Systems of Linear Equations and Matrix Rank: Problems with Solutions

Solution:
Denote the rows of [tex]A[/tex] by [tex]R_1, R_2, R_3, R_4[/tex]. The last two rows are the sum and the difference of the first two:
[tex]R_1+R_2=\left(3, 2, 0, -2\right)=R_3,\qquad R_1-R_2=\left(1, 4, -2, 0\right)=R_4[/tex]
The elementary row operations [tex]R_3-R_1-R_2[/tex] and [tex]R_4-R_1+R_2[/tex] do not change the rank and turn the last two rows into zero rows:
[tex]A\sim\begin{bmatrix} 2 & 3 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}[/tex]
Every minor of order [tex]3[/tex] or [tex]4[/tex] of this matrix contains a zero row and vanishes, so the rank is at most [tex]2[/tex]. On the other hand, there is a non-zero minor of order [tex]2[/tex]:
[tex]\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}=-2-3=-5\ne 0\ \Rightarrow\ \operatorname{rank}A=2[/tex]

Solution:
The determinant of the system, expanded along the first row, is
[tex]\Delta=\begin{vmatrix} 2 & -1 & -1 \\ 3 & 4 & -2 \\ 3 & -2 & 4 \end{vmatrix}=2(16-4)+(12+6)-(-6-12)=24+18+18=60[/tex]
Since [tex]\Delta\ne 0[/tex], Cramer's rule applies and the solution is unique. Replacing in turn the column of each unknown by the column of free terms, we get
[tex]\Delta_1=\begin{vmatrix} 4 & -1 & -1 \\ 11 & 4 & -2 \\ 11 & -2 & 4 \end{vmatrix}=4(16-4)+(44+22)-(-22-44)=48+66+66=180[/tex]
[tex]\Delta_2=\begin{vmatrix} 2 & 4 & -1 \\ 3 & 11 & -2 \\ 3 & 11 & 4 \end{vmatrix}=2(44+22)-4(12+6)-(33-33)=132-72=60[/tex]
[tex]\Delta_3=\begin{vmatrix} 2 & -1 & 4 \\ 3 & 4 & 11 \\ 3 & -2 & 11 \end{vmatrix}=2(44+22)+(33-33)+4(-6-12)=132-72=60[/tex]
Hence
[tex]x_1=\frac{\Delta_1}{\Delta}=\frac{180}{60}=3,\qquad x_2=\frac{\Delta_2}{\Delta}=\frac{60}{60}=1,\qquad x_3=\frac{\Delta_3}{\Delta}=\frac{60}{60}=1[/tex]
Check: [tex]6-1-1=4[/tex], [tex]9+4-2=11[/tex], [tex]9-2+4=11[/tex].

Solution:
By the Kronecker-Capelli theorem we compare the rank of the coefficient matrix [tex]A[/tex] with the rank of the augmented matrix [tex]\bar{A}[/tex]. First,
[tex]\det A=\begin{vmatrix}4&2&3\\2&8&-1\\9&1&8\end{vmatrix}=4(64+1)-2(16+9)+3(2-72)=260-50-210=0[/tex]
and the first two rows of [tex]A[/tex] are not proportional, so [tex]\operatorname{rank}A=2[/tex]. Indeed, the rows of [tex]A[/tex] satisfy the relation
[tex]5\left(4, 2, 3\right)-\left(2, 8, -1\right)-2\left(9, 1, 8\right)=\left(0, 0, 0\right)[/tex]
Denote the equations by [tex]E_1, E_2, E_3[/tex] and take the same combination [tex]5E_1-E_2-2E_3[/tex]. Its left-hand side vanishes, while its right-hand side equals
[tex]5\times (-2)-8-2\times 0=-18[/tex]
We get the contradiction [tex]0=-18[/tex]. Hence [tex]\operatorname{rank}\bar{A}=3\ne\operatorname{rank}A=2[/tex], the system is inconsistent and has no solution.

Solution:
The vectors are linearly dependent exactly when the matrix with rows [tex]\vec{a}_1, \vec{a}_2, \vec{a}_3[/tex] has rank less than [tex]3[/tex], i.e. when its determinant is zero. Expanding along the first row:
[tex]\begin{vmatrix}1&2&1\\-1&3&1\\-1&13&5\end{vmatrix}=1\times (15-13)-2\times (-5+1)+1\times (-13+3)=2+8-10=0[/tex]
Since [tex]\vec{a}_1[/tex] and [tex]\vec{a}_2[/tex] are not proportional, the rank equals [tex]2[/tex], so [tex]\vec{a}_3[/tex] is a linear combination of [tex]\vec{a}_1[/tex] and [tex]\vec{a}_2[/tex]. Let [tex]\vec{a}_3=x\vec{a}_1+y\vec{a}_2[/tex]. Comparing coordinates, we get
[tex]\begin{cases} x-y=-1 \\ 2x+3y=13 \\ x+y=5 \end{cases}[/tex]
The first and the third equation give [tex]x=2[/tex], [tex]y=3[/tex], and the second one is also satisfied: [tex]2\times 2+3\times 3=13[/tex]. Hence
[tex]2\left(1, 2, 1\right)+3\left(-1, 3, 1\right)=\left(-1, 13, 5\right)\ \Rightarrow\ \vec{a}_3=2\vec{a}_1+3\vec{a}_2[/tex]

Solution:
We apply Gaussian elimination to the coefficient matrix: [tex]R_2-3R_1[/tex] and [tex]R_3-4R_1[/tex], then we divide the new second and third rows by [tex]7[/tex] and [tex]9[/tex] and subtract them:
[tex]\begin{bmatrix} 1 & -4 & -2 \\ 3 & -5 & 1 \\ 4 & -7 & 1 \end{bmatrix}\sim\begin{bmatrix} 1 & -4 & -2 \\ 0 & 7 & 7 \\ 0 & 9 & 9 \end{bmatrix}\sim\begin{bmatrix} 1 & -4 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}[/tex]
The rank is [tex]2[/tex], less than the number of unknowns [tex]3[/tex], so the system has non-zero solutions depending on [tex]3-2=1[/tex] parameter. Put [tex]x_3=t[/tex]. From the second row and then from the first equation,
[tex]x_2+x_3=0\ \Rightarrow\ x_2=-t,\qquad x_1=4x_2+2x_3=-4t+2t=-2t[/tex]
Check in the third equation: [tex]4(-2t)-7(-t)+t=-8t+7t+t=0[/tex]. Hence the solutions are
[tex]\left(x_1, x_2, x_3\right)=\left(-2t, -t, t\right),\ t\in\mathbb{R}[/tex]

Solution:
Denote the equations by [tex]E_1, E_2, E_3, E_4[/tex] and compare them together with their right-hand sides. The third equation is the sum of the first two, and the fourth is the sum of the second and the third:
[tex]E_1+E_2:\ 3x-13y-z=-1\ \left(=E_3\right),\qquad E_2+E_3:\ 5x-22y-5z=1\ \left(=E_4\right)[/tex]
So the last two equations can be dropped. The first two equations are not proportional, hence [tex]\operatorname{rank}A=\operatorname{rank}\bar{A}=2\lt 3[/tex], and by the Kronecker-Capelli theorem the system is consistent and has infinitely many solutions depending on [tex]3-2=1[/tex] free unknown. Let [tex]z=t[/tex]. From the first equation [tex]x=4y-3t-3[/tex]; substituting into the second, we get
[tex]2(4y-3t-3)-9y-4t=2\ \Rightarrow\ -y-10t-6=2\ \Rightarrow\ y=-8-10t[/tex]
[tex]x=4(-8-10t)-3t-3=-35-43t[/tex]
Check:
[tex]x-4y+3z=-35-43t+32+40t+3t=-3,\qquad 2x-9y-4z=-70-86t+72+90t-4t=2[/tex]
Hence the solutions are
[tex]\left(x, y, z\right)=\left(-35-43t, -8-10t, t\right),\ t\in\mathbb{R}[/tex]

Solution:
The rank of a system of vectors equals the rank of the matrix whose rows are the coordinates of the vectors, and elementary row operations do not change it. With [tex]R_2-2R_1[/tex], [tex]R_3+R_1[/tex], [tex]R_4+2R_1[/tex], [tex]R_5-3R_1[/tex] we get
[tex]\begin{bmatrix} 1 & 2 & 3 & -4 \\ 2 & 3 & -4 & 1 \\ -1 & -9 & 19 & -9 \\ -2 & 4 & -18 & 12 \\ 3 & -4 & 1 & 2 \end{bmatrix}\sim\begin{bmatrix} 1 & 2 & 3 & -4 \\ 0 & -1 & -10 & 9 \\ 0 & -7 & 22 & -13 \\ 0 & 8 & -12 & 4 \\ 0 & -10 & -8 & 14 \end{bmatrix}[/tex]
Then [tex]R_3-7R_2[/tex], [tex]R_4+8R_2[/tex], [tex]R_5-10R_2[/tex], and finally [tex]R_4+R_3[/tex], [tex]R_5-R_3[/tex]:
[tex]\sim\begin{bmatrix} 1 & 2 & 3 & -4 \\ 0 & -1 & -10 & 9 \\ 0 & 0 & 92 & -76 \\ 0 & 0 & -92 & 76 \\ 0 & 0 & 92 & -76 \end{bmatrix}\sim\begin{bmatrix} 1 & 2 & 3 & -4 \\ 0 & -1 & -10 & 9 \\ 0 & 0 & 92 & -76 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}[/tex]
The echelon form has three non-zero rows: every minor of order [tex]4[/tex] contains a zero row and vanishes, while
[tex]\begin{vmatrix} 1 & 2 & 3 \\ 0 & -1 & -10 \\ 0 & 0 & 92 \end{vmatrix}=1\times (-1)\times 92=-92\ne 0[/tex]
Therefore the rank of the system is [tex]3[/tex]. Indeed, [tex]d=-(a+b+c)[/tex] and [tex]e=-2a+3b+c[/tex], while [tex]a, b, c[/tex] are linearly independent.