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Chapter 2: Q18Q (page 93)

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.

Short Answer

Expert verified

\(B = {P^{ - 1}}AP\)

Step by step solution

01

Condition for an invertible matrix

Theorem 5states that Ais an invertible \(n \times n\) matrix, then for each b in \({\mathbb{R}^n}\), the equation \(Ax = b\) has a unique solution \(x = {A^{ - 1}}b\).

02

Solve for B in terms of A

Multiply both sides of the equation \(A = PB{P^{ - 1}}\) by \({P^{ - 1}}\):

\(\begin{aligned}{c}{P^{ - 1}}A = {P^{ - 1}}PB{P^{ - 1}}\\{P^{ - 1}}A = IB{P^{ - 1}}\\{P^{ - 1}}A = B{P^{ - 1}}\end{aligned}\)

Multiply both sides of theobtainedequation by P:

\(\begin{aligned}{c}{P^{ - 1}}AP = B{P^{ - 1}}P\\{P^{ - 1}}AP = BI\\{P^{ - 1}}AP = B\end{aligned}\)

Thus, \(B = {P^{ - 1}}AP\).

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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?

Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

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 \({A^{ - 1}}B\).

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)
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