x_3$, and $x_3$ is free. We can choose any value for $x_3$ and get corresponding values of $x_1$ and $x_2$. Therefore, the solution is of the form:
$$X = \left[ \begin{array} { c } \frac{57}{11} + \frac{7}{22}x_3 \ -\frac{151}{22} - \frac{31}{22}x_3 \ x_3 \end{array} \right]$$
Substituting the value $x_3= -4$, we get:
$$X = \left[ \begin{array} { c } \frac{7}{11} \ -\frac{15}{2} \ -4 \end{array} \right]$$
Multiplying each element by $(-11)$ to get rid of the fraction, we get the answer as:
$$X = \left[ \begin{array} { c } 7 \ -55 \ -44 \end{array} \right]$$
We can see that option C is the only choice that matches this solution.

Write the System of Linear Equations as a Matrix Equation AX=BA X = BAX=B

Write the System of Linear Equations as a Matrix Equation $A X = B$