Finite Math Examples

$[\begin{array}{c} 3 & 5 \\ 7 & 12 \end{array}] x = [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$

Step 1

Find the inverse of $[\begin{array}{c} 3 & 5 \\ 7 & 12 \end{array}]$ .

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Step 1.1

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

Step 1.2

Find the determinant.

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Step 1.2.1

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

$3 \cdot 12 - 7 \cdot 5$

Step 1.2.2

Simplify the determinant.

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Step 1.2.2.1

Simplify each term.

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Step 1.2.2.1.1

Multiply $3$ by $12$ .

$36 - 7 \cdot 5$

Step 1.2.2.1.2

Multiply $- 7$ by $5$ .

$36 - 35$

Step 1.2.2.2

Subtract $35$ from $36$ .

$1$

Step 1.3

Since the determinant is non-zero, the inverse exists.

Step 1.4

Substitute the known values into the formula for the inverse.

$\frac{1}{1} [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}]$

Step 1.5

Divide $1$ by $1$ .

$1 [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}]$

Step 1.6

Multiply $1$ by each element of the matrix.

$[\begin{array}{c} 1 \cdot 12 & 1 \cdot - 5 \\ 1 \cdot - 7 & 1 \cdot 3 \end{array}]$

Step 1.7

Simplify each element in the matrix.

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Step 1.7.1

Multiply $12$ by $1$ .

$[\begin{array}{c} 12 & 1 \cdot - 5 \\ 1 \cdot - 7 & 1 \cdot 3 \end{array}]$

Step 1.7.2

Multiply $- 5$ by $1$ .

$[\begin{array}{c} 12 & - 5 \\ 1 \cdot - 7 & 1 \cdot 3 \end{array}]$

Step 1.7.3

Multiply $- 7$ by $1$ .

$[\begin{array}{c} 12 & - 5 \\ - 7 & 1 \cdot 3 \end{array}]$

Step 1.7.4

Multiply $3$ by $1$ .

$[\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}]$

Step 2

Multiply both sides by the inverse of $[\begin{array}{c} 3 & 5 \\ 7 & 12 \end{array}]$ .

$[\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 3 & 5 \\ 7 & 12 \end{array}] x = [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$

Step 3

Simplify the equation.

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Step 3.1

Multiply $[\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 3 & 5 \\ 7 & 12 \end{array}]$ .

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Step 3.1.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 2$ .

Step 3.1.2

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

$[\begin{array}{c} 12 \cdot 3 - 5 \cdot 7 & 12 \cdot 5 - 5 \cdot 12 \\ - 7 \cdot 3 + 3 \cdot 7 & - 7 \cdot 5 + 3 \cdot 12 \end{array}] x = [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$

Step 3.1.3

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

$[\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] x = [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$

Step 3.2

Multiplying the identity matrix by any matrix $A$ is the matrix $A$ itself.

$x = [\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$

Step 3.3

Multiply $[\begin{array}{c} 12 & - 5 \\ - 7 & 3 \end{array}] [\begin{array}{c} 2 & - 1 \\ 3 & 2 \end{array}]$ .

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Step 3.3.1

Step 3.3.2

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

$x = [\begin{array}{c} 12 \cdot 2 - 5 \cdot 3 & 12 \cdot - 1 - 5 \cdot 2 \\ - 7 \cdot 2 + 3 \cdot 3 & - 7 \cdot - 1 + 3 \cdot 2 \end{array}]$

Step 3.3.3

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

$x = [\begin{array}{c} 9 & - 22 \\ - 5 & 13 \end{array}]$