Homework 1

1. Solve the following systems of linear equations by Gaussian elimination:

(a) {2y−8z=8x−2y+z=0−4x+5y+9z=−9\left\{ \begin{array}{l} 2y -8z = 8 \\ x -2y + z = 0 \\ -4x + 5y + 9z = -9 \end{array}\right.

{2y−8z=8x−2y+z=0−4x+5y+9z=−9  ⟺  [02−881−210−459−9]→R2+R112R1[01−4410−78−459−9]→R3+4R2[01−4410−7805−1923]→R1↔R2[10−7801−4405−1923]→R3−5R2[10−7801−440013]→R2+4R3R1+7R3[10029010160013]∴x=29,y=16,z=3\begin{array}{ccl} \left\{ \begin{array}{l} 2y -8z = 8 \\ x -2y + z = 0 \\ -4x + 5y + 9z = -9 \end{array} \right. &\iff &\left[ \begin{array}{ccc|c} 0 & 2 & -8 & 8 \\ 1 & -2 & 1 & 0 \\ -4 & 5 & 9 & -9 \end{array} \right] \\ &\xrightarrow[R_2 + R_1]{\frac{1}{2}R_1} &\left[ \begin{array}{ccc|c} 0 & 1 & -4 & 4 \\ 1 & 0 & -7 & 8 \\ -4 & 5 & 9 & -9 \end{array} \right] \\ &\xrightarrow{R_3 + 4R_2} &\left[ \begin{array}{ccc|c} 0 & 1 & -4 & 4 \\ 1 & 0 & -7 & 8 \\ 0 & 5 & -19 & 23 \end{array} \right] \\ &\xrightarrow{R_1 \leftrightarrow R_2} &\left[ \begin{array}{ccc|c} 1 & 0 & -7 & 8 \\ 0 & 1 & -4 & 4 \\ 0 & 5 & -19 & 23 \end{array} \right] \\ &\xrightarrow{R_3 - 5R_2} &\left[ \begin{array}{ccc|c} 1 & 0 & -7 & 8 \\ 0 & 1 & -4 & 4 \\ 0 & 0 & 1 & 3 \end{array} \right] \\ &\xrightarrow[R_2 + 4R_3]{R_1 + 7R_3} &\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 29 \\ 0 & 1 & 0 & 16 \\ 0 & 0 & 1 & 3 \end{array} \right] \end{array} \\[2em] \therefore x = 29, y=16, z=3

(b) {x1−2x3=−1x2−x4=2−3x2+2x3=0−4x1+7x4=−5\left\{ \begin{array}{l} x_1 - 2x_3 = -1 \\ x_2 - x_4 = 2 \\ -3x_2 + 2x_3 = 0 \\ -4x_1 + 7x_4 = -5 \end{array} \right.

{x1−2x3=−1x2−x4=2−3x2+2x3=0−4x1+7x4=−5  ⟺  [10−20−1010−120−3200−4007−5]→R3+3R2R4+4R1[10−20−1010−12002−3600−87−9]→R4+4R3[10−20−1010−12002−36000−515]→−15R412R3[10−20−1010−12001−3230001−3]→R2+R3R3+32R4[10−20−10100−10010−320001−3]→R1+2R3[1000−40100−10010−320001−3]∴x1=−4,x2=−1,x3=−32,x4=−3\begin{array}{ccl} \left\{ \begin{array}{l} x_1 - 2x_3 = -1 \\ x_2 - x_4 = 2 \\ -3x_2 + 2x_3 = 0 \\ -4x_1 + 7x_4 = -5 \end{array} \right. &\iff &\left[ \begin{array}{cccc|c} 1 & 0 & -2 & 0 & -1 \\ 0 & 1 & 0 & -1 & 2 \\ 0 & -3 & 2 & 0 & 0 \\ -4 & 0 & 0 & 7 & -5 \end{array} \right] \\ &\xrightarrow[R_3 + 3R_2]{R_4 + 4R_1} &\left[ \begin{array}{cccc|c} 1 & 0 & -2 & 0 & -1 \\ 0 & 1 & 0 & -1 & 2 \\ 0 & 0 & 2 & -3 & 6 \\ 0 & 0 & -8 & 7 & -9 \end{array} \right] \\ &\xrightarrow{R_4 + 4R_3} &\left[ \begin{array}{cccc|c} 1 & 0 & -2 & 0 & -1 \\ 0 & 1 & 0 & -1 & 2 \\ 0 & 0 & 2 & -3 & 6 \\ 0 & 0 & 0 & -5 & 15 \end{array} \right] \\ &\xrightarrow[-\frac{1}{5}R_4]{\frac{1}{2}R_3} &\left[ \begin{array}{cccc|c} 1 & 0 & -2 & 0 & -1 \\ 0 & 1 & 0 & -1 & 2 \\ 0 & 0 & 1 & -\frac{3}{2} & 3 \\ 0 & 0 & 0 & 1 & -3 \end{array} \right] \\ &\xrightarrow[R_2 + R_3]{R_3 + \frac{3}{2}R_4} &\left[ \begin{array}{cccc|c} 1 & 0 & -2 & 0 & -1 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & -\frac{3}{2} \\ 0 & 0 & 0 & 1 & -3 \end{array} \right] \\ &\xrightarrow{R_1 + 2R_3} &\left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & -4 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 & -\frac{3}{2} \\ 0 & 0 & 0 & 1 & -3 \end{array} \right] \end{array} \\[2em] \therefore x_1 = -4, x_2 = -1, x_3 = -\frac{3}{2}, x_4 = -3

2. The sum of any two of three real numbers are 2424, 2828, 3030. Find these three numbers.

Let x,y,z∈Rx,y,z \in \R. Then:
{x+y=24y+z=28z+x=30  ⟺  [110240112810130]→R1−R2R3−R1[10−1−4011280−116]→R3+R2[10−1−40112800234]→12R3[10−1−40112800117]→R1+R3R2−R3[100130101100117]∴x=13,y=11,z=17\begin{array}{c} \left\{ \begin{array}{l} x+y = 24 \\ y+z = 28 \\ z+x = 30 \end{array} \right. &\iff &\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 24 \\ 0 & 1 & 1 & 28 \\ 1 & 0 & 1 & 30 \end{array} \right] \\ &\xrightarrow[R_1 - R_2]{R_3 - R_1} &\left[ \begin{array}{ccc|c} 1 & 0 & -1 & -4 \\ 0 & 1 & 1 & 28 \\ 0 & -1 & 1 & 6 \end{array} \right] \\ &\xrightarrow{R_3 + R_2} &\left[ \begin{array}{ccc|c} 1 & 0 & -1 & -4 \\ 0 & 1 & 1 & 28 \\ 0 & 0 & 2 & 34 \end{array} \right] \\ &\xrightarrow{\frac{1}{2}R_3} &\left[ \begin{array}{ccc|c} 1 & 0 & -1 & -4 \\ 0 & 1 & 1 & 28 \\ 0 & 0 & 1 & 17 \end{array} \right] \\ &\xrightarrow[R_1 + R_3]{R_2 - R_3} &\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 13 \\ 0 & 1 & 0 & 11 \\ 0 & 0 & 1 & 17 \end{array} \right] \end{array} \\[2em] \therefore x = 13, y = 11, z = 17

As such, the three real numbers are 1313, 1111, and 1717.

3. Find the polynomial of degree 2 f(t)=a+bt+ct2f(t) = a+bt+ct^2 whose graph passes through (1,−1)(1, -1), (2,3)(2, 3) and (3,13)(3, 13).

{a+b(1)+c(1)2=−1a+b(2)+c(2)2=3a+b(3)+c(3)2=13  ⟺  {a+b+c=−1a+2b+4c=3a+3b+9c=13  ⟺  [111−1124313913]→R3−R1R2−R1[111−1013402814]→R3−2R2[111−101340026]→12R3[111−101340013]→R2−3R3R1−R3[110−4010−50013]→R1−R2[1001010−50013]\begin{array}{c} &\left\{ \begin{array}{ll} a + b(1) + c(1)^2 &= -1 \\ a + b(2) + c(2)^2 &= 3 \\ a + b(3) + c(3)^2 &= 13 \end{array} \right. &\iff &\left\{ \begin{array}{ll} a + b + c &= -1 \\ a + 2b + 4c &= 3 \\ a + 3b + 9c &= 13 \end{array} \right. \\ \iff &\left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 1 & 2 & 4 & 3 \\ 1 & 3 & 9 & 13 \end{array} \right] &\xrightarrow[R_3 - R_1]{R_2 - R_1} &\left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 2 & 8 & 14 \end{array} \right] \\ &&\xrightarrow{R_3 - 2R_2} &\left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 2 & 6 \end{array} \right] \\ &&\xrightarrow{\frac{1}{2}R_3} &\left[ \begin{array}{ccc|c} 1 & 1 & 1 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \end{array} \right] \\ &&\xrightarrow[R_2 - 3R_3]{R_1 - R_3} &\left[ \begin{array}{ccc|c} 1 & 1 & 0 & -4 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{array} \right] \\ &&\xrightarrow{R_1 - R_2} &\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \end{array} \right] \end{array}

As such, the function
f(t)=1−5t+3t2f(t) = 1 -5t + 3t^2

is a polynomial of degree two that passes through the points (1,−1)(1, -1), (2,3)(2, 3), and (3,13)(3, 13).

4. Use some online program, write down the echelon form of the following system and solve the system as well. {x−2y+3z−4w+5v=−12x+3y+4z+5w−6v=22x−2y+3z−3w+6v=0x+y−z−w+3v=23x+4y+5z−6w−4v=0\left\{\begin{array}{l}x-2y+3z-4w+5v=-1\\2x+3y+4z+5w-6v=2\\2x-2y+3z-3w+6v=0\\x+y-z-w+3v=2\\3x+4y+5z-6w-4v=0\end{array}\right.

 
{x−2y+3z−4w+5v=−12x+3y+4z+5w−6v=22x−2y+3z−3w+6v=0x+y−z−w+3v=23x+4y+5z−6w−4v=0  ⟺  [1−23−45−12345−622−23−36011−1−132345−6−40]→careful calculations[100001716640100054366400100−5166400010229664000013383]∴x=171664,y=543664,z=−51664,w=229664,v=3383\begin{array}{rl} \left\{ \begin{array}{l} x-2y+3z-4w+5v=-1\\ 2x+3y+4z+5w-6v=2\\ 2x-2y+3z-3w+6v=0\\ x+y-z-w+3v=2\\ 3x+4y+5z-6w-4v=0 \end{array} \right. \iff &\left[ \begin{array}{ccccc|c} 1 & -2 & 3 & -4 & 5 & -1\\ 2 & 3 & 4 & 5 & -6 & 2\\ 2 & -2 & 3 & -3 & 6 & 0\\ 1 & 1 & -1 & -1 & 3 & 2\\ 3 & 4 & 5 & -6 & -4 & 0 \end{array} \right] \\[3em] \xrightarrow{\text{careful calculations}} &\left[ \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0 & \frac{171}{664} \\[.5em] 0 & 1 & 0 & 0 & 0 & \frac{543}{664} \\[.5em] 0 & 0 & 1 & 0 & 0 & -\frac{51}{664} \\[.5em] 0 & 0 & 0 & 1 & 0 & \frac{229}{664} \\[.5em] 0 & 0 & 0 & 0 & 1 & \frac{33}{83} \\[.5em] \end{array} \right] \end{array} \\[2em] \therefore x = \frac{171}{664}, y = \frac{543}{664}, z = -\frac{51}{664}, w = \frac{229}{664}, v = \frac{33}{83}

5. Find the following products. Explain why if it is undefined.

(a) [0132][2−3]\begin{bmatrix} 0 & 1 \\ 3 & 2\end{bmatrix}\begin{bmatrix} 2 \\ -3\end{bmatrix}

[0132][2−3]=[0(2)+1(−3)3(2)+2(−3)]=[−30]\begin{bmatrix} 0 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ -3 \end{bmatrix} = \begin{bmatrix} 0(2) + 1(-3) \\ 3(2) + 2(-3) \end{bmatrix} = \begin{bmatrix} -3 \\ 0 \end{bmatrix}

(b) [123456][78]\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\end{bmatrix}\begin{bmatrix} 7 \\ 8\end{bmatrix}

[123456][78]\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 \\ 8 \end{bmatrix} is undefined because the number of columns in the first matrix does not match the number of rows in the second.

(c) [013256][21]\begin{bmatrix} 0 & 1 \\ 3 & 2 \\ 5 & 6\end{bmatrix}\begin{bmatrix} 2 \\ 1\end{bmatrix}

[013256][21]=[0(2)+1(1)3(2)+2(1)5(2)+6(1)]=[1816]\begin{bmatrix} 0 & 1 \\ 3 & 2 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0(2) + 1(1) \\ 3(2) + 2(1) \\ 5(2) + 6(1) \end{bmatrix} = \begin{bmatrix} 1 \\ 8 \\ 16 \end{bmatrix}

(d) [0134][21−14]\begin{bmatrix} 0 & 1 & 3 & 4\end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ -1 \\ 4\end{bmatrix}

[0134][21−14]=[0(2)+1(1)+3(−1)+4(4)]=[14]\begin{bmatrix} 0 & 1 & 3 & 4 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ -1 \\ 4 \end{bmatrix} = \begin{bmatrix} 0(2) + 1(1) + 3(-1) +4(4) \end{bmatrix} = \begin{bmatrix} 14 \end{bmatrix}

(e) [0134][21−14]\begin{bmatrix} 0 \\ 1 \\ 3 \\ 4\end{bmatrix}\begin{bmatrix} 2 \\ 1 \\ -1 \\ 4\end{bmatrix}

[0134][21−14]\begin{bmatrix} 0 \\ 1 \\ 3 \\ 4 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ -1 \\ 4 \end{bmatrix} is undefined because the number of columns in the first matrix does not match the number of rows in the second.

6. Express the vector b=[212]\mathbf{b} = \begin{bmatrix}2\\1\\2\end{bmatrix} as a linear combination of v1=[1−11],v2=[−121],v3=[23−1]\mathbf{v_1} = \begin{bmatrix} 1\\-1\\1 \end{bmatrix}, \mathbf{v_2}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}, \mathbf{v_3}=\begin{bmatrix} 2\\3\\-1 \end{bmatrix}.

[1−122−123111−12]→R3+R1R2+R1[1−12201532014]→R3−2R1[1−122015302−30]→R3−2R2[1−122015300−13−6]→−113R3[1−1220153001613]→R2−5R3[1−122010913001613]→R1+R2[1023513010913001613]→R1−2R3[1002313010913001613]∴b=2313v1+913v2+613v3\begin{array}{c} \left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ -1 & 2 & 3 & 1 \\ 1 & 1 & -1 & 2 \end{array} \right] &\xrightarrow[R_3 + R_1]{R_2 + R_1} &\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 0 & 1 & 5 & 3 \\ 2 & 0 & 1 & 4 \end{array} \right] \\[2em] &\xrightarrow{R_3 - 2R_1} &\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 0 & 1 & 5 & 3 \\ 0 & 2 & -3 & 0 \end{array} \right] \\[2em] &\xrightarrow{R_3 - 2R_2} &\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 0 & 1 & 5 & 3 \\ 0 & 0 & -13 & -6 \end{array} \right] \\[2em] &\xrightarrow{-\frac{1}{13}R_3} &\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 0 & 1 & 5 & 3 \\ 0 & 0 & 1 & \frac{6}{13} \end{array} \right] \\[2em] &\xrightarrow{R_2 - 5R_3} &\left[ \begin{array}{ccc|c} 1 & -1 & 2 & 2 \\ 0 & 1 & 0 & \frac{9}{13} \\ 0 & 0 & 1 & \frac{6}{13} \end{array} \right] \\[2em] &\xrightarrow{R_1 + R_2} &\left[ \begin{array}{ccc|c} 1 & 0 & 2 & \frac{35}{13} \\ 0 & 1 & 0 & \frac{9}{13} \\ 0 & 0 & 1 & \frac{6}{13} \end{array} \right] \\[2em] &\xrightarrow{R_1 - 2R_3} &\left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{23}{13} \\ 0 & 1 & 0 & \frac{9}{13} \\ 0 & 0 & 1 & \frac{6}{13} \end{array} \right] \end{array} \\[2em] \therefore \mathbf{b} = \frac{23}{13}\mathbf{v_1} +\frac{9}{13}\mathbf{v_2} +\frac{6}{13}\mathbf{v_3}

7. Can the vector b=[212]\mathbf{b} = \begin{bmatrix}2\\1\\2\end{bmatrix} be expressed as a linear combination of v1=[123],v2=[456],v3=[789]\mathbf{v_1} = \begin{bmatrix} 1\\2\\3 \end{bmatrix}, \mathbf{v_2}=\begin{bmatrix} 4\\5\\6 \end{bmatrix}, \mathbf{v_3}=\begin{bmatrix} 7\\8\\9 \end{bmatrix}? Explain.

[147225813692]→R3−3R1R2−2R1[14720−3−6−30−6−12−4]→R3−2R2[14720−3−6−30002]\begin{array}{c} \left[ \begin{array}{ccc|c} 1 & 4 & 7 & 2 \\ 2 & 5 & 8 & 1 \\ 3 & 6 & 9 & 2\\ \end{array} \right] &\xrightarrow[R_3 - 3R_1]{R_2 - 2R_1} &\left[ \begin{array}{ccc|c} 1 & 4 & 7 & 2 \\ 0 & -3 & -6 & -3 \\ 0 & -6 & -12 & -4\\ \end{array} \right] \\ &\xrightarrow{R_3 - 2R_2} &\left[ \begin{array}{ccc|c} 1 & 4 & 7 & 2 \\ 0 & -3 & -6 & -3 \\ 0 & 0 & 0 & 2\\ \end{array} \right] \end{array}

No. The vector b=[212]\mathbf{b} = \begin{bmatrix}2\\1\\2\end{bmatrix} cannot be expressed as a linear combination of v1=[123],v2=[456],v3=[789]\mathbf{v_1} = \begin{bmatrix} 1\\2\\3 \end{bmatrix}, \mathbf{v_2}=\begin{bmatrix} 4\\5\\6 \end{bmatrix}, \mathbf{v_3}=\begin{bmatrix} 7\\8\\9 \end{bmatrix} because the system does not have a solution.

Homework 1
  1. Solve the following systems of linear equations by Gaussian elimination:
  1. The sum of any two of three real numbers are 2424, 2828, 3030. Find these three numbers.
  1. Use some online program, write down the echelon form of the following system and solve the system as well. {x−2y+3z−4w+5v=−12x+3y+4z+5w−6v=22x−2y+3z−3w+6v=0x+y−z−w+3v=23x+4y+5z−6w−4v=0\left\{\begin{array}{l}x-2y+3z-4w+5v=-1\\2x+3y+4z+5w-6v=2\\2x-2y+3z-3w+6v=0\\x+y-z-w+3v=2\\3x+4y+5z-6w-4v=0\end{array}\right.
  1. Find the following products. Explain why if it is undefined.
  1. Express the vector b=[212]\mathbf{b} = \begin{bmatrix}2\\1\\2\end{bmatrix} as a linear combination of v1=[1−11],v2=[−121],v3=[23−1]\mathbf{v_1} = \begin{bmatrix} 1\\-1\\1 \end{bmatrix}, \mathbf{v_2}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}, \mathbf{v_3}=\begin{bmatrix} 2\\3\\-1 \end{bmatrix}.
  1. Can the vector b=[212]\mathbf{b} = \begin{bmatrix}2\\1\\2\end{bmatrix} be expressed as a linear combination of v1=[123],v2=[456],v3=[789]\mathbf{v_1} = \begin{bmatrix} 1\\2\\3 \end{bmatrix}, \mathbf{v_2}=\begin{bmatrix} 4\\5\\6 \end{bmatrix}, \mathbf{v_3}=\begin{bmatrix} 7\\8\\9 \end{bmatrix}? Explain.
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