Linear Algebra Examples

$[\begin{array}{c} a & b & d \end{array}] [\begin{array}{c} 1 & 5 & 0 \\ 5 & 3 & 6 \\ 0 & 6 & 2 \end{array}] [\begin{array}{c} a \\ b \\ d \end{array}]$

Step 1

Multiply $[\begin{array}{c} a & b & d \end{array}] [\begin{array}{c} 1 & 5 & 0 \\ 5 & 3 & 6 \\ 0 & 6 & 2 \end{array}]$ .

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Step 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 $1 \times 3$ and the second matrix is $3 \times 3$ .

Step 1.2

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

$[\begin{array}{c} a \cdot 1 + b \cdot 5 + d \cdot 0 & a \cdot 5 + b \cdot 3 + d \cdot 6 & a \cdot 0 + b \cdot 6 + d \cdot 2 \end{array}] [\begin{array}{c} a \\ b \\ d \end{array}]$

Step 1.3

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

$[\begin{array}{c} a + 5 b & 5 a + 3 b + 6 d & 6 b + 2 d \end{array}] [\begin{array}{c} a \\ b \\ d \end{array}]$

Step 2

Multiply $[\begin{array}{c} a + 5 b & 5 a + 3 b + 6 d & 6 b + 2 d \end{array}] [\begin{array}{c} a \\ b \\ d \end{array}]$ .

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Step 2.1

Step 2.2

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

$[\begin{array}{c} (a + 5 b) a + (5 a + 3 b + 6 d) b + (6 b + 2 d) d \end{array}]$

Step 2.3

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

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Step 2.3.1

Move $d$ .

$[\begin{array}{c} a^{2} + 10 a b + 3 b^{2} + 6 b d + 6 b d + 2 d^{2} \end{array}]$

Step 2.3.2

Add $6 b d$ and $6 b d$ .

$[\begin{array}{c} a^{2} + 10 a b + 3 b^{2} + 12 b d + 2 d^{2} \end{array}]$

Step 3

The determinant of a $1 \times 1$ matrix is the element itself.

$a^{2} + 10 a b + 3 b^{2} + 12 b d + 2 d^{2}$

Step 4

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

Step 5

The inverse of a $1 \times 1$ matrix is a $1 \times 1$ matrix with the reciprocal of the original element.

$[\begin{array}{c} \frac{1}{a^{2} + 10 a b + 3 b^{2} + 12 b d + 2 d^{2}} \end{array}]$