Systems of Linear Equations and Matrix Rank: Very Difficult Problems with Solutions

Solution:
The determinant of the system is
[tex]\Delta=\begin{vmatrix}\lambda&1&1\\1&\mu&1\\1&2\mu&1\end{vmatrix}=\mu(1-\lambda)[/tex]
so for [tex]\mu\ne 0[/tex] and [tex]\lambda\ne 1[/tex] the solution is unique (Cramer's rule). The remaining cases are studied by elimination. Subtracting the second equation from the third gives
[tex]\mu x_2=1[/tex]
If [tex]\mu=0[/tex], this reads [tex]0=1[/tex] and the system is inconsistent. If [tex]\mu\ne 0[/tex], then [tex]x_2=\frac{1}{\mu}[/tex], and the second and the first equation become
[tex]x_1+x_3=2,\qquad \lambda x_1+x_3=4-\frac{1}{\mu}\ \Rightarrow\ (\lambda-1)x_1=2-\frac{1}{\mu}[/tex]
For [tex]\lambda\ne 1[/tex] (and [tex]\mu\ne 0[/tex]) the solution is unique. For [tex]\lambda=1[/tex] the last equation becomes [tex]0\times x_1=2-\frac{1}{\mu}[/tex]: if [tex]\mu\ne\frac{1}{2}[/tex], there is no solution, and if [tex]\mu=\frac{1}{2}[/tex], it holds for every [tex]x_1[/tex], so the system has infinitely many solutions:
[tex](x_1, x_2, x_3)=(t, 2, 2-t),\qquad t\in\mathbb{R}[/tex]
(here the matrix of the system and the augmented matrix both have rank [tex]2[/tex]). Note that [tex]\Delta=0[/tex] alone does not decide between "no solution" and "infinitely many solutions". Answer:
[tex]\lambda=1,\ \mu=\frac{1}{2}[/tex]

Solution:
Let [tex]A[/tex] be the matrix of the system and [tex]\bar{A}[/tex] its augmented matrix. First we compute the determinant of the system. Adding the second, third and fourth columns to the first one makes every entry of the first column equal to [tex]k+6[/tex]; we factor it out and subtract the first row from the others:
[tex]\Delta=(k+6)\begin{vmatrix} 1 & 3 & 2 & 1 \\ 1 & k & 3 & 2 \\ 1 & 1 & k & 3 \\ 1 & 2 & 1 & k \end{vmatrix}=(k+6)\begin{vmatrix} k-3 & 1 & 1 \\ -2 & k-2 & 2 \\ -1 & -1 & k-1 \end{vmatrix}[/tex]
Expanding the last determinant along its first row:
[tex](k-3)\left((k-2)(k-1)+2\right)-\left(-2(k-1)+2\right)+\left(2+(k-2)\right)=k^3-6k^2+16k-16=(k-2)(k^2-4k+8)[/tex]
Therefore
[tex]\Delta=(k+6)(k-2)\left((k-2)^2+4\right)[/tex]
and since [tex](k-2)^2+4\gt 0[/tex], we have [tex]\Delta=0[/tex] only for [tex]k=-6[/tex] and [tex]k=2[/tex]. If [tex]k\ne -6[/tex] and [tex]k\ne 2[/tex], Cramer's rule gives a unique solution.
For [tex]k=-6[/tex]: the sum of the coefficients in every column is [tex]k+6=0[/tex] and the sum of the free terms is [tex]5+k+1+0=k+6=0[/tex], so the sum of the four rows of [tex]\bar{A}[/tex] is the zero row and the fourth equation is a consequence of the other three. Moreover
[tex]\begin{vmatrix} -6 & 3 & 2 \\ 1 & -6 & 3 \\ 2 & 1 & -6 \end{vmatrix}=-136\ne 0[/tex]
hence [tex]\operatorname{rank}A=\operatorname{rank}\bar{A}=3\lt 4[/tex], and by the Kronecker-Capelli theorem the system has infinitely many solutions (depending on one parameter).
For [tex]k=2[/tex]: denote the equations by [tex]E_1,E_2,E_3,E_4[/tex]. The combination [tex]E_1-E_2+E_3-E_4[/tex] has coefficients [tex]\left(k-2 , 2-k , k-2 , 2-k\right)[/tex] and free term [tex]6-k[/tex]; for [tex]k=2[/tex] it reads
[tex]0=4[/tex]
so the system has no solutions. (Indeed, [tex]\operatorname{rank}A=3[/tex], while [tex]\operatorname{rank}\bar{A}=4[/tex].)
[tex]\begin{vmatrix} 2 & 3 & 2 \\ 1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}=8\ne 0\ \Rightarrow\ \operatorname{rank}A=3[/tex]

Solution:
Elementary row operations do not change the rank. With the third row [tex]R_3=(1, 5, 5, 1, -1)[/tex] we clear the first column:
[tex]R_1-5R_3=(0, 3, \lambda, 3, -1),\qquad R_2+9R_3=(0, -9, -6, \lambda-6, 3)[/tex]
[tex]R_4-4R_3=(0, \lambda+1, 2, 3, -1),\qquad R_5-2R_3=(0, 3, 2, 3, -1)=P[/tex]
Now we subtract multiples of [tex]P[/tex] from the other new rows:
[tex](0, 3, \lambda, 3, -1)-P=(0, 0, \lambda-2, 0, 0)[/tex]
[tex](0, -9, -6, \lambda-6, 3)+3P=(0, 0, 0, \lambda+3, 0)[/tex]
[tex](0, \lambda+1, 2, 3, -1)-P=(0, \lambda-2, 0, 0, 0)[/tex]
Therefore
[tex]A\sim\begin{bmatrix}1&5&5&1&-1\\0&3&2&3&-1\\0&\lambda-2&0&0&0\\0&0&\lambda-2&0&0\\0&0&0&\lambda+3&0\end{bmatrix}[/tex]
The first two rows together with the non-zero rows among the last three are linearly independent: in a vanishing linear combination the coefficient of the first row is [tex]0[/tex] (look at the first column), then the coefficient of the second row is [tex]0[/tex] (look at the last column), and the remaining rows are multiples of different unit vectors. Hence [tex]\operatorname{rank}A[/tex] equals [tex]2[/tex] plus the number of non-zero numbers among [tex]\lambda-2,\ \lambda-2,\ \lambda+3[/tex]:
for [tex]\lambda\ne 2,\ \lambda\ne -3[/tex] the rank is [tex]5[/tex] (indeed [tex]\det A=-(\lambda-2)^2(\lambda+3)\ne 0[/tex]); for [tex]\lambda=-3[/tex] it is [tex]2+2=4[/tex]; for [tex]\lambda=2[/tex] it is [tex]2+1=3[/tex].
[tex]\operatorname{rank}A=\begin{cases}3, & \lambda=2\\4, & \lambda=-3\\5, & \lambda\ne 2,\ \lambda\ne -3\end{cases}[/tex]

Solution:
The equality [tex]\vec{v}=c_1\vec{v}_1+c_2\vec{v}_2+c_3\vec{v}_3+c_4\vec{v}_4[/tex] is a linear system for [tex]c_1,c_2,c_3,c_4[/tex] whose matrix [tex]A[/tex] has the vectors [tex]\vec{v}_1,\ldots,\vec{v}_4[/tex] as columns. There is more than one representation exactly when this system is consistent and its solution is not unique, i.e. (by the Kronecker-Capelli theorem) when [tex]\operatorname{rank}A=\operatorname{rank}\bar{A}\lt 4[/tex]. We write the coordinates, starting with the third one:
[tex]\begin{cases}c_1+3c_2+3c_3+3c_4=-2 & (1)\\-c_1-c_2-2c_3-2c_4=3 & (2)\\-2c_1-10c_2-7c_3+(\mu+8)c_4=3\lambda-1 & (3)\\-c_1-6c_2-4c_3+2c_4=\lambda & (4)\end{cases}[/tex]
We eliminate [tex]c_1[/tex], adding [tex](1)[/tex] to [tex](2)[/tex] and to [tex](4)[/tex], and twice [tex](1)[/tex] to [tex](3)[/tex]:
[tex]\begin{cases}2c_2+c_3+c_4=1\\-4c_2-c_3+(\mu+14)c_4=3\lambda-5\\-3c_2-c_3+5c_4=\lambda-2\end{cases}[/tex]
We eliminate [tex]c_2[/tex], adding twice the first of these equations to the second, and three times the first to twice the third:
[tex]c_3+(\mu+16)c_4=3\lambda-3,\qquad c_3+13c_4=2\lambda-1[/tex]
Subtracting these two equations:
[tex](\mu+3)c_4=\lambda-2[/tex]
If [tex]\mu\ne -3[/tex], the system has a unique solution for every [tex]\lambda[/tex] (exactly one representation). If [tex]\mu=-3[/tex] and [tex]\lambda\ne 2[/tex], we get [tex]0=\lambda-2\ne 0[/tex], so there is no representation. If [tex]\mu=-3[/tex] and [tex]\lambda=2[/tex], then [tex]\operatorname{rank}A=\operatorname{rank}\bar{A}=3\lt 4[/tex] and there are infinitely many representations:
[tex](c_1, c_2, c_3, c_4)=(18t-8, 6t-1, 3-13t, t),\qquad t\in\mathbb{R}[/tex]
For example, [tex]t=0[/tex] gives
[tex]-8\vec{v}_1-\vec{v}_2+3\vec{v}_3=(5, 3, -2, 2)=\vec{v}[/tex]

Solution:
First we find [tex]W=\ell(\vec{a}_1,\vec{a}_2,\vec{a}_3,\vec{a}_4)[/tex]. We have
[tex]2\vec{a}_1-\vec{a}_2=(1, -3, -1, 8, 1)=\vec{a}_3,\qquad 2\vec{a}_2-\vec{a}_1=(4, 3, 2, -7, 1)=\vec{a}_4[/tex]
while [tex]\vec{a}_1,\vec{a}_2[/tex] are not proportional; hence [tex]\dim W=2[/tex] and [tex]W=\ell(\vec{a}_1,\vec{a}_2)[/tex].
For a homogeneous system in [tex]5[/tex] unknowns whose matrix has rank [tex]r[/tex], the solution space has dimension [tex]5-r[/tex] (a fundamental system of solutions consists of [tex]5-r[/tex] solutions). To obtain [tex]W[/tex] we need [tex]5-r=2[/tex], i.e. [tex]r=3[/tex]. Since the rank does not exceed the number of equations, the system has at least [tex]3[/tex] equations.
Three equations suffice. The coefficient vector [tex]\vec{b}[/tex] of each equation must satisfy [tex]\langle\vec{b},\vec{a}_1\rangle=\langle\vec{b},\vec{a}_2\rangle=0[/tex], i.e.
[tex]2b_1-b_2+3b_4+b_5=0,\qquad 3b_1+b_2+b_3-2b_4+b_5=0[/tex]
This system has rank [tex]2[/tex] and a fundamental system of solutions
[tex](1, 2, -5, 0, 0),\qquad (1, -13, 0, -5, 0),\qquad (2, -1, 0, 0, -5)[/tex]
These vectors give the system
[tex]\begin{cases}x_1+2x_2-5x_3=0\\x_1-13x_2-5x_4=0\\2x_1-x_2-5x_5=0\end{cases}[/tex]
whose rank is [tex]3[/tex] (look at the columns of [tex]x_3,x_4,x_5[/tex]). Its solution space has dimension [tex]5-3=2[/tex] and contains [tex]\vec{a}_1[/tex] and [tex]\vec{a}_2[/tex] (e.g. for [tex]\vec{a}_1[/tex]: [tex]2-2-0=0,\ 2+13-15=0,\ 4+1-5=0[/tex]), so it coincides with [tex]W[/tex]. The smallest possible number of equations is [tex]3[/tex].

Solution:
If [tex]x_1,\ldots,x_n\gt 0[/tex], every left-hand side is a product of positive numbers, hence positive; so a positive solution can exist only if [tex]c_1,\ldots,c_n\gt 0[/tex]. Assume this. The substitution [tex]x_j=e^{y_j}[/tex] is a bijection between positive [tex]x_j[/tex] and real [tex]y_j[/tex], and taking logarithms turns the system into the linear system
[tex]\sum_{j=1}^{n}a_{ij}y_j=\ln c_i,\qquad i=1,2,\ldots,n[/tex]
with matrix [tex](a_{ij})[/tex]. So the original system has a unique positive solution if and only if this linear system has a unique solution.
If [tex]\det(a_{ij})\ne 0[/tex], Cramer's rule gives exactly one solution [tex]y[/tex], hence exactly one positive solution [tex]x_j=e^{y_j}[/tex].
If [tex]\det(a_{ij})=0[/tex], then [tex]\operatorname{rank}(a_{ij})\lt n[/tex], and by the Kronecker-Capelli theorem the linear system is either inconsistent or has infinitely many solutions (with [tex]n-\operatorname{rank}(a_{ij})\ge 1[/tex] free unknowns); in neither case is the positive solution unique.
Therefore the condition is
[tex]\det(a_{ij})\ne 0,\quad c_1\gt 0,\ c_2\gt 0,\ \ldots,\ c_n\gt 0[/tex]

Solution:
We use the rank of a matrix: the maximal number of linearly independent columns (equal to the highest order of a non-zero minor).
Step [tex]1[/tex]: [tex]\operatorname{rank}(AX)\le\operatorname{rank}A[/tex]. Let [tex]A^{(1)},\ldots,A^{(n)}[/tex] be the columns of [tex]A[/tex] and [tex]x_{ij}[/tex] the entries of [tex]X[/tex]. The [tex]j[/tex]-th column of [tex]AX[/tex] is
[tex]x_{1j}A^{(1)}+x_{2j}A^{(2)}+\ldots+x_{nj}A^{(n)}[/tex]
so every column of [tex]AX[/tex] lies in the linear span of the columns of [tex]A[/tex], whose dimension is [tex]\operatorname{rank}A[/tex]. Hence [tex]AX[/tex] cannot have more than [tex]\operatorname{rank}A[/tex] linearly independent columns, i.e. [tex]\operatorname{rank}(AX)\le\operatorname{rank}A[/tex].
Step [tex]2[/tex]: [tex]\operatorname{rank}A\le n[/tex], because [tex]A[/tex] has only [tex]n[/tex] columns.
Step [tex]3[/tex]: [tex]\operatorname{rank}E_m=m[/tex], because [tex]\det E_m=1\ne 0[/tex].
Combining the three steps:
[tex]m=\operatorname{rank}E_m=\operatorname{rank}(AX)\le\operatorname{rank}A\le n[/tex]
Hence [tex]m\le n[/tex]. (Moreover, since also [tex]\operatorname{rank}A\le m[/tex], we get [tex]\operatorname{rank}A=m[/tex].)

Solution:
The matrix [tex]A^{T}A[/tex] is square of order [tex]n[/tex], so it is invertible if and only if the homogeneous system [tex]A^{T}A\vec{x}=\vec{0}[/tex], [tex]\vec{x}\in\mathbb{R}^n[/tex], has only the zero solution (then its rank is [tex]n[/tex] and its determinant is non-zero). Let [tex]A^{T}A\vec{x}=\vec{0}[/tex] and put [tex]\vec{y}=A\vec{x}=(y_1, \ldots, y_m)^{T}[/tex]. Multiplying on the left by [tex]\vec{x}^{T}[/tex]:
[tex]0=\vec{x}^{T}A^{T}A\vec{x}=(A\vec{x})^{T}(A\vec{x})=y_1^2+y_2^2+\ldots+y_m^2[/tex]
Since the entries are real, a sum of squares equals [tex]0[/tex] only if all [tex]y_i=0[/tex], so [tex]A\vec{x}=\vec{0}[/tex]. But
[tex]A\vec{x}=x_1\vec{A}_1+x_2\vec{A}_2+\ldots+x_n\vec{A}_n[/tex]
where [tex]\vec{A}_j[/tex] are the columns of [tex]A[/tex], and [tex]\operatorname{rank}A=n[/tex] means that these [tex]n[/tex] columns are linearly independent. Hence [tex]x_1=x_2=\ldots=x_n=0[/tex]. Thus [tex]A^{T}A\vec{x}=\vec{0}[/tex] has only the zero solution, and [tex]A^{T}A[/tex] is invertible.
Remark: the realness is essential. Over [tex]\mathbb{C}[/tex] the matrix [tex]A=\begin{bmatrix}1\\i\end{bmatrix}[/tex] has rank [tex]1[/tex], but [tex]A^{T}A=1+i^2=0[/tex].