## Linear Algebra and Its Applications, exercise 1.4.2

Exercise 1.4.2. Multiply the matrices below: $\left[ \begin{array}{rr} 4&1 \\ 5&1 \\ 6&1 \end{array} \right] \left[ \begin{array}{r} 1 \\ 3 \end{array} \right] \rm and \left[ \begin{array}{rrr} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{array} \right] \left[ \begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right] \rm and \left[ \begin{array}{rr} 4&3 \\ 6&6 \\ 8&9\end{array} \right] \left[ \begin{array}{r} \frac{1}{2} \\ \frac{1}{3} \end{array} \right]$

Work by columns instead of by rows.

Answer: The first example multiplies a 3×2 matrix by a 2×1 matrix (column vector), producing a 3×1 matrix (column vector). Working by rows gives us the following: $\left[ \begin{array}{rr} 4&1 \\ 5&1 \\ 6&1 \end{array} \right] \left[ \begin{array}{r} 1 \\ 3 \end{array} \right] = \left[ \begin{array}{r} 4 \cdot 1 + 1 \cdot 3 \\ 5 \cdot 1 + 1 \cdot 3 \\ 6 \cdot 1 + 1 \cdot 3 \end{array} \right] = \left[ \begin{array}{r} 7 \\ 8 \\ 9 \end{array} \right]$

Working by columns is done as follows: $\begin{bmatrix} 4&1 \\ 5&1 \\ 6&1 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} \cdot 1 + \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \cdot 3 = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} + \begin{bmatrix} 3 \\ 3 \\ 3 \end{bmatrix} = \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix}$

For the second example we’re multiplying a 3×3 matrix times a 3×1 matrix to produce a 3×1 matrix (column vector). Working by rows gives us the following: $\begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 0 \\ 4 \cdot 0 + 5 \cdot 1 + 6 \cdot 0 \\ 7 \cdot 0 + 8 \cdot 1 + 9 \cdot 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix}$

Working by columns is done as follows: $\begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ 7 \end{bmatrix} \cdot 0 + \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix} \cdot 1 + \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \cdot 0 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ 8 \end{bmatrix}$

Note that this is a special case of the general case that multiplying an nxn matrix by a column vector of the form (0, …, 0, 1, 0, …, 0) (where the jth entry is 1 and the others zero) produces a column vector corresponding to the jth column of the matrix.

The third example multiplies a 3×2 matrix by a 2×1 matrix to form a 3×1 matrix. Working by rows gives us the following: $\begin{bmatrix} 4&3 \\ 6&6 \\ 8&9 \end{bmatrix} \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 4 \cdot \frac{1}{2} + 3 \cdot \frac{1}{3} \\ 6 \cdot \frac{1}{2} + 6 \cdot \frac{1}{3} \\ 8 \cdot \frac{1}{2} + 9 \cdot \frac{1}{3} \end{bmatrix} = \begin{bmatrix} 2 + 1 \\ 3 + 2 \\ 4 + 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$

Working by columns is done as follows: $\begin{bmatrix} 4 \\ 6 \\ 8 \end{bmatrix} \cdot \frac{1}{2} + \begin{bmatrix} 3 \\ 6 \\ 9 \end{bmatrix} \cdot \frac{1}{3} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix} + \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 5 \\ 7 \end{bmatrix}$

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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