Linear algebra and its applications 5th edition answers

Answers to even exercises

Section 1.1

2. The solution is (x1, x2) = (12, –7), or simply (12, –7).

4. The point of intersection is (x1, x2) = (9/4, 1/4).

8. The solution set contains one solution: (0, 0, 0).

10. The solution set contains one solution: (–3, –5, 6, –3).

12. The solution set is empty.

14. The solution is (2, –1, 1).

16. The system is now in triangular form and has a solution.

18. The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point

in common.

Section 1.2

2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c.

4.

8.

10

12.

14.

16.

a. The system is consistent, with a unique solution.

b. The system is consistent. There are many solutions because x2 is a free variable.

Section 1.3

2.

,

See our solution for Question 3E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.

Step 1
Given equations of lines:
\[\begin{array}{l}{x_1} + 5{x_2} = 7\\\\{x_1} - 2{x_2} = - 2\end{array}\]To find the point of intersection, we have to Solve the system of equations given above by using elementary row operations on the equations or on the augmented matrix. We will prefer the augmented matrix method.

Step 2
Convert the given system of Equations into augmented matrix form . \[\begin{array}{*{20}{l}}{\left[ {\begin{array}{*{20}{c}}1&5\\1&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}7\\{ - 2}\end{array}} \right]}\\{}\\{\left[ {\begin{array}{*{20}{c}}1&5&7\\1&{ - 2}&{ - 2}\end{array}} \right]}\end{array}\]

Step 3: $R_2 = R_2 - R_1$
\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&5&7\\1&{ - 2}&{ - 2}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&5&7\\{1 - 1}&{ - 2 - 5}&{ - 2 - 7}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&5&7\\0&{ - 7}&{ - 9}\end{array}} \right]\end{array}\]

Step 4: ${R_2} = -{R_2}/7 $
\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&5&7\\0&{ - 7/ - 7}&{ - 9/ - 7}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&5&7\\0&1&{9/7}\end{array}} \right]\end{array}\]

Step 5: Write the reduced form in system of equations
\[\begin{array}{l}{x_1} + 5{x_2} = 7\\\\{x_2} = \dfrac{9}{7}\end{array}\]

Step 6
From the last row, we have\[{x_2} = \dfrac{9}{7}\]Now put the value of $x_2$ in first row
\[\begin{array}{l}{x_1} + 5\left( {\dfrac{9}{7}} \right) = 7\\\\{x_1} = 7 - 5\left( {\dfrac{9}{7}} \right)\\\\{x_1} = \dfrac{4}{7}\end{array}\]

ANSWERS

\[\left[ {\begin{array}{*{20}{l}}{{x_1} = \dfrac{4}{7}}\\{{x_2} = \dfrac{9}{7}}\end{array}} \right]\]

See our solution for Question 3E from Chapter 1.9 from Lay's Linear Algebra and Its Applications, 5th Edition.

Step 1
Let the Vectors are:\[{{\bf{e}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}0\\1\end{array}} \right];\,\,\,{{\bf{e}}_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]\]Given Transformation (see Figure 1 in the section 9.1 of textbook):\[T:{{\rm{R}}^2} \to {{\rm{R}}^2}::T\left( {{{\bf{e}}_1}} \right) = \left[ {\begin{array}{*{20}{c}}{\cos \phi }\\{\sin \phi }\end{array}} \right],\,\,:T\left( {{{\bf{e}}_2}} \right) = \left[ {\begin{array}{*{20}{c}}{ - \sin \phi }\\{\cos \phi }\end{array}} \right]\,\]We have to find the transformation matrix for the above transformation.

Step 2: The Rotation Angle
\[\begin{array}{l}\phi = \dfrac{{3\pi }}{2}\\\\\phi = \dfrac{{3\pi }}{2}{\kern 1pt} \dfrac{{180^\circ }}{\pi }\\\\\phi = 270^\circ \end{array}\]

Step 3: The Transformation Matrix
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}{\cos \phi }&{ - \sin \phi }\\{\sin \phi }&{\cos \phi }\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}{\cos 270^\circ }&{ - \sin 270^\circ }\\{\sin 270^\circ }&{\cos 270^\circ }\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\end{array}\]

ANSWER

\[A = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\]