## Section 1.1: Linear Systems

1. Which of the following are linear equations? If an equation is not a linear equation, tell why.
${\bf (a)} \quad 6\,x + x\,y + 3\, z = 2; \qquad {\bf (b)} \quad 5\,a -\pi + 3\,b =0;$
${\bf (c)} \quad \cos \left( \frac{\pi}{2} \right) x^2 -\sin (\pi )\, x\,y + 2\, z = 6; \qquad {\bf (d)} \quad \sqrt{x} -3\,y -z =1.$
2. Solve each system of equations.
${\bf (a)} \quad \begin{cases} 2\,x + 3\,y &= 2, \\ -4\,x-y &= 1; \end{cases} \qquad {\bf (b)} \quad \begin{cases} 2\,x + 4\,y &= -4, \\ 3\,x+ 5\,y &= 1; \end{cases}$
-1/2, 1
m = {{2, 4}, {3, 5}}; b = {-4, 1};
LinearSolve[m, b]
{12, -7}
3. Find the point of intersection of the lines x −4y = −2 and 3x + 2y = 8.
m = {{1, -4}, {3, 2}}; b = {-2, 8};
LinearSolve[m, b]
{2, 1}
4. Solve systems of equations:
${\bf (a)} \quad \begin{split} \phantom{x} - 3\,y +4\,z&= \phantom{-}28, \\ 2\,x-8\,y + 6\,z &= -42 , \\ -5\,x +5\,y -2\,z &= \phantom{-}56; \end{split} \qquad {\bf (b)} \quad \begin{split} 2\,x - 2\,y + z&= 145, \\ 3\,x -y + 10\,z &= 203, \\ x+ 2\,y + 6\,z &= \phantom{1}58; \end{split}$
m = {{0, -4, 3}, {2, -8, 6}, {-5, 5, -2}}; b = {28, -42, 56};
LinearSolve[m, b]
{-49, -73, -88}
m = {{2, -2, 1}, {3, -1, 10}, {1, 2, 6}}; b = {145, 203, 58};
LinearSolve[m, b]
{70, -3, -1}
${\bf (c)} \quad \begin{split} 2\,x - 2\,y + z &= 75, \\ \phantom{x+} 2y + 6\,z &= 30 , \\ x + 2y +2z &= 0; \end{split} \qquad {\bf (d)} \quad \begin{split} 4\,x + 5\,y -z &= 310, \\ 2x+ 3\,y + 2z &= 0, \\ 8x+ 5\,y + 4z &= 186 . \end{split}$
m = {{2, -2, 1}, {0, 2, 6}, {1, 2, 2}}; b = {75, 30, 0};
LinearSolve[m, b]
{14, -18, 11}
m = {{4, 5, -1}, {2, 3, 2}, {8, 5, 4}}; b = {310, 0, 186};
LinearSolve[m, b]
{49, 10, -64}
5. In the study of heat transfer, it is important to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known (usually from an experiment or other measurment). Let us consider the plate shown in Figure below representing a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let t₁, t₂, t₃, t₄ denote the temperatures at the four interior nodes of the mesh in Figure. The temperature at a particular node is approximately equal to the average of the four nearest nodes---to the left, above, to the right, and below.

Write a system of four equations whose solution gives estimates for the temperatures t₁, t₂, t₃, t₄.

## Section 4: Gaussian Elimination

1. Add -2 times the first row to the second row.
2. Exchange row two and row three.
3. Add -3 times the third row to the first row.
1. The system of equations can be represented as the following matrix.
$\left[ \begin{array}{cc|c} 1 & 2 & -3\\ 2 & 3 & 4 \end{array} \right] .$
Multiply the first line by -2 and add it to the second line.
$-2*\left( 1\ ~2\ \vert\ -3\right) = \left(-2\ -4\ \vert\ 6\right) .$
Upon adding to the last row, we get
$\left[ \begin{array}{cc|c} 1 & 2 & -3\\ 0 & -1 & 10 \end{array} \right] .$
From the reduced matrix, the value of x2 appears to be
$x_{2}=-10 .$
Substitution can be used to solve the first equation.
\begin{eqnarray*} x_{1}+2\,x_{2}&=& -3 , \\ x_{1}+2*(-10)&=& -3 , \\ x_{1}&=& -3+20=17 \end{eqnarray*}
Final solution: x1 = 17, x2 = -10.
2. The system of equations can be represented as the following matrix.
$\left[ \begin{array}{cc|c} 3 & 1 & 6\\ 3 & 3 & 2 \end{array} \right] .$
Multiply the first line by -1 and add it to the second line, which leads to
$-1*\left( 3\ ~1\ \vert \ 6\right) = \left( -3\ -1\ \vert \ -6\right) .$
So
$\left[ \begin{array}{cc|c} 3 & 1 & 6\\ 0 & 2 & -4 \end{array} \right] .$
From the reduced matrix, the value of x2 appears to be x2 = -2. Substitution can be used to solve the first equation.
\begin{eqnarray*} 3x_{1}+x_{2}&=&6 , \\ 3x_{1}+(-2)&=&6 , \\ 3x_{1}&=&8 , \\ x_{1}&=&8/3 \end{eqnarray*}
Final solution: x1 = 8/3 and x2 = -2.
1. From the reduced matrix, the system of equations can be written as
$\begin{split} x_{1}-x_{2}+2x_{3}-1x_{4}&=5 , \\ 5x_{2}-x_{3}+3x_{4}&=4, \\ 3x_{3}-x_{4}&=5, \\ x_{4}&=3. \end{split}$
Back substitution yields
Final solution: x1=2, x2=-1, x3=3, x4=3.
2. The matrix indicates that 0x2=3, which is not possible. Therefore, the system is inconsistent.
1. The system of equations can be represented by the following matrix.
$\left[ \begin{array}{ccc|c} 3 & -2 & -1 & -1\\ 2 & -1 & 2 & 4\\ 0 & 1 & -2 & -6\\ \end{array} \right] .$
Multiply the first line by -⅔ and add it to the second row.
$-\frac{2}{3}*\left( 3\ -2\ -1 \ \vert \ -1\right) = \left( -2\ \frac{4}{3}\ \frac{2}{3}\ \big\vert \ \frac{2}{3} \right) .$
Then
$\left[ \begin{array}{ccc|c} 3 & -2 & -1 & -1\\ 0 & \frac{1}{3} & \frac{8}{3} & \frac{14}{3}\\ 0 & 1 & -2 & -6\\ \end{array} \right] .$
Multiply the second line by 3 to convert the fractions to whole numbers.
$\left[ \begin{array}{ccc|c} 3 & -2 & -1 & -1\\ 0 & 1 & 8 & 14\\ 0 & 1 & -2 & -6 \end{array} \right] .$
Multiply the second line by -1 and add it to the third line.
$-1*\left( 0\ -1\ -8 \ \big\vert \ 14 \right) = \left( 0\ 1\ 8\ \big\vert \ -14 \right) .$
Then
$\left[ \begin{array}{ccc|c} 3 & -2 & -1 & -1\\ 0 & 1 & 8 & 14\\ 0 & 0 & -10 & -20\\ \end{array} \right] .$
From the reduced matrix the value of x2 appears.
$-10x_{3}=-20 \qquad \Longrightarrow \qquad x_{3}=2$
Back substitution can be used to solve the system of equations.
$\begin{split} x_{2}+8x_{3}&=14, \\ x_{2}+8(2)&=14, \\ x_{2}&=14-8(2)=14-16, \\ x_{2}&=-2 \end{split}$
$\begin{split} 3x_{1}+-2x_{2}-1x_{3}&=-1, \\ 3x_{1}+-2(-2)-1(2)&=-1, \\ 3x_{1}&=-1+2(-2)+1(2)=-1+-4+2=-3, \\ 3x_{1}&=-3, \\ x_{1}&=-1 \end{split}$
Final solution: x1 = -1, x2 = -2, x3 = 2 .
2. The system of equations can be represented by the following matrix.
$\left[ \begin{array}{ccc|c} 0 & 2 & 1 & 3\\ 2 & 1 & -2 & -3\\ 4 & 4 & -3 & -4\\ \end{array} \right] .$
Multiply the second row by -2 and add it to the third row.
$-2*\left( 2 \ ~1\ -2\ |\ -3\right) = \left( -4\ -2\ 4\ |\ 6\right) .$
$\left[ \begin{array}{ccc|c} 0 & 2 & 1 & 3\\ 2 & 1 & -2 & -3\\ 0 & 2 & 1 & 2\\ \end{array} \right] .$
Multiply the first row by -1 and add it to the third row.
$\left[ \begin{array}{ccc|c} 0 & 2 & 1 & 3\\ 2 & 1 & -2 & -3\\ 0 & 0 & 0 & -1 \end{array} \right] .$
The matrix indicates that 0x3 = -1, which is not possible. Therefore, the system is inconsistent.
1. Multiplying the first row by 2 and adding it to the second row transforms A to B.
${\bf A} = \begin{bmatrix} 2 & f & 3\\ -4 & 8 & -6\\ \end{bmatrix} , \qquad {\bf B} = \begin{bmatrix} 2 & f & 3\\ 0 & -8+2f & 0 \end{bmatrix}$
Therefore, matrix A is consistent for all values of f.
2. Multiplying the first row by -2 and adding it to the second row transforms matrix A to B.
${\bf A} = \begin{bmatrix} 1 & f & -3\\ 2 & 4 & -8 \end{bmatrix}, \qquad {\bf B} = \begin{bmatrix} 1 & f & -3\\ 0 & 4-2f & -2 \end{bmatrix}$
Therefore, the matrix is consistent if f≠2.
3. Multiplying the first row by 2 and adding it to the second row transforms A to B.
${\bf A} = \begin{bmatrix} 2 & -1 & f\\ -4 & 2 & 6\\ \end{bmatrix} , \qquad {\bf B} = \begin{bmatrix} 2 & -1 & f\\ 0 & 0 & 6+2f\\ \end{bmatrix} .$
Therefore, the matrix is inconsistent for all values of f ≠ -3. When f = -3, the system has infinite many solutions.

## Section 5: Reduced Row-Echelon Form

1. $A= \begin{bmatrix} 4 & 4 & 1 & 24\\ 2 & -4 & 1 & 0\\ 5 & -4 & -5 & 12\\ \end{bmatrix}$ Multiply row 1 by 1/4. Add -2 times row 1 to row 2. $A= \begin{bmatrix} 1 & 1 & \frac{1}{4} & 6\\ 0 & -6 & \frac{1}{2} & -12\\ 5 & -4 & -5 & 12\\ \end{bmatrix}$ Add -5 times row 1 to row 3. Multiply row 2 by -1/6. $A= \begin{bmatrix} 1 & 1 & \frac{1}{4} & 6\\ 0 & 1 & -\frac{1}{12} & 2\\ 0 & -9 & -\frac{25}{4} & -18\\ \end{bmatrix}$ Add 9 times row 2 to row 3. Multiply row 3 by -1/7. $A= \begin{bmatrix} 1 & 1 & \frac{1}{4} & 6\\ 0 & 1 & -\frac{1}{12} & 2\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Add 1/12 times row 3 to row 2. $A= \begin{bmatrix} 1 & 1 & \frac{1}{4} & 6\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Add -1/4 times row 3 to row 1. $A= \begin{bmatrix} 1 & 1 & 0 & 6\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Add -1 times row 2 to row 1. $A= \begin{bmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$
2. $A= \begin{bmatrix} 2 & 3 & -1 & 24\\ 1 & 1 & 4 & 13\\ 3 & -2 & 1 & -7\\ \end{bmatrix}$ Divide row 1 by 2. Add -1 times row 1 to row 2. Add -3 times row 1 to row 3. Multiply row 2 by -2. $A= \begin{bmatrix} 1 & \frac{3}{2} & \frac{-1}{2} & 12\\ 0 & 1 & -9 & -2\\ 0 & \frac{-13}{2} & \frac{5}{2} & -43\\ \end{bmatrix}$ Add 13/2 times row 2 tor row 3. Multiply row 3 by -1/56. $A= \begin{bmatrix} 1 & \frac{3}{2} & -\frac{1}{2} & 12\\ 0 & 1 & 0 & 7\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$ Add 9 times row 3 to row 2. Add 1/2 times row 3 to row 1. $A= \begin{bmatrix} 1 & \frac{3}{2} & 0 & \frac{25}{2}\\ 0 & 1 & 0 & 7\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$ Add -3/2 times row 2 to row 1. $A= \begin{bmatrix} 1 & 0 & 0 & 2\\ 0 & 1 & 0 & 7\\ 0 & 0 & 1 & 1\\ \end{bmatrix}$
1. $A= \begin{bmatrix} -1 & -5 & -5 & 22\\ 4 & -5 & 4 & 19\\ 1 & 5 & -1 & -10\\ \end{bmatrix}$ Multiply row 1 by -1. Add -4 times row 1 to row 2. Add -1 times row 1 to row 3. $A= \begin{bmatrix} 1 & 5 & 5 & -22\\ 0 & -25 & -16 & 107\\ 0 & 0 & -6 & 12\\ \end{bmatrix}$ Multiply row 22 by -1/25. Multiply row 3 by -1/6. Add -16/25 times row 3 to row 2. $A= \begin{bmatrix} 1 & 5 & 5 & -22\\ 0 & 1 & 0 & -3\\ 0 & 0 & 1 & -2 \end{bmatrix}$ Add -5 times row 3 to row 1. Add -5 times row 2 to row 1. $A= \begin{bmatrix} 1 & 0 & 0 & 3\\ 0 & 1 & 0 & -3\\ 0 & 0 & 1 & -2 \end{bmatrix}$ Then, x1 = 3, x2 = -3, and x3 = -2.
2. $A= \begin{bmatrix} 1 & 2 & 2 & 2\\ 1 & 1 & 1 & 3\\ 1 & 0 & 2 & 4\\ \end{bmatrix}$ Add -1 times row 1 to row 2. Add -1 tomes row 1 to row 3. Multiply row 2 by -1. $A= \begin{bmatrix} 1 & 2 & 2 & 2\\ 0 & 1 & 1 & -1\\ 0 & -2 & 0 & 2 \end{bmatrix}$ Add 2 times row 2 to row 3. Multiply row 3 by 1/2. $A= \begin{bmatrix} 1 & 2 & 2 & 2\\ 0 & 1 & 1 & -1\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Add -1 times row 3 to row 2. Add -2 times row 3 to row 1. $A= \begin{bmatrix} 1 & 2 & 0 & 2\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Add -2 times row 2 to row 1. $A= \begin{bmatrix} 1 & 0 & 0 & 4\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$ Then, x1 = 4, x2 = -1, and x3 is undefined.

## Section 11: Matrix Transformations

1. $\mbox{(a)}\quad \begin{bmatrix} 3 & -4 \\ 8 & 15 \\ \end{bmatrix}, \qquad \mbox{(b)}\quad \begin{bmatrix} 1 & -3 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -1 \\ -1 & 4 & -3 \\ \end{bmatrix} , \qquad \mbox{(c)}\quad \begin{bmatrix} 4 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & -8 \\ \end{bmatrix} ,$ $\mbox{(d)}\quad \begin{bmatrix} 8 & 4 & -3 \\ 10 & -9 & 12 \\ -2 & 0 & -1 \\ \end{bmatrix}, \qquad \mbox{(e)}\quad \begin{bmatrix} 2 & 0 & 2 & -2 \\ 0 & 0 & 0 & -1 \\ 2 & 0 & -1 & 0 \\ 0 & -1 & 3 & 0 \\ \end{bmatrix}.$
2. (a) T(A + B) ≠ T(A) + T(B);   (b) T(A + B) = T(A) + T(B).
3. (a) T(cA) = cT(A), (b) T(cA) ≠ cT(A)
4. Since the transformation satisfies both axioms, the transformation is linear.
5. $\mbox{(a)}\quad S \circ T = \begin{bmatrix} 3 & -2 & 1 \\ 5 & 2 & -7 \\ 1 & 7 & -4 \\ \end{bmatrix} \begin{bmatrix} 1 & -4 \\ 6 & 3 \\ -2 & 4 \\ \end{bmatrix} = \begin{bmatrix} -11 & -14 \\ 31 & -42 \\ 35 & 33 \\ \end{bmatrix};$ $\mbox{(b)}\quad S \circ T = \begin{bmatrix} 2 & 1 & 4 \\ 0 & 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 6 & 3 & -1 & 0 \\ 1 & 1 & 0 & 4 \\ -2 & 5 & 0 & 2 \\ \end{bmatrix} = \begin{bmatrix} 5 & 27 & -2 & 12 \\ -1 & 6 & 0 & 6 \\ \end{bmatrix},$ $\mbox{(c)}\quad S \circ T = \begin{bmatrix} 1 & -2 & 3 \\ 3 & -2 & 1 \\ 2 & 1 & -3 \\ \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 3 \\ 2 & 4 \\ \end{bmatrix} = \begin{bmatrix} -1 & 8 \\ -3 & 4 \\ 0 & -5 \\ \end{bmatrix}.$
6. $$\begin{bmatrix} 1&3&-5 \\ 2&-4&6 \end{bmatrix} .$$

## Section 12: Plane Transformations

1. Let T1 be a reflection through the horizontal x1-axis. Then it has the following matrix form: $$\begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix} .$$ The reflection with respect to the line x1 = x2 has matrix form $$\begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} .$$ Their composition gives the required matrix $$T = \begin{bmatrix} 0&-1 \\ 1&0 \end{bmatrix} .$$
2. Let T1 be a rotation by -3&ri;/4 radian. With unit vectors i = (1,0) and j = (0,1), we have T1(i) = -(1,1)/2½ and T1(j) = (1,-1)/2½. Therefore, $$T_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} -1&1 \\ -1&-1 \end{bmatrix} .$$ For reflection T2, we have T2(i) = -i and T2(j) =j, so this matrix becomes $$T_2 = \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} .$$ Their composition gives the required matrix: $$T = T_2 T_1 = \begin{bmatrix} -1&0 \\ 0&1 \end{bmatrix} \, \frac{1}{\sqrt{2}} \begin{bmatrix} -1&1 \\ -1&-1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1&-1 \\ -1&-1 \end{bmatrix} .$$
3. $$T = \begin{bmatrix} 1&3 \\ 0&2 \end{bmatrix} .$$
4. Because T(i) = -j and T(j) = i, the matrix is $$T = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix} .$$
5. $$\frac{1}{\sqrt{2}} \begin{bmatrix} 1&1 \\ -1&1 \end{bmatrix} .$$

## Review problems

Gaussian Elimination

1. a ≠ 0, a ≠ 1.
2. a ≠ 0, a ≠ 1, and a ≠ 3.

Matrix transformations:
1. $\mbox{(a)} \quad \begin{bmatrix} 1 & 4 & -5 \\ -3 & 3 & -2 \\ 1 & -2 & 6 \end{bmatrix}; \qquad \mbox{(b)} \quad \begin{bmatrix} 1 & -1 & 1 \\ 3 & 1 & -1 \\ 2 & 2 & 1 \\ \end{bmatrix};$ $\mbox{(c)} \quad \begin{bmatrix} 1 & 3 & 2 & 0 & 1 & 0 \\ -1 & -1 & -1 & 1 & 0 & 1 \\ 0 & 4 & 2 & 4 & 3 & 3 \\ 1 & 3 & 2 & -2 & 0 & 0 \\ \end{bmatrix}; \qquad \mbox{(d)} \quad \begin{bmatrix} 1 & 3 & 2 & 0 & 1 & 0 \\ -1 & -1 & -1 & 1 & 0 & 1 \\ 0 & 4 & 2 & 4 & 3 & 3 \\ 1 & 3 & 2 & -2 & 0 & 0 \\ \end{bmatrix} .$
2. (a) Since the transformation satisfies both axioms, the transformation is linear;
(b) The transformation must satisfy both axioms to be linear. Since the transformation does not satisfy either axioms, the transformation is non-linear.
(c) As the transformation satisfies both theorems, the transformation is linear.
3. $\mbox{(a)} \quad \begin{bmatrix} 40 & 74 \\ 5 & 35 \\ 23 & 49 \\ \end{bmatrix}; \qquad \mbox{(b)} \quad \begin{bmatrix} 7 & 6 \\ 2 & 30 \\ 13 & 7 \\ \end{bmatrix}.$