Linear Algebra and Its Applications, Exercise 1.6.4

Exercise 1.6.4. (a) Given AB = AC, show that B = C if A is invertible.

(b) Given

$A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$

find B and C such that AB = AC but $B \ne C$ .

Answer: (a) If A is invertible then

$AB = AC \rightarrow A^{-1}AB = A^{-1}AC \rightarrow B = C$

(b) If B is the identity matrix I then we have AB = AI = A. If C is the matrix

$C = \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix}$

then we have

$AC = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} = A = AB$

So AB = AC even though B and C are not equal.

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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