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Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

Short Answer

Expert verified

The solutions are \({x_1} = 2\) and \({x_2} = - 5\).

Step by step solution

01

Write the matrix form

The given system is equivalent to\(Ax = b\).

Here, \(A = \left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right),{\rm{ }}x = \left( {\begin{aligned}{*{20}{c}}{{x_1}}\\{{x_2}}\end{aligned}} \right),\) and \(b = \left( {\begin{aligned}{*{20}{c}}{ - 9}\\{11}\end{aligned}} \right)\).

02

Write the inverse obtained in Exercise 3

From Exercise 3, \({\left( {\begin{aligned}{*{20}{c}}8&5\\{ - 7}&{ - 5}\end{aligned}} \right)^{ - 1}} = \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}5&5\\{ - 7}&{ - 8}\end{aligned}} \right)\).

03

Express the solution

\(\begin{aligned}{c}x = {A^{ - 1}}b\\ = \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}5&5\\{ - 7}&{ - 8}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 9}\\{11}\end{aligned}} \right)\\ = \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}{ - 40 + 55}\\{63 - 88}\end{aligned}} \right)\\ = \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}{10}\\{ - 25}\end{aligned}} \right)\\x = \left( {\begin{aligned}{*{20}{c}}2\\{ - 5}\end{aligned}} \right)\end{aligned}\)

Thus, \({x_1} = 2\) and \({x_2} = - 5\).

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Most popular questions from this chapter

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

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 \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.

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.

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