Question

use cramer’s rule to solve each system o

1. $4x - 2y + 7z = 26$

$5x + 3y - 5z = -50$
$-7x - 8y - 3z = 49$

Explanation:

Step1: Find the coefficient matrix determinant ($D$)

The system is:

$$\begin{cases} 4x - 2y + 7z = 26 \\ 5x + 3y - 5z = -50 \\ -7x - 8y - 3z = 49 \end{cases}$$

The coefficient matrix $A$ is:

$$ A=
LATEXBLOCK8
$$

Calculate $\det(A)$ (denoted as $D$):

$$\begin{align*} D&=4\begin{vmatrix}3&-5\\-8&-3\end{vmatrix}-(-2)\begin{vmatrix}5&-5\\-7&-3\end{vmatrix}+7\begin{vmatrix}5&3\\-7&-8\end{vmatrix}\\ &=4(3\times(-3)-(-5)\times(-8)) + 2(5\times(-3)-(-5)\times(-7)) + 7(5\times(-8)-3\times(-7))\\ &=4(-9 - 40)+2(-15 - 35)+7(-40 + 21)\\ &=4(-49)+2(-50)+7(-19)\\ &=-196 - 100 - 133\\ &=-429 \end{align*}$$

Step2: Find $D_x$ (replace $x$-column with constants)

Matrix for $D_x$:

$$\begin{bmatrix} 26 & -2 & 7 \\ -50 & 3 & -5 \\ 49 & -8 & -3 \end{bmatrix}$$

Calculate $\det(D_x)$:

$$\begin{align*} D_x&=26\begin{vmatrix}3&-5\\-8&-3\end{vmatrix}-(-2)\begin{vmatrix}-50&-5\\49&-3\end{vmatrix}+7\begin{vmatrix}-50&3\\49&-8\end{vmatrix}\\ &=26(3\times(-3)-(-5)\times(-8)) + 2((-50)\times(-3)-(-5)\times49) + 7((-50)\times(-8)-3\times49)\\ &=26(-9 - 40)+2(150 + 245)+7(400 - 147)\\ &=26(-49)+2(395)+7(253)\\ &=-1274 + 790 + 1771\\ &=1287 \end{align*}$$

Step3: Find $D_y$ (replace $y$-column with constants)

Matrix for $D_y$:

$$\begin{bmatrix} 4 & 26 & 7 \\ 5 & -50 & -5 \\ -7 & 49 & -3 \end{bmatrix}$$

Calculate $\det(D_y)$:

$$\begin{align*} D_y&=4\begin{vmatrix}-50&-5\\49&-3\end{vmatrix}-26\begin{vmatrix}5&-5\\-7&-3\end{vmatrix}+7\begin{vmatrix}5&-50\\-7&49\end{vmatrix}\\ &=4((-50)\times(-3)-(-5)\times49) - 26(5\times(-3)-(-5)\times(-7)) + 7(5\times49-(-50)\times(-7))\\ &=4(150 + 245)-26(-15 - 35)+7(245 - 350)\\ &=4(395)-26(-50)+7(-105)\\ &=1580 + 1300 - 735\\ &=2145 \end{align*}$$

Step4: Find $D_z$ (replace $z$-column with constants)

Matrix for $D_z$:

$$\begin{bmatrix} 4 & -2 & 26 \\ 5 & 3 & -50 \\ -7 & -8 & 49 \end{bmatrix}$$

Calculate $\det(D_z)$:

$$\begin{align*} D_z&=4\begin{vmatrix}3&-50\\-8&49\end{vmatrix}-(-2)\begin{vmatrix}5&-50\\-7&49\end{vmatrix}+26\begin{vmatrix}5&3\\-7&-8\end{vmatrix}\\ &=4(3\times49-(-50)\times(-8)) + 2(5\times49-(-50)\times(-7)) + 26(5\times(-8)-3\times(-7))\\ &=4(147 - 400)+2(245 - 350)+26(-40 + 21)\\ &=4(-253)+2(-105)+26(-19)\\ &=-1012 - 210 - 494\\ &=-1716 \end{align*}$$

Step5: Solve for $x$, $y$, $z$

By Cramer's Rule:

$$ x = \frac{D_x}{D}=\frac{1287}{-429}=-3 $$

$$ y = \frac{D_y}{D}=\frac{2145}{-429}=-5 $$

$$ z = \frac{D_z}{D}=\frac{-1716}{-429}=4 $$

Answer:

$x = -3$, $y = -5$, $z = 4$