Suppose \(C = {E_3}{E_2}{E_1}B\), where \({E_1},{E_2},\) , and \({E_3}\) are elementary matrices. Explain why C is row equivalent to B.

Short Answer

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\(C\)is row equivalent to B.

Step by step solution

01

Explain that \(C\) is row equivalent to B

Multiply the left side of the elementary matrix to produce an elementary row operation, as shown below.

\(B \sim {E_1}B \sim {E_2}{E_1}B \sim {E_3}{E_2}{E_1}B = C\)

Therefore, B is equivalent to C. Also, \(C\) is row equivalent to B because row operation can be reversed.

Alternatively, you can use the inverse of \({E_i}\) to show the transformation of C into B by row operation.

Thus, it is proved that \(C\) is row equivalent to B.

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Most popular questions from this chapter

Suppose \({A_{{\bf{11}}}}\) is invertible. Find \(X\) and \(Y\) such that

\[\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{\bf{0}}\\{\bf{0}}&S\end{array}} \right]\left[ {\begin{array}{*{20}{c}}I&Y\\{\bf{0}}&I\end{array}} \right]\]

Where \(S = {A_{{\bf{22}}}} - {A_{21}}A_{{\bf{11}}}^{ - {\bf{1}}}{A_{{\bf{12}}}}\). The matrix \(S\) is called the Schur complement of \({A_{{\bf{11}}}}\). Likewise, if \({A_{{\bf{22}}}}\) is invertible, the matrix \({A_{{\bf{11}}}} - {A_{{\bf{12}}}}A_{{\bf{22}}}^{ - {\bf{1}}}{A_{{\bf{21}}}}\) is called the Schur complement of \({A_{{\bf{22}}}}\). Such expressions occur frequently in the theory of systems engineering, and elsewhere.

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