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Chapter 1: Q10Q (page 1)

Let \({{\bf{a}}_1}\) \({{\bf{a}}_2}\), and b be the vectors in \({\mathbb{R}^{\bf{2}}}\) shown in the figure, and let \(A = \left( {\begin{aligned}{*{20}{c}}{{{\bf{a}}_1}}&{{{\bf{a}}_2}}\end{aligned}} \right)\). Does the equation \(A{\bf{x}} = {\bf{b}}\) have a solution? If so, is the solution unique? Explain.

Short Answer

Expert verified

The system of equations \(A{\bf{x}} = {\bf{b}}\) has a unique solution.

Step by step solution

01

Construct the graph with a grid

Consider the figure shown below:

On the \({x_1}{x_2}\)-plane, the lines between \({{\bf{a}}_1}\) and the origin and between \({{\bf{a}}_2}\) and the origin form a grid. Each point can be defined by using the grid.

02

Determine the solution

In the above figure,move some steps in the direction of the vectors to reach towards vectors\({{\bf{a}}_1}\),\({{\bf{a}}_2}\), and bfrom the origin.

There is always a unique way to reach these vectors. It means,\(A{\bf{x}} = {\bf{b}}\)has a solution.

Thus, the system of equations \(A{\bf{x}} = {\bf{b}}\) has a unique solution.

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Most popular questions from this chapter

In Exercises 31, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

31. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\4&{ - 1}&3&{ - 6}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\0&7&{ - 1}&{ - 6}\end{array}} \right]\)

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

The following equation describes a Givens rotation in \({\mathbb{R}^3}\). Find \(a\) and \(b\).

\(\left( {\begin{aligned}{*{20}{c}}a&0&{ - b}\\0&1&0\\b&0&a\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{3}}\\{\bf{4}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{\bf{2}}\sqrt {\bf{5}} }\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)

Question:Let A be the n x n matrix with 0's on the main diagonal, and 1's everywhere else. For an arbitrary vector b鈫�in 鈻�n, solve the linear system Ax鈫�=b鈬赌, expressing the components x1,.......,xnof x鈫�in terms of the components of b鈬赌. See Exercise 69 for the case n=3 .

Consider a dynamical system x鈫�(t+1)=Ax鈫�(t) with two components. The accompanying sketch shows the initial state vector x鈫�0and two eigen vectors 蠀1鈫�鈥�鈥�and鈥�鈥�蠀2鈫� of A (with eigen values 位1鈫�and位2鈫� respectively). For the given values of 位1鈫�and位2鈫�, draw a rough trajectory. Consider the future and the past of the system.

位1鈫�=1,位2鈫�=0.9
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