Linear Algebra Examples

$[\begin{array}{c} \sqrt{7} & 0 \\ 2 & \sqrt{28} \end{array}] [\begin{array}{c} 2 \sqrt{3} & - \sqrt{7} \\ 0 & - 6 \end{array}]$

Step 1

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 1.1

Factor $4$ out of $28$ .

$[\begin{array}{c} \sqrt{7} & 0 \\ 2 & \sqrt{4 (7)} \end{array}] [\begin{array}{c} 2 \sqrt{3} & - \sqrt{7} \\ 0 & - 6 \end{array}]$

Step 1.2

Rewrite $4$ as $2^{2}$ .

$[\begin{array}{c} \sqrt{7} & 0 \\ 2 & \sqrt{2^{2} \cdot 7} \end{array}] [\begin{array}{c} 2 \sqrt{3} & - \sqrt{7} \\ 0 & - 6 \end{array}]$

Step 2

Pull terms out from under the radical.

$[\begin{array}{c} \sqrt{7} & 0 \\ 2 & 2 \sqrt{7} \end{array}] [\begin{array}{c} 2 \sqrt{3} & - \sqrt{7} \\ 0 & - 6 \end{array}]$

Step 3

Multiply $[\begin{array}{c} \sqrt{7} & 0 \\ 2 & 2 \sqrt{7} \end{array}] [\begin{array}{c} 2 \sqrt{3} & - \sqrt{7} \\ 0 & - 6 \end{array}]$ .

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Step 3.1

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 3.2

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

$[\begin{array}{c} \sqrt{7} (2 \sqrt{3}) + 0 \cdot 0 & \sqrt{7} (- \sqrt{7}) + 0 \cdot - 6 \\ 2 (2 \sqrt{3}) + 2 \sqrt{7} \cdot 0 & 2 (- \sqrt{7}) + 2 \sqrt{7} \cdot - 6 \end{array}]$

Step 3.3

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

$[\begin{array}{c} 2 \sqrt{21} & - 7 \\ 4 \sqrt{3} & - 14 \sqrt{7} \end{array}]$