## 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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