10.2.1 Example of Matrix-Column-Vector Multiplication Using Dot Products

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Properties: 10.3 Properties of Matrix-Column-Vector Multiplication
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Example

Find the individual entries of vector $\vec{b} = A \cdot \vec{x}$ where

A = {[\begin{matrix} - 3 & 4 & - 3 \\ - 1 & 7 & 6 \\ 0 & 1 & 2 \\ 2 & - 5 & 2 \end{matrix}]}_{4 \times 3} and \vec{x} = {[\begin{matrix} 5 \\ 2 \\ - 2 \end{matrix}]}_{3 \times 1}

using 10.2 Matrix-Column-Vector Multiplication via Dot Products.

Solution. Using our definition of matrix-column-vector multiplication via dot products, we know

\begin{aligned} b_{i} & = [A (i, :)]^{T} \cdot \vec{x} \\ = [\begin{array}{c} a_{i 1} \\ a_{i 2} \\ ⋮ \\ a_{i n} \end{array}] \cdot [\begin{array}{c} x_{1} \\ x_{2} \\ ⋮ \\ x_{n} \end{array}] \\ = a_{i 1} x_{1} + a_{i 2} x_{2} + \dots + a_{i n} x_{n} \end{aligned}

Recall that the inner dimensions must always agree.

[\vec{b}]_{4 \times 1} = [A]_{4 \times 3} \cdot {\vec{x}}_{3 \times 1}

Therefore, the outer dimensions will be the dimensions of our output vector, which is $4 \times 1$ . With this information at hand, let's start with Entry 1 where $i = 1$ .

\begin{aligned} b_{1} & = [A (1, :)]^{T} \cdot \vec{x} \\ = {[\begin{array}{c} - 3 \\ 4 \\ - 3 \end{array}]}_{3 \times 1} \cdot {[\begin{array}{c} 5 \\ 2 \\ - 2 \end{array}]}_{3 \times 1} \\ = - 3 \cdot 5 + 4 \cdot 2 + - 3 \cdot - 2 \\ = - 15 + 8 + 6 \\ = - 7 + 6 \\ ⟹ b_{1} & = - 1 \end{aligned}

Let's continue with Entry 2 where $i = 2$ . Again, we are going to get the second row of $A$ , transpose it, and take the 5.1 Inner Product Between Vectors with $\vec{x}$ .

\begin{aligned} b_{2} & = [A (2, :)]^{T} \cdot \vec{x} \\ = {[\begin{array}{c} - 1 \\ 7 \\ 6 \end{array}]}_{3 \times 1} \cdot {[\begin{array}{c} 5 \\ 2 \\ - 2 \end{array}]}_{3 \times 1} \\ = - 1 \cdot 5 + 7 \cdot 2 + 6 \cdot - 2 \\ = - 5 + 14 + - 12 \\ = 9 - 12 \\ ⟹ b_{2} & = - 3 \end{aligned}

Now, let's find Entry 3 where $i = 3$ .

\begin{aligned} b_{3} & = [A (3, :)]^{T} \cdot \vec{x} \\ = {[\begin{array}{c} 0 \\ 1 \\ 2 \end{array}]}_{3 \times 1} \cdot {[\begin{array}{c} 5 \\ 2 \\ - 2 \end{array}]}_{3 \times 1} \\ = 0 \cdot 5 + 1 \cdot 2 + 2 \cdot - 2 \\ = 0 + 2 + - 4 \\ = 2 + - 4 \\ ⟹ b_{3} & = - 2 \end{aligned}

Let's finish our entries with Entry 4 where $i = 4$ .

\begin{aligned} b_{4} & = [A (4, :)]^{T} \cdot \vec{x} \\ = {[\begin{array}{c} 2 \\ - 5 \\ 2 \end{array}]}_{3 \times 1} \cdot {[\begin{array}{c} 5 \\ 2 \\ - 2 \end{array}]}_{3 \times 1} \\ = 2 \cdot 5 + - 5 \cdot 2 + 2 \cdot - 2 \\ = 10 + - 10 + - 4 \\ = 0 + - 4 \\ ⟹ b_{4} & = - 4 \end{aligned}

Finally, we can look at all 4 entries as the vector $\vec{b}$ .

\vec{b} = [\begin{matrix} - 3 & 4 & - 3 \\ - 1 & 7 & 6 \\ 0 & 1 & 2 \\ 2 & - 5 & 2 \end{matrix}] [\begin{matrix} 5 \\ 2 \\ - 2 \end{matrix}] = [\begin{matrix} - 1 \\ - 3 \\ - 2 \\ - 4 \end{matrix}]

Remark. Notice the "faster" way to do this calculation by hand.

[\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{matrix}] = [\begin{matrix} - 3 \cdot 5 + 4 \cdot 2 + - 3 \cdot 5 \\ - 1 \cdot 5 + 7 \cdot 2 + 6 \cdot 5 \\ 0 \cdot 5 + 1 \cdot 2 + 2 \cdot - 2 \\ 2 \cdot 5 + - 5 \cdot 2 + 2 \cdot - 2 \end{matrix}] = [\begin{matrix} - 1 \\ - 3 \\ - 2 \\ 4 \end{matrix}]

Although this is technically faster to write, this is not a great way to build deep understanding.
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