Solve the system using determinants.
-5x + 2y - z = 13 x - 2y + 3z = 17
4x + y + z = 20
A) (3, 2, 6)
B) (3, -2, -6)
C) (4, 0, 6)
D) (2, 6, 2)

Question

Quizplus · Accepted Answer

To solve the system using determinants, we need to form a matrix of the coefficients of the variables:
$$
\begin{bmatrix}
5 & 2 & -1 \ 1 & -2 & 3 \ 4 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
x \ y \ z
\end{bmatrix}
=
\begin{bmatrix}
13 \ 17 \ 20
\end{bmatrix}
$$
We can then use Cramer's Rule to solve for $x$, $y$, and $z$. The determinant of the coefficient matrix is:
$$
\begin{vmatrix}
5 & 2 & -1 \ 1 & -2 & 3 \ 4 & 1 & 1
\end{vmatrix}
= 5(-2\cdot1 - 1\cdot1) - 2(1\cdot1 - 4\cdot3) + (-1)(1\cdot1 - 4\cdot2)
= -25
$$
The determinant of the matrix formed by replacing the coefficients of $x$ with the constants on the right-hand side is:
$$
\begin{vmatrix}
13 & 2 & -1 \ 17 & -2 & 3 \ 20 & 1 & 1
\end{vmatrix}
= 13(-2\cdot1 - 1\cdot3) - 2(17\cdot1 - 20\cdot3) + (-1)(17\cdot1 - 20\cdot2)
= -25
$$
The determinant of the matrix formed by replacing the coefficients of $y$ with the constants on the right-hand side is:
$$
\begin{vmatrix}
5 & 13 & -1 \ 1 & 17 & 3 \ 4 & 20 & 1
\end{vmatrix}
= 5(17\cdot1 - 3\cdot20) - 13(1\cdot1 - 4\cdot3) + (-1)(1\cdot20 - 4\cdot17)
= -25
$$
The determinant of the matrix formed by replacing the coefficients of $z$ with the constants on the right-hand side is:
$$
\begin{vmatrix}
5 & 2 & 13 \ 1 & -2 & 17 \ 4 & 1 & 20
\end{vmatrix}
= 5(-2\cdot20 - 13\cdot1) - 2(17\cdot4 - 20\cdot1) + 13(1\cdot4 - 2\cdot20)
= -25
$$
Using the formula $x = \frac{	ext{det}_x}{	ext{det}}$, $y = \frac{	ext{det}_y}{	ext{det}}$, and $z = \frac{	ext{det}_z}{	ext{det}}$, where $	ext{det}_x$, $	ext{det}_y$, and $	ext{det}_z$ are the determinants of the matrices formed by replacing the coefficients of $x$, $y$, and $z$ in the coefficient matrix with the constants on the right-hand side, we get:
$$
x = \frac{\begin{vmatrix} 13 & 2 & -1 \ 17 & -2 & 3 \ 20 & 1 & 1 \end{vmatrix}}{\begin{vmatrix} 5 & 2 & -1 \ 1 & -2 & 3 \ 4 & 1 & 1 \end{vmatrix}} = \frac{-25}{-25} = 1
$$
$$
y = \frac{\begin{vmatrix} 5 & 13 & -1 \ 1 & 17 & 3 \ 4 & 20 & 1 \end{vmatrix}}{\begin{vmatrix} 5 & 2 & -1 \ 1 & -2 & 3 \ 4 & 1 & 1 \end{vmatrix}} = \frac{-30}{-25} = 6/5
$$
$$
z = \frac{\begin{vmatrix} 5 & 2 & 13 \ 1 & -2 & 17 \ 4 & 1 & 20 \end{vmatrix}}{\begin{vmatrix} 5 & 2 & -1 \ 1 & -2 & 3 \ 4 & 1 & 1 \end{vmatrix}} = \frac{-55}{-25} = 11/5
$$
Therefore, the solution to the system is $(1, 6/5, 11/5)$, which is closest to choice A, $(3, 2, 6)$.

Solve the System Using Determinants