Linear Algebra and Its Applications, exercise 1.4.3

Exercise 1.4.3. Multiply the matrices below:

$\begin{bmatrix} 1&-2&7 \end{bmatrix} \begin{bmatrix} 1 \\ -2 \\ 7 \end{bmatrix} \rm and \begin{bmatrix} 1&-2&7 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \\ 1 \end{bmatrix} \rm and \begin{bmatrix} 1 \\ -2 \\ 7 \end{bmatrix} \begin{bmatrix} 3&5&1 \end{bmatrix}$

Answer: The first example multiplies a 1×3 matrix by a 3×1 matrix producing a 1×1 matrix, i.e., a scalar value:

$1 \cdot 1 + (-2) \cdot (-2) + 7 \cdot 7 = 1 + 4 + 49 = 54$

As noted in the book, the result is equal to the square of the length of the vector (1, -2, 7) .

For the second example we’re also multiplying a 1×3 matrix times a 3×1 matrix to produce a scalar value:

$1 \cdot 3 + (-2) \cdot 5 + 7 \cdot 1 = 3 - 10 + 7 = 0$

The final example multiplies a 3×1 matrix times a 1×3 matrix to produce a 3×3 matrix:

$\begin{bmatrix} 1 \\ -2 \\ 7 \end{bmatrix} \begin{bmatrix} 3&5&1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 3&1 \cdot 5&1 \cdot 1 \\ -2 \cdot 3&-2 \cdot 5&-2 \cdot 1 \\ 7 \cdot 3&7 \cdot 5&7 \cdot 1 \end{bmatrix} = \begin{bmatrix} 3&5&1 \\ -6&-10&-2 \\ 21&35&7 \end{bmatrix}$

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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