1-05 Multiply Matrices (12.2)

Example 1

Simplify \(\left[\begin{matrix}2&-1\\3&0\end{matrix}\right]\left[\begin{matrix}-2&4\\-3&5\end{matrix}\right]\).

Solution

This a 2×2 · 2×2. The middle numbers are the same, so the number of columns of the first matrix equals the number of rows of the second matrix. They can be multiplied.

Choose a row and column such as the 1st row and 1st column.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}2&-1\end{matrix}$} \\ \begin{matrix}3&0\end{matrix} \end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-2\\-3\\\end{matrix}$}&\begin{matrix}4\\5\end{matrix}\end{matrix}\right] $$

Multiply the first elements of that row and column. It goes in the 1st row 1st column of the answer.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}\mathbf{2}&-1\end{matrix}$} \\ \begin{matrix}3&0\end{matrix}\end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-\mathbf{2}\\-3 \end{matrix}$}&\begin{matrix}4\\5\end{matrix} \end{matrix}\right]=\left[\begin{matrix}-4&\\& \end{matrix}\right] $$

Put a + and multiply the second elements of that row and column.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}2&-\mathbf{1}\end{matrix}$} \\ \begin{matrix}3&0\end{matrix} \end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-2\\-\mathbf{3} \end{matrix}$}&\begin{matrix}4\\5 \end{matrix} \end{matrix}\right]=\left[\begin{matrix}-4+\mathbf{3}&\\&\\\end{matrix}\right] $$

That is the end of that row and column, so choose another row and column such as 1st row and 2nd column.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}2&-1\end{matrix}$} \\ \begin{matrix}3&0\end{matrix} \end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3\end{matrix}&\fbox{$\begin{matrix}4\\5\end{matrix}$} \end{matrix}\right]=\left[\begin{matrix}-4+3&\\&\end{matrix}\right] $$

Multiply the first elements of that row and column. The product goes in the 1st row 2nd column of the answer.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}\mathbf{2}&-1 \end{matrix}$} \\ \begin{matrix}3&0 \end{matrix} \end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3 \end{matrix}&\fbox{$\begin{matrix}\mathbf{4}\\5 \end{matrix}$} \end{matrix}\right]=\left[\begin{matrix}-4+3&\mathbf{8}\\& \end{matrix}\right] $$

Put a + and multiply the second elements of that row and column.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}2&-\mathbf{1}\\\end{matrix}$}\\\begin{matrix}3&0\\\end{matrix}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3\\\end{matrix}&\fbox{$\begin{matrix}4\\\mathbf{5}\\\end{matrix}$}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+(-\mathbf{5})\\&\\\end{matrix}\right] $$

That is the end of that row and column, so choose another row and column such as 2nd row and 1st column.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}3&0\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-2\\-3\\\end{matrix}$}&\begin{matrix}4\\5\\\end{matrix}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\&\\\end{matrix}\right] $$

Multiply the first elements of that row and column. The product goes in the 1st row 2nd column of the answer.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}\mathbf{3}&0\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-\mathbf{2}\\-3\\\end{matrix}$}&\begin{matrix}4\\5\\\end{matrix}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\-\mathbf{6}&\\\end{matrix}\right] $$

Put a + and multiply the second elements of that row and column.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}3&\mathbf{0}\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\fbox{$\begin{matrix}-2\\-\mathbf{3}\\\end{matrix}$}&\begin{matrix}4\\5\\\end{matrix}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\-6+\mathbf{0}&\\\end{matrix}\right] $$

That is the end of that row and column, so choose another row and column such as 2nd row and 2nd column.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}3&0\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3\\\end{matrix}&\fbox{$\begin{matrix}4\\5\\\end{matrix}$}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\-6+0&\\\end{matrix}\right] $$

Multiply the first elements of that row and column. The product goes in the 1st row 2nd column of the answer.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}\mathbf{3}&0\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3\\\end{matrix}&\fbox{$\begin{matrix}\mathbf{4}\\5\\\end{matrix}$}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\-6+0&\mathbf{12}\\\end{matrix}\right] $$

Put a + and multiply the second elements of that row and column.

$$ \left[\begin{matrix}\begin{matrix}2&-1\\\end{matrix}\\\fbox{$\begin{matrix}3&\mathbf{0}\\\end{matrix}$}\\\end{matrix}\right]\left[\begin{matrix}\begin{matrix}-2\\-3\\\end{matrix}&\fbox{$\begin{matrix}4\\\mathbf{5}\\\end{matrix}$}\\\end{matrix}\right]=\left[\begin{matrix}-4+3&8+\left(-5\right)\\-6+0&12+\mathbf{0}\\\end{matrix}\right] $$

That is the end of that row and column. It is also the end of all the combinations of a row of the first matrix and a column of the second matrix. Now just simplify the elements of the answer.

$$ =\left[\begin{matrix}-1&3\\-6&12\\\end{matrix}\right] $$

Example 2

Simplify \(\left[\begin{matrix}1&-2&0\\-1&4&5\\\end{matrix}\right]\left[\begin{matrix}3\\-5\\1\\\end{matrix}\right]\).

Solution

This a 2×3 · 3×1. The middle numbers are the same, so the number of columns of the first matrix equals the number of rows of the second matrix. They can be multiplied. The result will be a 2×1 matrix.

Choose a row and column such as the 1st row and 1st column and multiply them. The result goes in the 1st row 1st column.

$$ \left[\begin{matrix}\fbox{$\begin{matrix}1&-2&0\\\end{matrix}$}\\\begin{matrix}-1&4&5\\\end{matrix}\\\end{matrix}\right]\left[\fbox{$\begin{matrix}3\\-5\\1\\\end{matrix}$}\right]=\left[\begin{matrix}3+10+0\\\\\end{matrix}\right] $$

Choose another row and column such as the 2nd row and 1st column and multiply them. The result goes in the 2nd row 1st column.

$$ \left[\begin{matrix}\begin{matrix}1&-2&0\\\end{matrix}\\\fbox{$\begin{matrix}-1&4&5\\\end{matrix}$}\\\end{matrix}\right]\left[\fbox{$\begin{matrix}3\\-5\\1\\\end{matrix}$}\right]=\left[\begin{matrix}3+10+0\\-3+\left(-20\right)+5\\\end{matrix}\right] $$

That is the end of all the combinations of a row of the first matrix and a column of the second matrix. Now just simplify the elements of the answer.

$$ =\left[\begin{matrix}13\\-18\\\end{matrix}\right] $$

Example 3

Use the given matrices to evaluate (AB) − 2C.

$$ A=\left[\begin{matrix}1&4\\-2&-1\\\end{matrix}\right],\ B=\left[\begin{matrix}5\\3\\\end{matrix}\right],\ C=\left[\begin{matrix}0\\7\\\end{matrix}\right] $$

Solution

Substitute the matrices into the equation.

$$ \left(\left[\begin{matrix}1&4\\-2&-1\\\end{matrix}\right]\left[\begin{matrix}5\\3\\\end{matrix}\right]\right)-2\left[\begin{matrix}0\\7\\\end{matrix}\right] $$

Order of operations says groups first, so multiply the two matrices in parentheses.

$$ \left(\left[\begin{matrix} 1· 5+4·3 \\ (-2)·5+(-1)·3 \end{matrix} \right]\right) -2 \left[\begin{matrix} 0 \\ 7 \end{matrix} \right] $$

$$ \left[\begin{matrix}17\\-13\\\end{matrix}\right]-2\left[\begin{matrix}0\\7\\\end{matrix}\right] $$

Next do the scalar multiplication because multiplication comes before subtraction.

$$ \left[\begin{matrix}17\\-13\\\end{matrix}\right]-\left[\begin{matrix}0\\14\\\end{matrix}\right] $$

Finally, subtract.

$$ \left[\begin{matrix}17\\-27\\\end{matrix}\right] $$

Example 4

The members of two bowling leagues submit meal choices for an upcoming banquet as shown. Each pizza meal costs $12, each spaghetti meal costs $15, and each Sam's chicken meal costs $20. Use matrix multiplication to find the total cost of the meals for each league.

	Pizza	Spaghetti	Sam's Chicken
League A	19	6	2
League B	10	9	1

Solution

Write the table of purchases as a matrix and multiply it by a price matrix.

$$ \left[\begin{matrix}19&6&2\\10&9&1\\\end{matrix}\right]\left[\begin{matrix}12\\15\\20\\\end{matrix}\right] $$

$$ \left[\begin{matrix}19·12+6·15+2·20 \\ 10·12+9·15+1·20 \end{matrix}\right] $$

$$ \left[\begin{matrix}358\\275\\\end{matrix}\right] $$

League A: $358, League B: $275