## 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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