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Chapter 2: Q2.5-21Q (page 93)

21. Suppose\[A = BC\], where B is invertible. Show that any sequence of row operations that reduces B to I also reduces A to C. The converse is not true, since the zero matrix may be factored as \[{\bf{0}} = B \cdot {\bf{0}}\]. a

Short Answer

Expert verified

It is proven that the sequence of row operations that reduces B to I also reduces A to C.

Step by step solution

01

Describe the given data

Given, \[A = BC\] with B is invertible. Therefore, there exists a sequence of elementary matrices \[{E_1},{E_2},...,{E_n}\] as row operations that reduce B to I such that \[{E_n}{E_{n - 1}}...{E_1}B = I\].

02

Reduce A to C

Apply the same sequence of row operations to A as follows:

\[\begin{array}{c}{E_n}{E_{n - 1}}...{E_1}A = {E_n}{E_{n - 1}}...{E_1}\left( {BC} \right)\\ = \left( {{E_n}{E_{n - 1}}...{E_1}B} \right)C\\ = IC\\{E_n}{E_{n - 1}}...{E_1}A = C\end{array}\]

The matrix multiplication is associative.

03

Conclusion

Hence, the sequence of row operations that reduces B to I also reduces A to C.

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Most popular questions from this chapter

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