Homework 2

1. Find the reduced row echelon form of the following matrices and compute the rank.

(a) [1−111−12−110]\begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & 2 \\ -1 & 1 & 0 \\\end{bmatrix}

[1−111−12−110]→R3+R2R2−R1[1−11001002]→R3−2R1[1−11001000]∴rank⁡([1−111−12−110])=2\begin{array}{c} \begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & 2 \\ -1 & 1 & 0 \end{bmatrix} &\xrightarrow[R_3 + R_2]{R_2 - R_1} &\begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix} \\ &\xrightarrow{R_3 - 2R_1} &\begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix} \end{array} \\[2em] \therefore \operatorname{rank}\( \begin{bmatrix} 1 & -1 & 1 \\ 1 & -1 & 2 \\ -1 & 1 & 0 \end{bmatrix} \) = 2

(b) [1−12121−1012−3−14−13203−5−2]\begin{bmatrix} 1 & -1 & 2 & 1 \\ 2 & 1 & -1 & 0 \\ 1 & 2 & -3 & -1 \\ 4 & -1 & 3 & 2 \\ 0 & 3 & -5 & -2 \\\end{bmatrix}

[1−12121−1012−3−14−13203−5−2]→R3−R1R2−2R1[1−12103−5−203−5−24−13203−5−2]→R2↔R4R2−R3R3−R2R5−R2[1−12103−5−203−5−24−13203−5−2]→R3−R1R2−2R1[1−1214−132000000000000]→R2−4R1[1−12103−5−2000000000000]→13R2[1−12101−53−23000000000000]→R1+R2[10131301−53−23000000000000]∴rank⁡([1−12121−1012−3−14−13203−5−2])=2\begin{array}{c} \begin{bmatrix} 1 & -1 & 2 & 1 \\ 2 & 1 & -1 & 0 \\ 1 & 2 & -3 & -1 \\ 4 & -1 & 3 & 2 \\ 0 & 3 & -5 & -2 \end{bmatrix} &\xrightarrow[R_3 - R_1]{R_2 - 2R_1} &\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 3 & -5 & -2 \\ 0 & 3 & -5 & -2 \\ 4 & -1 & 3 & 2 \\ 0 & 3 & -5 & -2 \end{bmatrix} \\ &\xrightarrow[R_2 \leftrightarrow R_4] {\substack{ R_2 - R_3 \\[.2em] R_3 - R_2 \\[.2em] R_5 - R_2 }} &\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 3 & -5 & -2 \\ 0 & 3 & -5 & -2 \\ 4 & -1 & 3 & 2 \\ 0 & 3 & -5 & -2 \end{bmatrix} \\ &\xrightarrow[R_3 - R_1]{R_2 - 2R_1} &\begin{bmatrix} 1 & -1 & 2 & 1 \\ 4 & -1 & 3 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ &\xrightarrow{R_2 - 4R_1} &\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 3 & -5 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ &\xrightarrow{\frac{1}{3}R_2} &\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & -\frac{5}{3} & -\frac{2}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\ &\xrightarrow{R_1 + R_2} &\begin{bmatrix} 1 & 0 & \frac{1}{3} & \frac{1}{3} \\ 0 & 1 & -\frac{5}{3} & -\frac{2}{3} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{array} \\[2em] \therefore\operatorname{rank}\( \begin{bmatrix} 1 & -1 & 2 & 1 \\ 2 & 1 & -1 & 0 \\ 1 & 2 & -3 & -1 \\ 4 & -1 & 3 & 2 \\ 0 & 3 & -5 & -2 \end{bmatrix} \) = 2

(c) [1−11−11−1111−113]\begin{bmatrix} 1 & -1 & 1 & -1 \\ 1 & -1 & 1 & 1 \\ 1 & -1 & 1 & 3 \\\end{bmatrix}

[1−11−11−1111−113]→R2−R1R3−R2[1−11−100020002]→12R312R2[1−11−100010001]→R3−R2R1+R2[1−11000010000]∴rank⁡([1−11−11−1111−113])=2\begin{array}{c} \begin{bmatrix} 1 & -1 & 1 & -1 \\ 1 & -1 & 1 & 1 \\ 1 & -1 & 1 & 3 \\ \end{bmatrix} &\xrightarrow[R_2 - R_1]{R_3 - R_2} &\begin{bmatrix} 1 & -1 & 1 & -1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \\ &\xrightarrow[\frac{1}{2}R_3]{\frac{1}{2}R_2} &\begin{bmatrix} 1 & -1 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ &\xrightarrow[R_3 - R_2]{R_1 + R_2} &\begin{bmatrix} 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \end{array} \\[2em] \therefore\operatorname{rank}\( \begin{bmatrix} 1 & -1 & 1 & -1 \\ 1 & -1 & 1 & 1 \\ 1 & -1 & 1 & 3 \\ \end{bmatrix} \) = 2

(d) [30−2]\begin{bmatrix} 3 \\ 0 \\ -2\end{bmatrix}

[30−2]→−12R3R1+32R3[001]→R1↔R2[100]∴rank⁡([30−2])=1\begin{array}{c} \begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix} &\xrightarrow[-\frac{1}{2}R_3]{R_1 + \frac{3}{2}R_3} &\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \\ &\xrightarrow{R_1 \leftrightarrow R_2} &\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \end{array} \\[2em] \therefore\operatorname{rank}\( \begin{bmatrix} 3 \\ 0 \\ -2 \end{bmatrix} \) = 1

2. For which values of a,b,c,d,ea, b, c, d, e is the following matrix in reduced row echelon form? [1ab30−200c1d300e011]\begin{bmatrix} 1 & a & b & 3 & 0 & -2 \\ 0 & 0 & c & 1 & d & 3 \\ 0 & 0 & e & 0 & 1 & 1 \\\end{bmatrix}

[1ab30−200c1d300e011]\begin{bmatrix} 1 & a & b & 3 & 0 & -2 \\ 0 & 0 & c & 1 & d & 3 \\ 0 & 0 & e & 0 & 1 & 1 \\ \end{bmatrix} is in reduced row echelon form if c=1c=1 and b=d=e=0b=d=e=0.

3. If the rank of a 4×44\times4 matrix AA is 44, what is its rref⁡(A)\operatorname{rref}(A)?

AA must have a full row rank, therefore rref⁡(A)\operatorname{rref}(A) must be:
[1000010000100001]\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

4. Find all the possible solutions of the following systems.

(i) {x−2y+2z−w=33x+y+6z+11w=162x−y+4z+w=9\left\{\begin{array}{c} x - 2y + 2z - w = 3 \\ 3x + y + 6z + 11w = 16 \\ 2x - y + 4z + w = 9 \\\end{array}\right.

{x−2y+2z−w=33x+y+6z+11w=162x−y+4z+w=9  ⟺  [1−22−1331611162−1419]→R3−2R1R2−3R1[1−22−1307014703033]→17R213R3[1−22−130102101011]→R1+R3R2−R3[1−12040001001011]→R3−R2R1+R3[102050001001001]→R2↔R3[102050100100010]∴{x+2z=5y=1w=0\begin{array}{c} \left\{ \begin{array}{c} x - 2y + 2z - w = 3 \\ 3x + y + 6z + 11w = 16 \\ 2x - y + 4z + w = 9 \\ \end{array} \right. &\iff& \left[ \begin{array}{cccc|c} 1 & - 2 & 2 & - 1 & 3 \\ 3 & 1 & 6 & 11 & 16 \\ 2 & - 1 & 4 & 1 & 9 \end{array} \right] \\ &\xrightarrow[R_3 - 2R_1]{R_2 - 3R_1} &\left[ \begin{array}{cccc|c} 1 & - 2 & 2 & - 1 & 3 \\ 0 & 7 & 0 & 14 & 7\\ 0 & 3 & 0 & 3 & 3 \end{array} \right] \\ &\xrightarrow[\frac{1}{7}R_2]{\frac{1}{3}R_3} &\left[ \begin{array}{cccc|c} 1 & - 2 & 2 & - 1 & 3 \\ 0 & 1 & 0 & 2 & 1 \\ 0 & 1 & 0 & 1 & 1 \end{array} \right] \\ &\xrightarrow[R_1 + R_3]{R_2 - R_3} &\left[ \begin{array}{cccc|c} 1 & -1 & 2 & 0 & 4 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \end{array} \right] \\ &\xrightarrow[R_3 - R_2]{R_1 + R_3} &\left[ \begin{array}{cccc|c} 1 & 0 & 2 & 0 & 5 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \end{array} \right] \\ &\xrightarrow{R_2 \leftrightarrow R_3} &\left[ \begin{array}{cccc|c} 1 & 0 & 2 & 0 & 5 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ \end{array} \right] \end{array} \\[2em] \therefore \left\{ \begin{array}{c} x + 2z &= 5 \\ y &= 1 \\ w &= 0 \end{array} \right.

Let z∈Rz\in\R. Then, x=5−2zx= 5-2z. As such, the solution set is:
{(5−2z1z0):z∈R}\left\{ \begin{pmatrix} 5 - 2z \\ 1 \\ z \\ 0 \end{pmatrix} : z \in\R \right\}

(ii) {x+y−2z=−32x−y+3z=7x−2y+5z=1\left\{\begin{array}{c} x + y - 2z = -3 \\ 2x - y + 3z = 7 \\ x - 2y + 5z = 1 \\\end{array}\right.

{x+y−2z=−32x−y+3z=7x−2y+5z=1  ⟺  [11−2−32−1371−251]→R3−R1R2−2R1[11−2−30−37130−374]→R3−R2[11−2−30−3713000−9]\begin{array}{c} \left\{ \begin{array}{c} x + y - 2z = -3 \\ 2x - y + 3z = 7 \\ x - 2y + 5z = 1 \end{array} \right. &\iff &\left[ \begin{array}{ccc|c} 1 & 1 & -2 & -3 \\ 2 & - 1 & 3 & 7 \\ 1 & - 2 & 5 & 1 \end{array} \right] \\ &\xrightarrow[R_3 - R_1]{R_2 - 2R_1} &\left[ \begin{array}{ccc|c} 1 & 1 & -2 & -3 \\ 0 & -3 & 7 & 13 \\ 0 & -3 & 7 & 4 \end{array} \right] \\ &\xrightarrow{R_3 - R_2} &\left[ \begin{array}{ccc|c} 1 & 1 & -2 & -3 \\ 0 & -3 & 7 & 13 \\ 0 & 0 & 0 & -9 \end{array} \right] \end{array}

No solutions.

(iii) {x+2y=12x+5y=23x+6y=3\left\{\begin{array}{c} x + 2y = 1 \\ 2x + 5y = 2 \\ 3x + 6y = 3 \\\end{array}\right.

{x+2y=12x+5y=23x+6y=3  ⟺  [121252363]→R3−2R1R2−2R1[121010000]→R1−2R2[101010000]∴x=1,y=0\begin{array}{c} \left\{ \begin{array}{c} x + 2y = 1 \\ 2x + 5y = 2 \\ 3x + 6y = 3 \end{array} \right. &\iff &\left[ \begin{array}{cc|c} 1 & 2 & 1 \\ 2 & 5 & 2 \\ 3 & 6 & 3 \end{array} \right] \\ &\xrightarrow[R_3 - 2R_1]{R_2 - 2R_1} &\left[ \begin{array}{cc|c} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \\ &\xrightarrow{R_1 - 2R_2} &\left[ \begin{array}{cc|c} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \end{array} \\[2em] \therefore x=1, y=0

(iv) {x1+x2+x3+9x4=8x2+2x3+8x4=7−3x1+x3−7x4=9\left\{\begin{array}{c} x_1 + x_2 + x_3 + 9x_4 = 8 \\ x_2 + 2x_3 + 8x_4 = 7 \\ -3x_1 + x_3 - 7x_4 = 9 \\\end{array}\right.

{x1+x2+x3+9x4=8x2+2x3+8x4=7−3x1+x3−7x4=9  ⟺  [1119801287−301−79]→R3+3R1[11198012870342033]→R3−3R2[111980128700−2−412]→−12R3[11198012870012−6]→R1−R3R2−2R3[1107140104190012−6]→R1−R2[1003−50104190012−6]∴{x1+3x4=−5x2+4x4=19x3+2x4=−6\begin{array}{c} \left\{ \begin{array}{c} x_1 + x_2 + x_3 + 9x_4 = 8 \\ x_2 + 2x_3 + 8x_4 = 7 \\ -3x_1 + x_3 - 7x_4 = 9 \end{array} \right. &\iff &\left[ \begin{array}{cccc|c} 1 & 1 & 1 & 9 & 8 \\ 0 & 1 & 2 & 8 & 7 \\ -3 & 0 & 1 & -7 & 9 \end{array} \right] \\ &\xrightarrow{R_3 + 3R_1} &\left[ \begin{array}{cccc|c} 1 & 1 & 1 & 9 & 8 \\ 0 & 1 & 2 & 8 & 7 \\ 0 & 3 & 4 & 20 & 33 \end{array} \right] \\ &\xrightarrow{R_3 - 3R_2} &\left[ \begin{array}{cccc|c} 1 & 1 & 1 & 9 & 8 \\ 0 & 1 & 2 & 8 & 7 \\ 0 & 0 & -2 & -4 & 12 \end{array} \right] \\ &\xrightarrow{-\frac{1}{2}R_3} &\left[ \begin{array}{cccc|c} 1 & 1 & 1 & 9 & 8 \\ 0 & 1 & 2 & 8 & 7 \\ 0 & 0 & 1 & 2 & -6 \end{array} \right] \\ &\xrightarrow[R_1 - R_3]{R_2 - 2R_3} &\left[ \begin{array}{cccc|c} 1 & 1 & 0 & 7 & 14 \\ 0 & 1 & 0 & 4 & 19 \\ 0 & 0 & 1 & 2 & -6 \end{array} \right] \\ &\xrightarrow{R_1 - R_2} &\left[ \begin{array}{cccc|c} 1 & 0 & 0 & 3 & -5 \\ 0 & 1 & 0 & 4 & 19 \\ 0 & 0 & 1 & 2 & -6 \end{array} \right] \end{array} \\[2em] \therefore \left\{ \begin{array}{c} x_1 + 3x_4 &= -5 \\ x_2 + 4x_4 &= 19 \\ x_3 + 2x_4 &= -6 \end{array} \right.

Let x4∈Rx_4\in\R. Then:
x1=−5−3x4x2=19−4x4x3=−6−2x4x_1 = -5-3x_4 \\ x_2 = 19-4x_4 \\ x_3 = -6-2x_4

As such, the solution set is:
{(−5−3x419−4x4−6−2x4x4):x4∈R}={(−519−60)+(−3−4−21)x4:x4∈R}\left\{ \begin{pmatrix} -5-3x_4 \\ 19-4x_4 \\ -6-2x_4 \\ x_4 \end{pmatrix} : x_4 \in \R \right\} = \left\{ \begin{pmatrix} -5 \\ 19 \\ -6 \\ 0 \end{pmatrix} + \begin{pmatrix} - 3 \\ - 4 \\ - 2 \\ 1 \end{pmatrix} x_4 : x_4 \in \R \right\}

5. Determine kk for which the following system has infinitely many solutions. {x+y=02y+2kz=1y+kz=2k\left\{\begin{array}{c} x + y = 0 \\ 2y + 2kz = 1 \\ y + kz = 2k \\\end{array}\right.

{x+y=02y+2kz=1y+kz=2k  ⟺  [1100022k101k2k]→R2−2R3[11000001−4k01k2k]→R2↔R3[110001k2k0001−4k]∴rank⁡([110001k2k0001−4k])<3  ⟺  k=14\begin{array}{c} \left\{ \begin{array}{c} x + y = 0 \\ 2y + 2kz = 1 \\ y + kz = 2k \\ \end{array} \right. &\iff &\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 2 & 2k & 1 \\ 0 & 1 & k & 2k \end{array} \right] \\ &\xrightarrow{R_2 - 2R_3} &\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1-4k \\ 0 & 1 & k & 2k \end{array} \right] \\ &\xrightarrow{R_2 \leftrightarrow R_3} &\left[ \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 1 & k & 2k \\ 0 & 0 & 0 & 1-4k \end{array} \right] \end{array} \\[2em] \therefore\operatorname{rank}\left( \left[ \begin{array}{ccc|c} 1 & 1 & 0 & 0 \\ 0 & 1 & k & 2k \\ 0 & 0 & 0 & 1-4k \end{array} \right] \right) < 3 \iff k=\frac{1}{4}

The system will have infinitely many solutions for k=14k=\displaystyle\frac{1}{4}.

6. (True or False) Determine if the following statements are true or false. If it is true, explain and prove it. If it is false, give a counterexample.

Let AA be an 3×53\times5 matrix, then:

(i) Ax=bAx = b always has a solution.

False. Let A=[100000100000100]A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{bmatrix} and b⃗=[110]\vec{b} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}. Clearly, [100001010001001000]\left[ \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 \end{array} \right] has no solutions.

(ii) Ax=0Ax = 0 always has a solution.

True. Ax⃗=0⃗  ⟺  x⃗=0⃗A\vec{x} = \vec{0} \iff \vec{x} = \vec{0}.

(iii) If a system Ax=bAx = b has no solution, then rank⁡(A)<3\operatorname{rank}(A) < 3.

True. Consider the inverse, where rank⁡(A)=3\operatorname{rank}(A) = 3 (which means it has a full row rank). Then, there must be at least one solution.

(iv) There are always infinitely many solutions to the system Ax=0Ax = 0.

True. rank⁡(A)≤3\operatorname{rank}(A)\leq3 and the number of columns n=5n=5. By definition, for a matrix AA with nn columns where rank⁡(A)<n\operatorname{rank}(A) < n, Ax⃗=0⃗A\vec{x}=\vec{0} will have infinitely many solutions.

(v) It is possible that the system Ax=bAx = b has a unique solution.

True. Sure, it’s possible. Let A=[100000100000100]A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{bmatrix} and b⃗=[111]\vec{b} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}. Then clearly, [100001010001001001]\left[ \begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array} \right] has a unique solution.

Homework 2
  1. Find the reduced row echelon form of the following matrices and compute the rank.
  1. For which values of a,b,c,d,ea, b, c, d, e is the following matrix in reduced row echelon form? [1ab30−200c1d300e011]\begin{bmatrix} 1 & a & b & 3 & 0 & -2 \\ 0 & 0 & c & 1 & d & 3 \\ 0 & 0 & e & 0 & 1 & 1 \\\end{bmatrix}
  1. If the rank of a 4×44\times4 matrix AA is 44, what is its rref⁡(A)\operatorname{rref}(A)?
  1. Find all the possible solutions of the following systems.

(i) {x−2y+2z−w=33x+y+6z+11w=162x−y+4z+w=9\left\{\begin{array}{c} x - 2y + 2z - w = 3 \\ 3x + y + 6z + 11w = 16 \\ 2x - y + 4z + w = 9 \\\end{array}\right.

(ii) {x+y−2z=−32x−y+3z=7x−2y+5z=1\left\{\begin{array}{c} x + y - 2z = -3 \\ 2x - y + 3z = 7 \\ x - 2y + 5z = 1 \\\end{array}\right.

(iii) {x+2y=12x+5y=23x+6y=3\left\{\begin{array}{c} x + 2y = 1 \\ 2x + 5y = 2 \\ 3x + 6y = 3 \\\end{array}\right.

(iv) {x1+x2+x3+9x4=8x2+2x3+8x4=7−3x1+x3−7x4=9\left\{\begin{array}{c} x_1 + x_2 + x_3 + 9x_4 = 8 \\ x_2 + 2x_3 + 8x_4 = 7 \\ -3x_1 + x_3 - 7x_4 = 9 \\\end{array}\right.

  1. Determine kk for which the following system has infinitely many solutions. {x+y=02y+2kz=1y+kz=2k\left\{\begin{array}{c} x + y = 0 \\ 2y + 2kz = 1 \\ y + kz = 2k \\\end{array}\right.
  1. (True or False) Determine if the following statements are true or false. If it is true, explain and prove it. If it is false, give a counterexample.
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