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Chapter 7: Problem 29

Let \\[ A=\left(\begin{array}{rr} 1 & -0.99 \\ -1 & 1 \end{array}\right) \\] Find \(A^{-1}\) and \(\operatorname{cond}_{\infty}(A)\)

Short Answer

Expert verified
The inverse of matrix A is \(A^{-1} = \left(\begin{array}{rr} 100 & 99 \\ 100 & 100 \end{array}\right)\), and its infinity condition number is \(\operatorname{cond}_{\infty}(A) = 2\).

Step by step solution

01

Check if the matrix A is invertible

In order for a matrix to be invertible, its determinant must be non-zero. Let's compute the determinant of matrix A: \\[ \operatorname{det}(A) = 1 \times 1 - (-1) \times (-0.99) = 1 - 0.99 = 0.01 \\] Since the determinant is not equal to zero, the matrix A is invertible.
02

Find the inverse of matrix A

To find the inverse of A, use the following formula: \\[ A^{-1} = (1/\operatorname{det}(A)) \times \operatorname{adj}(A) \\] Where \(\operatorname{adj}(A)\) is the adjugate matrix of A, obtained by swapping the diagonal elements and changing the signs of the off-diagonal elements: \\[ \operatorname{adj}(A) = \left(\begin{array}{rr} 1 & 0.99 \\ 1 & 1 \end{array}\right) \\] Now, we can calculate the inverse of A: \\[ A^{-1} = \frac{1}{0.01} \times \left(\begin{array}{rr} 1 & 0.99 \\ 1 & 1 \end{array}\right) = \left(\begin{array}{rr} 100 & 99 \\ 100 & 100 \end{array}\right) \\] So, the inverse of A is: \\[ A^{-1} = \left(\begin{array}{rr} 100 & 99 \\ 100 & 100 \end{array}\right) \\]
03

Calculate the infinity norm of A and A^{-1}

The infinity norm of a matrix is the maximum absolute row sum. Calculate the infinity norm of both A and A^{-1}: \\[ \|A\|_{\infty} = \max\{|1+(-0.99)|, |-1+1|\} = \max\{0.01, 0\} = 0.01 \\] \\[ \|A^{-1}\|_{\infty} = \max\{|100+99|, |100+100|\} = \max\{199, 200\} = 200 \\]
04

Calculate the infinity condition number of A

The infinity condition number of A is the product of the infinity norm of A and A^{-1}: \\[ \operatorname{cond}_{\infty}(A) = \|A\|_{\infty}\|A^{-1}\|_{\infty} = 0.01 \times 200 = 2 \\] The infinity condition number of matrix A is 2.

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Most popular questions from this chapter

Suppose that \(A^{-1}\) and the \(L U\) factorization of \(A\) have already been determined. How many scalar additions and multiplications are necessary to compute \(A^{-1} \mathbf{b} ?\) Compare this number with the number of operations required to solve \(L U \mathbf{x}=\mathbf{b}\) using \(\mathrm{Al}\) gorithm \(7.2 .2 .\) Suppose that we have a number of systems to solve with the same coefficient matrix \(A .\) Is it worthwhile to compute \(A^{-1} ?\) Explain.

Let \(A\) be a symmetric \(n \times n\) matrix with eigenvalues \(\lambda_{1}, \ldots, \lambda_{n}\) and orthonormal eigenvectors \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{n} .\) Let \(\mathbf{x} \in \mathbb{R}^{n}\) and let \(c_{i}=\mathbf{u}_{i}^{T} \mathbf{x}\) for \(i=1,2, \ldots, n .\) Show that (a) \(\|A \mathbf{x}\|_{2}^{2}=\sum_{i=1}^{n}\left(\lambda_{i} c_{i}\right)^{2}\) (b) If \(\mathbf{x} \neq \mathbf{0},\) then \\[ \min _{1 \leq i \leq n}\left|\lambda_{i}\right| \leq \frac{\|A \mathbf{x}\|_{2}}{\|\mathbf{x}\|_{2}} \leq \max _{1 \leq i \leq n}\left|\lambda_{i}\right| \\] (c) \(\|A\|_{2}=\max _{1 \leq i \leq n}\left|\lambda_{i}\right|\)

Let \\[ A=\left(\begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{c} 5.00 \\ 1.02 \\ 1.04 \\ 1.10 \end{array}\right) \\] An approximate solution of \(A \mathbf{x}=\mathbf{b}\) is calculated by rounding the entries of \(\mathbf{b}\) to the nearest integer and then solving the rounded system with integer arithmetic. The calculated solution is \(\mathbf{x}^{\prime}=\) \((12,4,2,1)^{T}\). Let \(\mathbf{r}\) denote the residual vector. (a) Determine the values of \(\|\mathbf{r}\|_{\infty}\) and \(\operatorname{cond}_{\infty}(A)\) (b) Use your answer to part (a) to find an upper bound for the relative error in the solution. (c) Compute the exact solution \(\mathbf{x}\) and determine the relative error \(\frac{\left\|\mathbf{x}-\mathbf{x}^{\prime}\right\|_{\infty}}{\|\mathbf{x}\|_{\infty}}\)

Let \(A\) be an \(n \times n\) matrix with distinct real eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} .\) Let \(\lambda\) be a scalar that is not an eigenvalue of \(A\) and let \(B=(A-\lambda I)^{-1}\). Show that (a) the scalars \(\mu_{j}=1 /\left(\lambda_{j}-\lambda\right), j=1, \ldots, n\) are the eigenvalues of \(B\) (b) if \(\mathbf{x}_{j}\) is an eigenvector of \(B\) belonging to \(\mu_{j}\) then \(\mathbf{x}_{j}\) is an eigenvector of \(A\) belonging to \(\lambda_{j}\) (c) if the power method is applied to \(B\), then the sequence of vectors will converge to an eigenvector of \(A\) belonging to the eigenvalue that is closest to \(\lambda\). [The convergence will be rapid if \(\lambda\) is much closer to one \(\lambda_{i}\) than to any of the others. This method of computing eigenvectors by using powers of \((A-\lambda I)^{-1}\) is called the inverse power method.]

Let \(A=\mathbf{w y}^{T},\) where \(\mathbf{w} \in \mathbb{R}^{m}\) and \(\mathbf{y} \in \mathbb{R}^{n} .\) Show that (a) \(\frac{\|A \mathbf{x}\|_{2}}{\|\mathbf{x}\|_{2}} \leq\|\mathbf{y}\|_{2}\|\mathbf{w}\|_{2}\) for all \(\mathbf{x} \neq \mathbf{0}\) in \(\mathbb{R}^{n}\) (b) \(\|A\|_{2}=\|\mathbf{y}\|_{2}\|\mathbf{w}\|_{2}\)
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