## Linear Algebra and Its Applications, exercise 1.4.1

Exercise 1.4.1. Multiply the matrices below:

$\left[ \begin{array}{rrr} 4&0&1 \\ 0&1&0 \\ 4&0&1 \end{array} \right] \left[ \begin{array}{r} 3 \\ 4 \\ 5 \end{array} \right] \rm and \left[ \begin{array}{rrr} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array} \right] \left[ \begin{array}{r} 5 \\ -2 \\ 3 \end{array} \right] \rm and \left[ \begin{array}{rr} 2&0 \\ 1&3 \end{array} \right] \left[ \begin{array}{r} 1 \\ 1 \end{array} \right]$

(The exercise also asks you to draw a graph showing addition of the two vectors (2, 1) and (0, 3); I’m skipping that part.)

Answer: The first example multiplies a 3×3 matrix by a 3×1 matrix (column vector), producing a 3×1 matrix. The first entry in the resulting column vector is

$4 \cdot 3 + 0 \cdot 4 + 1 \cdot 5 = 12 + 0 + 5 = 17$

The second entry in the resulting column vector is

$0 \cdot 3 + 1 \cdot 4 + 0 \cdot 5 = 0 + 4 + 0 = 4$

The third entry in the resulting column vector is the same as the first entry:

$4 \cdot 3 + 0 \cdot 4 + 1 \cdot 5 = 12 + 0 + 5 = 17$

The column vector that is the product of the multiplication is therefore (17, 4, 17):

$\left[ \begin{array}{rrr} 4&0&1 \\ 0&1&0 \\ 4&0&1 \end{array} \right] \left[ \begin{array}{r} 3 \\ 4 \\ 5 \end{array} \right] = \left[ \begin{array}{r} 17 \\ 4 \\ 17 \end{array} \right]$

For the second example we’re again multiplying a 3×3 matrix times a 3×1 matrix to produce 3×1 matrix (column vector). The three entries in the column vector are computed as follows:

$1 \cdot 5 + 0 \cdot (-2) + 0 \cdot 3 = 5 + 0 + 0 = 5$

$0 \cdot 5 + 1 \cdot (-2) + 0 \cdot 3 = 0 - 2 + 0 = -2$

$0 \cdot 5 + 0 \cdot (-2) + 1 \cdot 3 = 0 + 0 + 3 = 3$

The product is thus the column vector (5, -2, 3), and this is a special case of the general case IA = A where I is the identity matrix.

The third example multiplies a 2×2 matrix by a 2×1 matrix to form a 2×1 matrix, with entries as follows:

$2 \cdot 1 + 0 \cdot 1 = 2 + 0 = 2$

$1 \cdot 1 + 3 \cdot 1 = 1 + 3 = 4$

We therefore have

$\left[ \begin{array}{rr} 2&0 \\ 1&3 \end{array} \right] \left[ \begin{array}{r} 1 \\ 1 \end{array} \right] = \left[ \begin{array}{r} 2 \\ 4 \end{array} \right]$

Note that we also have

$\left[ \begin{array}{r} 2 \\ 1 \end{array} \right] + \left[ \begin{array}{r} 0 \\ 3 \end{array} \right] = \left[ \begin{array}{r} 2 \\ 4 \end{array} \right]$

This is a special case of the general case that multiplying a matrix by a column vector of the form (1, 1, 1, …, 1) is equivalent to adding the column vectors of the original matrix.

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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