Solve Using a Matrix with Cramer's Rule 2x+y+z=4 , 6y+5z=-33x , 5y-z=27

$2 x + y + z = 4$ , $6 y + 5 z = - 33 x$ , $5 y - z = 27$

Step 1

Add $33 x$ to both sides of the equation.

$6 y + 5 z + 33 x = 0$

$2 x + y + z = 4$

$5 y - z = 27$

Step 2

Represent the system of equations in matrix format.

$[\begin{array}{c} 2 & 1 & 1 \\ 33 & 6 & 5 \\ 0 & 5 & - 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 4 \\ 0 \\ 27 \end{array}]$

Step 3

Find the determinant of $[\begin{array}{c} 2 & 1 & 1 \\ 33 & 6 & 5 \\ 0 & 5 & - 1 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D = 2 | \begin{array}{c} 6 & 5 \\ 5 & - 1 \end{array} | - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Find the determinant of $2 | \begin{array}{c} 6 & 5 \\ 5 & - 1 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = 2 ((6) (- 1) - 5 \cdot 5) - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $- 1$ .

$D = 2 (- 6 - 5 \cdot 5) - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $- 5$ by $5$ .

$D = 2 (- 6 - 25) - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the expression.

Subtract $25$ from $- 6$ .

$D = 2 \cdot - 31 - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $2$ by $- 31$ .

$D = - 62 - 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Find the determinant of $- 33 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D = - 62 - 33 ((1) (- 1) - 5 \cdot 1) + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $1$ .

$D = - 62 - 33 (- 1 - 5 \cdot 1) + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $- 5$ by $1$ .

$D = - 62 - 33 (- 1 - 5) + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the expression.

Subtract $5$ from $- 1$ .

$D = - 62 - 33 \cdot - 6 + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $- 33$ by $- 6$ .

$D = - 62 + 198 + 0 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D = - 62 + 198 + 0$

Add $- 62$ and $198$ .

$D = 136 + 0$

Add $136$ and $0$ .

$D = 136$

Step 4

Find the determinant of $[\begin{array}{c} 4 & 1 & 1 \\ 0 & 6 & 5 \\ 27 & 5 & - 1 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{x} = 4 | \begin{array}{c} 6 & 5 \\ 5 & - 1 \end{array} | + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Find the determinant of $4 | \begin{array}{c} 6 & 5 \\ 5 & - 1 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = 4 ((6) (- 1) - 5 \cdot 5) + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $- 1$ .

$D_{x} = 4 (- 6 - 5 \cdot 5) + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $- 5$ by $5$ .

$D_{x} = 4 (- 6 - 25) + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Simplify the expression.

Subtract $25$ from $- 6$ .

$D_{x} = 4 \cdot - 31 + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Multiply $4$ by $- 31$ .

$D_{x} = - 124 + 0 | \begin{array}{c} 1 & 1 \\ 5 & - 1 \end{array} | + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{x} = - 124 + 0 + 27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$

Find the determinant of $27 | \begin{array}{c} 1 & 1 \\ 6 & 5 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{x} = - 124 + 0 + 27 ((1) (5) - 6 \cdot 1)$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $1$ .

$D_{x} = - 124 + 0 + 27 (5 - 6 \cdot 1)$

Multiply $- 6$ by $1$ .

$D_{x} = - 124 + 0 + 27 (5 - 6)$

Simplify the expression.

Subtract $6$ from $5$ .

$D_{x} = - 124 + 0 + 27 \cdot - 1$

Multiply $27$ by $- 1$ .

$D_{x} = - 124 + 0 - 27$

Add $- 124$ and $0$ .

$D_{x} = - 124 - 27$

Subtract $27$ from $- 124$ .

$D_{x} = - 151$

Step 5

Find the determinant of $[\begin{array}{c} 2 & 4 & 1 \\ 33 & 0 & 5 \\ 0 & 27 & - 1 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{y} = 2 | \begin{array}{c} 0 & 5 \\ 27 & - 1 \end{array} | - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Find the determinant of $2 | \begin{array}{c} 0 & 5 \\ 27 & - 1 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = 2 ((0) (- 1) - 27 \cdot 5) - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $0$ by $- 1$ .

$D_{y} = 2 (0 - 27 \cdot 5) - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Multiply $- 27$ by $5$ .

$D_{y} = 2 (0 - 135) - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Simplify the expression.

Subtract $135$ from $0$ .

$D_{y} = 2 \cdot - 135 - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Multiply $2$ by $- 135$ .

$D_{y} = - 270 - 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} | + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Find the determinant of $- 33 | \begin{array}{c} 4 & 1 \\ 27 & - 1 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{y} = - 270 - 33 ((4) (- 1) - 27 \cdot 1) + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $4$ by $- 1$ .

$D_{y} = - 270 - 33 (- 4 - 27 \cdot 1) + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Multiply $- 27$ by $1$ .

$D_{y} = - 270 - 33 (- 4 - 27) + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Simplify the expression.

Subtract $27$ from $- 4$ .

$D_{y} = - 270 - 33 \cdot - 31 + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Multiply $- 33$ by $- 31$ .

$D_{y} = - 270 + 1023 + 0 | \begin{array}{c} 4 & 1 \\ 0 & 5 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{y} = - 270 + 1023 + 0$

Add $- 270$ and $1023$ .

$D_{y} = 753 + 0$

Add $753$ and $0$ .

$D_{y} = 753$

Step 6

Find the determinant of $[\begin{array}{c} 2 & 1 & 4 \\ 33 & 6 & 0 \\ 0 & 5 & 27 \end{array}]$ .

Set up the determinant by breaking it into smaller components.

$D_{z} = 2 | \begin{array}{c} 6 & 0 \\ 5 & 27 \end{array} | - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Find the determinant of $2 | \begin{array}{c} 6 & 0 \\ 5 & 27 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{z} = 2 ((6) (27) - 5 \cdot 0) - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $6$ by $27$ .

$D_{z} = 2 (162 - 5 \cdot 0) - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Multiply $- 5$ by $0$ .

$D_{z} = 2 (162 + 0) - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Simplify the expression.

Add $162$ and $0$ .

$D_{z} = 2 \cdot 162 - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Multiply $2$ by $162$ .

$D_{z} = 324 - 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} | + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Find the determinant of $- 33 | \begin{array}{c} 1 & 4 \\ 5 & 27 \end{array} |$ .

The determinant of a $2 \times 2$ matrix can be found using the formula $| \begin{array}{c} a & b \\ c & d \end{array} | = a d - c b$ .

$D_{z} = 324 - 33 ((1) (27) - 5 \cdot 4) + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Simplify the determinant.

Simplify each term.

Multiply $27$ by $1$ .

$D_{z} = 324 - 33 (27 - 5 \cdot 4) + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Multiply $- 5$ by $4$ .

$D_{z} = 324 - 33 (27 - 20) + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Simplify the expression.

Subtract $20$ from $27$ .

$D_{z} = 324 - 33 \cdot 7 + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Multiply $- 33$ by $7$ .

$D_{z} = 324 - 231 + 0 | \begin{array}{c} 1 & 4 \\ 6 & 0 \end{array} |$

Since the matrix is multiplied by $0$ , the determinant is $0$ .

$D_{z} = 324 - 231 + 0$

Subtract $231$ from $324$ .

$D_{z} = 93 + 0$

Add $93$ and $0$ .

$D_{z} = 93$

Step 7

Find the value of $x$ by Cramer\'s Rule, which states that $x = \frac{D_{x}}{D}$ . In this case, $x = \frac{- 151}{136}$ .

Remove parentheses.

$x = \frac{- 151}{136}$

Move the negative in front of the fraction.

$x = - \frac{151}{136}$

Step 8

Remove parentheses.

$y = \frac{753}{136}$

Step 9

Remove parentheses.

$z = \frac{93}{136}$

Step 10

The solution to the system of equations using Cramer\'s Rule.

$x = - \frac{151}{136}, y = \frac{753}{136}, z = \frac{93}{136}$

Solve Using a Matrix with Cramer's Rule 2x+y+z=4 , 6y+5z=-33x , 5y-z=27

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