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Chapter 2: Q4SE (page 93)

Suppose \({A^n} = {\bf{0}}\) for some \({\bf{n}} > {\bf{1}}\). Find an inverse for \(I - A\).

Short Answer

Expert verified

The inverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

Step by step solution

01

Use the matrix computation in Exercise 3

That is \(\left( {I - A} \right)\left( {I + A + {A^2}} \right) = I - {A^3}\). In general,

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I - {A^n}\)

02

Suppose \({A^n} = {\bf{0}}\)

Substitute \({A^n} = 0\) in the above equation to obtain:

\(\left( {I - A} \right)\left( {I + A + {A^2} + ... + {A^{n - 1}}} \right) = I\)

This implies \(I - A\) is invertibleand \({\left( {I - A} \right)^{ - 1}} = I + A + {A^2} + ... + {A^{n - 1}}\)

03

Conclusion

Hence, theinverse of \(I - A\) is \(I + A + {A^2} + ... + {A^{n - 1}}\).

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Most popular questions from this chapter

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

If A, B, and X are \(n \times n\) invertible matrices, does the equation \({C^{ - 1}}\left( {A + X} \right){B^{ - 1}} = {I_n}\) have a solution, X? If so, find it.

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.]

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

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 \(AB\).

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

How many rows does \(B\) have if \(BC\) is a \({\bf{3}} \times {\bf{4}}\) matrix?
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