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Chapter 4: Problem 343

If $$ \mathrm{A}=\left|\begin{array}{ll} -2 & 3 \\ -1 & 1 \end{array}\right| $$ then \(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\ldots \ldots \infty=\) (a) \(\left|\begin{array}{ll}0 & 3 \\ 1 & 3\end{array}\right|\) (b) \(\quad(1 / 3)\left|\begin{array}{cc}0 & -3 \\ -1 & 3\end{array}\right|\) (c) \((1 / 3)\left|\begin{array}{cc}0 & 3 \\ -1 & 3\end{array}\right|\) (d) undefined

Short Answer

Expert verified
The sum of the infinite geometric series I + A + A^2 + ... is: (d) undefined.

Step by step solution

01

Check if the series converges

To check for the convergence of the geometric series, we need to find the Spectral radius (\(\rho(A)\)) of the matrix A and check if it is smaller than 1. The spectral radius of a matrix is the largest absolute value of its eigenvalues. Let's find the eigenvalues of the matrix A.
02

Find the eigenvalues of matrix A

To find the eigenvalues of matrix A, we look for the values of λ that satisfy the following equation: \((A-\lambda I)=0\), where I is the identity matrix. \(det\begin{vmatrix}-2-\lambda & 3\\ -1 & 1-\lambda\end{vmatrix} = 0\) Expanding the determinant, we get: \((-2 - \lambda)(1 - \lambda) - (-1)(3) = 0\) \(\lambda^2 + \lambda - 2 - 3 + 3= 0\) \(\lambda^2 + \lambda - 2 = 0\) Now, we can solve the quadratic equation for λ: \(\lambda = \frac{-1 \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2}\) Solving for λ, we get two eigenvalues: \(\lambda_1 = 1\), \(\lambda_2 = -2\)
03

Check the spectral radius

Now, we have the eigenvalues, so we can determine the spectral radius as: \(\rho(A) = max\left| \lambda_1, \lambda_2 \right|\) \(\rho(A) = max\left| 1, -2 \right|\) \(\rho(A) = 2\) As the spectral radius \(\rho(A)\) is greater than 1, the series doesn't converge. Therefore, the sum of the series is undefined.
04

Answer

The sum of the infinite geometric series I + A + A^2 + ... is: (d) undefined

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Most popular questions from this chapter

If $$ A=\left|\begin{array}{ll} k & 3 \\ 3 & k \end{array}\right| $$ and \(\left|\mathrm{A}^{3}\right|=343\), then find the value of \(\mathrm{k}=\ldots\) \(\begin{array}{llll}\text { (a) } \pm 1 & \text { (b) } \pm 2 & \text { (c) } \pm 3 & \text { (d) } \pm 4\end{array}\)

If the equations \(x+3 y+z=0,2 x-y-z=0\) \(\mathrm{kx}+2 \mathrm{y}+3 \mathrm{z}=0\) have non-trivial solution then \(\mathrm{k}=\) (a) \((13 / 2)\) (b) \((9 / 2)\) (c) \(-(15 / 2)\) (d) \(-(13 / 2)\)

\(\left|\begin{array}{ccc}3 \mathrm{x}-8 & 3 & 3 \\ 3 & 3 \mathrm{x}-8 & 3 \\\ 3 & 3 & 3 \mathrm{x}-8\end{array}\right|=0\), then \(\mathrm{x}=\ldots\) (a) \((3 / 2),(3 / 11)\) (b) \((3 / 2),(11 / 3)\) (c) \((2 / 3),(11 / 3)\) (d) \((2 / 3),(3 / 11)\)

\(\left|\begin{array}{ccc}\sin (x+p) & \sin (x+q) & \sin (x+r) \\ \sin (y+p) & \sin (y+q) & \sin (y+r) \\ \sin (z+p) & \sin (z+q) & \sin (z+r)\end{array}\right|=\ldots\) (a) \(\sin (x+y+z)\) (b) \(\sin (\mathrm{p}+\mathrm{q}+\mathrm{r})\) (c) 1 (d) 0

Let \(\lambda\) and \(\alpha\) be real. The system of equations \(\lambda x+(\sin \alpha) y+(\cos \alpha) z=0\) \(x+(\cos \alpha) y+(\sin \alpha) z=0\) \(-x+(\sin \alpha) \mathrm{y}-(\cos \alpha) \mathrm{z}=0\) has no trivial solution. (i) The set of all values of \(\lambda\) is (a) \([-\sqrt{3}, \sqrt{3}]\) (b) \([-\sqrt{2}, \sqrt{2}]\) (c) \([-1,1]\) (d) \([0,(\pi / 2)]\) (ii) For \(\lambda=1, \alpha=\ldots .\) (a) \(\mathrm{n} \pi, \mathrm{n} \pi-(\pi / 4)\) (b) \(2 \mathrm{n} \pi, \mathrm{n} \pi-(\pi / 4)\) (c) \(\mathrm{n} \pi, \mathrm{n} \pi+(\pi / 4)\) (d) \(-\pi,-(3 \pi / 4)\)
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