Linear Algebra and Its Applications, Review Exercise 1.14

Review exercise 1.14. For each of the following find a 3 by 3 matrix B such that for any matrix A

(a) BA = 2A

(b) BA = 2B

(c) The first row of BA is the last row of A and the last row of BA is the first row of A.

(d) The first column of BA is the last column of A and the last column of BA is the first column of A.

Answer: (a) We have 2A = 2IA so that we can choose B = 2I

B = \begin{bmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&2 \end{bmatrix}

(b) For all A we have 0 \cdot A = 0 = 2 \cdot 0 so that we can choose B = 0

B = \begin{bmatrix} 0&0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}

(c) We can choose B to be the permutation matrix that reverses the order of rows in A

B = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix}

For example

BA = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} \begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} = \begin{bmatrix} 7&8&9 \\ 4&5&6 \\ 1&2&3 \end{bmatrix}

(d) The first column of BA is produced by multiplying the first row of B by the first column of A. Since this computation does not involve the last column of A in general it is impossible to find a matrix B such that the first column of BA is equal to the last column of A.

However note that we can reverse the order of columns in A by multiplying A on the right by the value of B from (c) above:

AB = \begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} = \begin{bmatrix} 3&2&1 \\ 6&5&4 \\ 9&8&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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