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Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

Short Answer

Expert verified

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since Ais an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\).

Step by step solution

01

The equation \(AX = B\) is equivalent to the p systems

It is known that \(A\)is a \(m \times n\) matrix. If Bis a \(n \times p\) matrix with columns \({{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _p}\), the product ABis the \(m \times p\) matrix whose columns are \(A{{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm b}\nolimits} _p}\). That is, \(AB = A\left( {\begin{aligned}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{A{b_1}}&{A{b_2}}&{A{b_3}}\end{aligned}} \right)\) .

Write the matrix \(B = \left( {\begin{aligned}{*{20}{c}}{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\) and \(X = \left( {\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\). \(AX = \left( {\begin{aligned}{*{20}{c}}{A{{\mathop{\rm u}\nolimits} _1}}&{...}&{A{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right)\) according to the matrix multiplication. Therefore, the equation \(AX = B\) is equivalent to the p system.

\(A{{\mathop{\rm u}\nolimits} _1} = {{\mathop{\rm b}\nolimits} _1},...,A{{\mathop{\rm u}\nolimits} _p} = {{\mathop{\rm b}\nolimits} _p}\)

Since Ais the coefficient matrix in each system, these systems can be solved simultaneously by placing the augmented columns of each system next to Asuch that \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\).

02

The row reduced matrix of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\)

If Ais an invertible \(n \times n\) matrix, for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has the unique solution \(x = {A^{ - 1}}b\).

The solution \(\begin{aligned}{*{20}{c}}{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}\) can be uniquely determined since Ais an invertible matrix, and \(\left( {\begin{aligned}{*{20}{c}}A&{{b_1}}&{...}&{{b_p}}\end{aligned}} \right)\)must row reduced to \(\left( {\begin{aligned}{*{20}{c}}I&{{{\mathop{\rm u}\nolimits} _1}}&{...}&{{{\mathop{\rm u}\nolimits} _p}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\). Thus, \(X\) is a unique solution \({A^{ - 1}}B\) of the equation \(AX = B\).

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Most popular questions from this chapter

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

8. \[\left[ {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\{\bf{0}}&{\bf{0}}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&I\end{array}} \right]\]

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).

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

Describe in words what happens when you compute \({A^{\bf{5}}}\), \({A^{{\bf{10}}}}\), \({A^{{\bf{20}}}}\), and \({A^{{\bf{30}}}}\) for \(A = \left( {\begin{aligned}{*{20}{c}}{1/6}&{1/2}&{1/3}\\{1/2}&{1/4}&{1/4}\\{1/3}&{1/4}&{5/12}\end{aligned}} \right)\).

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