Linear Algebra Examples

$2 x + 4 y = 25$ , $x + 5 y = 2$

Step 1

Find the $A X = B$ from the system of equations.

$[\begin{array}{c} 2 & 4 \\ 1 & 5 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 25 \\ 2 \end{array}]$

Step 2

Find the inverse of the coefficient matrix.

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The inverse of a $2 \times 2$ matrix can be found using the formula $\frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$ where $| A |$ is the determinant of $A$ .

If $A = [\begin{array}{c} a & b \\ c & d \end{array}]$ then $A^{- 1} = \frac{1}{| A |} [\begin{array}{c} d & - b \\ - c & a \end{array}]$

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

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

$determinant [\begin{array}{c} 2 & 4 \\ 1 & 5 \end{array}] = | \begin{array}{c} 2 & 4 \\ 1 & 5 \end{array} |$

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

$(2) (5) - 1 \cdot 4$

Simplify the determinant.

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Simplify each term.

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Multiply $2$ by $5$ .

$10 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$10 - 4$

Subtract $4$ from $10$ .

$6$

Substitute the known values into the formula for the inverse of a matrix.

$\frac{1}{6} [\begin{array}{c} 5 & - (4) \\ - (1) & 2 \end{array}]$

Simplify each element in the matrix.

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Rearrange $- (4)$ .

$\frac{1}{6} [\begin{array}{c} 5 & - 4 \\ - (1) & 2 \end{array}]$

Rearrange $- (1)$ .

$\frac{1}{6} [\begin{array}{c} 5 & - 4 \\ - 1 & 2 \end{array}]$

Multiply $\frac{1}{6}$ by each element of the matrix.

$[\begin{array}{c} \frac{1}{6} \cdot 5 & \frac{1}{6} \cdot - 4 \\ \frac{1}{6} \cdot - 1 & \frac{1}{6} \cdot 2 \end{array}]$

Simplify each element in the matrix.

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Rearrange $\frac{1}{6} \cdot 5$ .

$[\begin{array}{c} \frac{5}{6} & \frac{1}{6} \cdot - 4 \\ \frac{1}{6} \cdot - 1 & \frac{1}{6} \cdot 2 \end{array}]$

Rearrange $\frac{1}{6} \cdot - 4$ .

$[\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ \frac{1}{6} \cdot - 1 & \frac{1}{6} \cdot 2 \end{array}]$

Rearrange $\frac{1}{6} \cdot - 1$ .

$[\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ - \frac{1}{6} & \frac{1}{6} \cdot 2 \end{array}]$

Rearrange $\frac{1}{6} \cdot 2$ .

$[\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ - \frac{1}{6} & \frac{1}{3} \end{array}]$

Step 3

Left multiply both sides of the matrix equation by the inverse matrix.

$([\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ - \frac{1}{6} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 2 & 4 \\ 1 & 5 \end{array}]) \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ - \frac{1}{6} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 25 \\ 2 \end{array}]$

Step 4

Any matrix multiplied by its inverse is equal to $1$ all the time. $A \cdot A^{- 1} = 1$ .

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{5}{6} & - \frac{2}{3} \\ - \frac{1}{6} & \frac{1}{3} \end{array}] \cdot [\begin{array}{c} 25 \\ 2 \end{array}]$

Step 5

Simplify the right side of the equation.

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Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} \frac{5}{6} \cdot 25 - \frac{2}{3} \cdot 2 \\ - \frac{1}{6} \cdot 25 + \frac{1}{3} \cdot 2 \end{array}]$

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} \frac{39}{2} \\ - \frac{7}{2} \end{array}]$

Step 6

Simplify the left and right side.

$[\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{39}{2} \\ - \frac{7}{2} \end{array}]$

Step 7

Find the solution.

$x = \frac{39}{2}$

$y = - \frac{7}{2}$

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