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Chapter 2: Q17SE (page 93)

Let A be a \(6 \times 4\) matrix and B a \(4 \times 6\) matrix. Show that the \(6 \times 6\) matrix \(AB\) cannot be invertible.

Short Answer

Expert verified

It is proved that the \(6 \times 6\) matrix \(AB\) cannot be invertible.

Step by step solution

01

Show that the \(6 \times 6\)matrix \(AB\) cannot be invertible

Consider A as a \(6 \times 4\) matrix and B as a \(4 \times 6\) matrix. The six columns of B are linearly dependent. There is a nonzero x such that \(Bx = 0\) because \(B\) has more columns than rows. Then,

\(\begin{aligned}{c}ABx = A0\\ = 0.\end{aligned}\)

This demonstrates that matrix \(AB\) is not invertible according to the invertible matrix theorem.

Thus, it is proved that the \(6 \times 6\) matrix \(AB\) cannot be invertible.

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Most popular questions from this chapter

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\( - 2A\), \(B - 2A\), \(AC\), \(CD\).

Suppose Ais an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

In exercise 11 and 12, mark each statement True or False.Justify each answer.

a. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}&{{A_{\bf{2}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), with \({A_{\bf{1}}}\) and \({A_{\bf{2}}}\) the same sizes as \({B_{\bf{1}}}\) and \({B_{\bf{2}}}\), respectively then \(A + B = \left[ {\begin{array}{*{20}{c}}{{A_1} + {B_1}}&{{A_{\bf{2}}} + {B_{\bf{2}}}}\end{array}} \right]\).

b. If \(A = \left[ {\begin{array}{*{20}{c}}{{A_{{\bf{11}}}}}&{{A_{{\bf{12}}}}}\\{{A_{{\bf{21}}}}}&{{A_{{\bf{22}}}}}\end{array}} \right]\) and \(B = \left[ {\begin{array}{*{20}{c}}{{B_1}}\\{{B_{\bf{2}}}}\end{array}} \right]\), then the partitions of \(A\) and \(B\) are comfortable for block multiplication.

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

Suppose the last column of ABis entirely zero but Bitself has no column of zeros. What can you sayaboutthe columns of A?
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