Solve Using a Matrix with Cramer's Rule X+Y=5 X-Y=3

$X + Y = 5$ $X - Y = 3$

Step 1

Represent the system of equations in matrix format.

$[\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}] [\begin{array}{c} X \\ Y \end{array}] = [\begin{array}{c} 5 \\ 3 \end{array}]$

Step 2

Find the determinant of $[\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array}] = | \begin{array}{c} 1 & 1 \\ 1 & - 1 \end{array} |$

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

$D = (1) (- 1) - 1 \cdot 1$

Simplify the determinant.

Simplify each term.

Multiply $- 1$ by $1$ .

$D = - 1 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$D = - 1 - 1$

Subtract $1$ from $- 1$ .

$D = - 2$

Step 3

Find the determinant of $[\begin{array}{c} 5 & 1 \\ 3 & - 1 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 5 & 1 \\ 3 & - 1 \end{array}] = | \begin{array}{c} 5 & 1 \\ 3 & - 1 \end{array} |$

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

$D_{X} = (5) (- 1) - 3 \cdot 1$

Simplify the determinant.

Simplify each term.

Multiply $5$ by $- 1$ .

$D_{X} = - 5 - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$D_{X} = - 5 - 3$

Subtract $3$ from $- 5$ .

$D_{X} = - 8$

Step 4

Find the determinant of $[\begin{array}{c} 1 & 5 \\ 1 & 3 \end{array}]$ .

These are both valid notations for the determinant of a matrix.

$determinant [\begin{array}{c} 1 & 5 \\ 1 & 3 \end{array}] = | \begin{array}{c} 1 & 5 \\ 1 & 3 \end{array} |$

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

$D_{Y} = (1) (3) - 1 \cdot 5$

Simplify the determinant.

Simplify each term.

Multiply $3$ by $1$ .

$D_{Y} = 3 - 1 \cdot 5$

Multiply $- 1$ by $5$ .

$D_{Y} = 3 - 5$

Subtract $5$ from $3$ .

$D_{Y} = - 2$

Step 5

Find the value of $X$ by Cramer\'s Rule, which states that $X = \frac{D_{X}}{D}$ . In this case, $X = \frac{- 8}{- 2}$ .

Remove parentheses.

$X = \frac{- 8}{- 2}$

Divide $- 8$ by $- 2$ .

$X = 4$

Step 6

Find the value of $Y$ by Cramer\'s Rule, which states that $Y = \frac{D_{Y}}{D}$ . In this case, $Y = \frac{- 2}{- 2}$ .

Remove parentheses.

$Y = \frac{- 2}{- 2}$

Divide $- 2$ by $- 2$ .

$Y = 1$

Step 7

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

$X = 4, Y = 1$

Step 8

Solve Using a Matrix with Cramer's Rule X+Y=5 X-Y=3

Related Questions

Support LCMGCF.com by disabling your adblocker.