## Linear Algebra and Its Applications, Exercise 1.16.13

Exercise 1.6.13. Compute $A^TB$, $B^TA$, $AB^T$, and $BA^T$ for the following matrices: $A = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \quad B = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ $A^TB = \begin{bmatrix} 3&1 \end{bmatrix} \begin{bmatrix} 2 \\ 2 \end{bmatrix} = 8$ $B^TA = \begin{bmatrix} 2&2 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix} = 8$ $AB^T = \begin{bmatrix} 3 \\ 1 \end{bmatrix} \begin{bmatrix} 2&2 \end{bmatrix} = \begin{bmatrix} 6&6 \\ 2&2 \end{bmatrix}$ $BA^T = \begin{bmatrix} 2 \\ 2 \end{bmatrix} \begin{bmatrix} 3&1 \end{bmatrix} = \begin{bmatrix} 6&2 \\ 6&2 \end{bmatrix}$

Note that per Equation 1M(i) on page 47 we have $(AB)^T = B^TA^T$

This means that $(A^TB)^T = B^T(A^T)^T = B^TA$

and $(AB^T)^T = (B^T)^TA^T = BA^T$

as confirmed by the computation above.

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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