## 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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