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

Solution:
Denote the equations by [tex]E_1, E_2, E_3, E_4[/tex]. Keeping [tex]E_1[/tex] and replacing [tex]E_2, E_3, E_4[/tex] by the differences [tex]E_2-E_1[/tex], [tex]E_3-E_2[/tex], [tex]E_4-E_3[/tex], we get an equivalent system:
[tex]\begin{cases} x_1+2x_2+3x_3+4x_4=5 \\ x_1-x_2-x_3-x_4=-4 \\ x_1+x_2-x_3-x_4=0 \\ x_1+x_2+x_3-x_4=-6 \end{cases}[/tex]
Subtracting the second equation from the third gives [tex]2x_2=4[/tex], so [tex]x_2=2[/tex]; subtracting the third from the fourth gives [tex]2x_3=-6[/tex], so [tex]x_3=-3[/tex]. Substituting into the third and the first equation:
[tex]x_1+2+3-x_4=0\ \Rightarrow\ x_1-x_4=-5,\qquad x_1+4-9+4x_4=5\ \Rightarrow\ x_1+4x_4=10[/tex]
Subtracting, [tex]5x_4=15[/tex], so [tex]x_4=3[/tex] and [tex]x_1=-2[/tex]. All transformations were equivalent, so the solution is unique (in agreement with the determinant of the system, which equals [tex]-20\ne 0[/tex]):
[tex]x_1=-2,\ x_2=2,\ x_3=-3,\ x_4=3[/tex]
Check: [tex]-4+2-6+9=1[/tex], [tex]-6+4-3+6=1[/tex], [tex]-8+6-6+3=-5[/tex].

Solution:
Elementary row operations do not change the rank. To get a convenient pivot, we replace the first row [tex]R_1[/tex] by [tex]-2R_1-R_3[/tex] (multiplying a row by a non-zero number and adding another row to it is allowed):
[tex]-2R_1-R_3=\left(-28+29, 54-55, 98-96, -226+227\right)=\left(1, -1, 2, 1\right)[/tex]
Then [tex]R_2-43R_1[/tex], [tex]R_3+29R_1[/tex], [tex]R_4-128R_1[/tex] give
[tex]A\sim\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & -39 & -231 & 297 \\ 0 & 26 & 154 & -198 \\ 0 & -117 & -694 & 889 \end{bmatrix}[/tex]
Dividing the second row by [tex]-3[/tex] and the third row by [tex]2[/tex], we get the same row [tex]\left(0, 13, 77, -99\right)[/tex] twice. Now [tex]R_3-R_2[/tex] and [tex]R_4+9R_2[/tex]:
[tex]A\sim\begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 13 & 77 & -99 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -2 \end{bmatrix}[/tex]
The only minor of order [tex]4[/tex] of the last matrix contains the zero row, so it vanishes, while
[tex]\begin{vmatrix} 1 & -1 & 2 \\ 0 & 13 & 77 \\ 0 & 0 & -1 \end{vmatrix}=1\times 13\times (-1)=-13\ne 0[/tex]
Therefore [tex]\operatorname{rank}A=3[/tex].

Solution:
The rank of the system equals the rank of the matrix whose rows are [tex]a_1, a_2, a_3, a_4[/tex]. Replacing [tex]a_4[/tex] by [tex]a_4-a_2=\left(0 , 0 , 0 , p+1\right)[/tex] does not change the rank. We expand the determinant along the last row:
[tex]\begin{vmatrix} 1 & 2 & -1 & 3 \\ 1 & 0 & 1 & -1 \\ 2 & -1 & 1 & -2 \\ 0 & 0 & 0 & p+1 \end{vmatrix}=(p+1)\begin{vmatrix} 1 & 2 & -1 \\ 1 & 0 & 1 \\ 2 & -1 & 1 \end{vmatrix}[/tex]
[tex]\begin{vmatrix} 1 & 2 & -1 \\ 1 & 0 & 1 \\ 2 & -1 & 1 \end{vmatrix}=1\times (0+1)-2\times (1-2)+(-1)\times (-1-0)=1+2+1=4[/tex]
So the determinant equals [tex]4(p+1)[/tex]. If [tex]p\ne -1[/tex], it is non-zero, the four vectors are linearly independent and the rank is [tex]4[/tex]. If [tex]p=-1[/tex], then [tex]a_4=a_2[/tex], so the rank equals the rank of [tex]a_1, a_2, a_3[/tex], which is [tex]3[/tex], because their minor of order [tex]3[/tex] computed above equals [tex]4\ne 0[/tex]. Hence the rank is [tex]3[/tex] only for
[tex]p=-1[/tex]

Solution:
A square homogeneous system has a non-zero solution if and only if its determinant is zero (i.e. the rank of its matrix is less than the number of unknowns). Expanding along the first row:
[tex]\Delta=\lambda\left[2(4\lambda+1)-5(\lambda+1)\right]-\left[3(4\lambda+1)-25\right]+4\left[3(\lambda+1)-10\right][/tex]
[tex]\Delta=\lambda(3\lambda-3)-(12\lambda-22)+(12\lambda-28)=3\lambda^2-3\lambda-6=3(\lambda-2)(\lambda+1)[/tex]
So [tex]\Delta=0[/tex] exactly for [tex]\lambda=2[/tex] and [tex]\lambda=-1[/tex]. Indeed, for [tex]\lambda=2[/tex] the vector [tex]\left(-3 , 2 , 1\right)[/tex] is a non-zero solution, and for [tex]\lambda=-1[/tex] the vector [tex]\left(3 , -17 , 5\right)[/tex] is. For all other values of [tex]\lambda[/tex] the system has only the zero solution. Answer:
[tex]\lambda\in\left\{-1 , 2\right\}[/tex]

Solution:
The equality [tex]a=c_1a_1+c_2a_2+c_3a_3[/tex], written coordinatewise, is a linear system for [tex]c_1, c_2, c_3[/tex] whose columns are [tex]a_1, a_2, a_3[/tex] and whose free terms are the coordinates of [tex]a[/tex]. So [tex]a[/tex] is a linear combination of [tex]a_1, a_2, a_3[/tex] exactly when this system is consistent. In its augmented matrix we add [tex]6[/tex] times the first row to the second row and subtract the first row from the third:
[tex]\left(\begin{array}{ccc|c} 2 & 3 & 1 & 7 \\ 3 & 7 & -6 & -2 \\ 5 & 8 & 1 & \lambda \end{array}\right)\sim\left(\begin{array}{ccc|c} 2 & 3 & 1 & 7 \\ 15 & 25 & 0 & 40 \\ 3 & 5 & 0 & \lambda-7 \end{array}\right)[/tex]
The second row equals [tex]5\left(3 , 5 , 0 , 8\right)[/tex]; we divide it by [tex]5[/tex] and subtract it from the third row:
[tex]\sim\left(\begin{array}{ccc|c} 2 & 3 & 1 & 7 \\ 3 & 5 & 0 & 8 \\ 0 & 0 & 0 & \lambda-15 \end{array}\right)[/tex]
The coefficient matrix has rank [tex]2[/tex] (its third row is zero and [tex]\begin{vmatrix} 2 & 3 \\ 3 & 5 \end{vmatrix}=1\ne 0[/tex]), while the augmented matrix has rank [tex]2[/tex] for [tex]\lambda=15[/tex] and rank [tex]3[/tex] otherwise. By the Kronecker-Capelli theorem the system is consistent only for [tex]\lambda=15[/tex]. Then, for example, [tex]a=a_1+a_2+2a_3=\left(7 , -2 , 15\right)[/tex]. Answer:
[tex]\lambda=15[/tex]

Solution:
Adding all four equations, and subtracting the [tex]j[/tex]-th equation from the [tex]i[/tex]-th one, we get
[tex](\lambda+3)(x_1+x_2+x_3+x_4)=4,\qquad (\lambda-1)(x_i-x_j)=0[/tex]
If [tex]\lambda\ne 1[/tex] and [tex]\lambda\ne -3[/tex], all [tex]x_i[/tex] are equal and [tex]4(\lambda+3)x_1=4[/tex], so the system has the unique solution [tex]x_1=x_2=x_3=x_4=\frac{1}{\lambda+3}[/tex] (its determinant is [tex](\lambda-1)^3(\lambda+3)\ne 0[/tex]).
If [tex]\lambda=1[/tex], all four equations become [tex]x_1+x_2+x_3+x_4=1[/tex]; the system is consistent and has infinitely many solutions.
If [tex]\lambda=-3[/tex], the sum of the equations gives [tex]0=4[/tex], a contradiction, so the system is inconsistent (here [tex]\operatorname{rank}A=3[/tex], [tex]\operatorname{rank}\bar{A}=4[/tex]). Hence the only such value is
[tex]\lambda=-3[/tex]

Solution:
The conditions form a linear system for [tex]f_0, f_1, f_2, f_3[/tex]:
[tex]\begin{cases} f_0-2f_1+4f_2-8f_3=-3 \\ f_0-f_1+f_2-f_3=-6 \\ f_0+f_1+f_2+f_3=6 \\ f_0+2f_1+4f_2+8f_3=-3 \end{cases}[/tex]
Its matrix is a Vandermonde matrix with the distinct nodes [tex]-2[/tex], [tex]-1[/tex], [tex]1[/tex], [tex]2[/tex], so its determinant is non-zero (it equals [tex]72[/tex]) and by Cramer's rule the system has a unique solution. Adding the second and third equations, and also the first and fourth:
[tex]2f_0+2f_2=0,\qquad 2f_0+8f_2=-6\ \Rightarrow\ f_2=-1,\ f_0=1[/tex]
Subtracting instead (the third minus the second, the fourth minus the first):
[tex]2f_1+2f_3=12,\qquad 4f_1+16f_3=0\ \Rightarrow\ f_3=-2,\ f_1=8[/tex]
Thus [tex]f(t)=1+8t-t^2-2t^3[/tex]. Check: [tex]f(-2)=1-16-4+16=-3[/tex], [tex]f(-1)=1-8-1+2=-6[/tex], [tex]f(1)=1+8-1-2=6[/tex], [tex]f(2)=1+16-4-16=-3[/tex]. Therefore
[tex]f(0)=f_0=1[/tex]

Solution:
Denote the given matrix by [tex]A[/tex]. It has [tex]3[/tex] rows and [tex]4[/tex] columns, and the right-hand side has [tex]1[/tex] row and [tex]4[/tex] columns, so [tex]X[/tex] has [tex]1[/tex] row and [tex]3[/tex] columns: [tex]X=\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}[/tex]. Then [tex]XA=x_1\left(1 , 1 , 1 , 1\right)+x_2\left(1 , 2 , 2 , 2\right)+x_3\left(1 , 1 , 3 , 2\right)[/tex], and comparing the entries we get four equations in three unknowns:
[tex]\begin{cases} x_1+x_2+x_3=1 \\ x_1+2x_2+x_3=1 \\ x_1+2x_2+3x_3=5 \\ x_1+2x_2+2x_3=3 \end{cases}[/tex]
The second minus the first equation gives [tex]x_2=0[/tex]; the third minus the second gives [tex]2x_3=4[/tex], so [tex]x_3=2[/tex]; the fourth minus the second gives [tex]x_3=2[/tex] again, so this overdetermined system is consistent. From the first equation [tex]x_1=1-0-2=-1[/tex]. Check:
[tex]-\left(1 , 1 , 1 , 1\right)+2\left(1 , 1 , 3 , 2\right)=\left(1 , 1 , 5 , 3\right)[/tex]
Hence
[tex]X=\begin{bmatrix} -1 & 0 & 2 \end{bmatrix}[/tex]

Solution:
In the augmented matrix we subtract twice the first row from the second and three times the first row from the third:
[tex]\left(\begin{array}{cccc|c} 3 & -2 & 5 & 4 & 2 \\ 6 & -4 & 4 & 3 & 3 \\ 9 & -6 & 3 & 2 & 4 \end{array}\right)\sim\left(\begin{array}{cccc|c} 3 & -2 & 5 & 4 & 2 \\ 0 & 0 & -6 & -5 & -1 \\ 0 & 0 & -12 & -10 & -2 \end{array}\right)[/tex]
The new third row is twice the new second row, so we subtract it; we also multiply the second row by [tex]-1[/tex]:
[tex]\sim\left(\begin{array}{cccc|c} 3 & -2 & 5 & 4 & 2 \\ 0 & 0 & 6 & 5 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)[/tex]
Hence [tex]\operatorname{rank}A=\operatorname{rank}\bar{A}=2\lt 4[/tex]: by the Kronecker-Capelli theorem the system is consistent and has [tex]4-2=2[/tex] free unknowns. The minor formed by the columns of [tex]x_3[/tex] and [tex]x_4[/tex] is [tex]\begin{vmatrix} 5 & 4 \\ 6 & 5 \end{vmatrix}=1\ne 0[/tex], so we take [tex]x_1=p[/tex], [tex]x_2=q[/tex] as parameters and solve
[tex]\begin{cases} 5x_3+4x_4=2-3p+2q \\ 6x_3+5x_4=1 \end{cases}[/tex]
By Cramer's rule (the determinant is [tex]1[/tex]):
[tex]x_3=5(2-3p+2q)-4=6-15p+10q,\qquad x_4=5-6(2-3p+2q)=18p-12q-7[/tex]
Check in the first equation: [tex]3p-2q+5(6-15p+10q)+4(18p-12q-7)=2[/tex]. So the general solution is
[tex]x_1=p,\ x_2=q,\ x_3=6-15p+10q,\ x_4=18p-12q-7,\quad p, q\in\mathbb{R}[/tex]

Solution:
Denote the equations by [tex]E_1, E_2, E_3, E_4[/tex]. Subtracting the second equation from the first gives the convenient equation
[tex]F_1:\ x_1+x_2+x_3+x_4+2x_5=0[/tex]
With it we eliminate [tex]x_1[/tex] from the other equations:
[tex]F_2=E_2-2F_1:\ x_2-2x_3-x_4-3x_5=0[/tex]
[tex]E_3-4F_1:\ x_2-2x_3-3x_4-7x_5=0[/tex]
[tex]E_4-3F_1:\ 2x_2-4x_3-2x_4-6x_5=0[/tex]
The last equation is [tex]2F_2[/tex], and subtracting [tex]F_2[/tex] from the middle one gives [tex]-2x_4-4x_5=0[/tex], i.e. [tex]x_4=-2x_5[/tex]. Hence the rank of the coefficient matrix is [tex]r=3[/tex], and a fundamental system consists of [tex]n-r=5-3=2[/tex] solutions. Taking [tex]x_3[/tex] and [tex]x_5[/tex] as free unknowns:
[tex]x_4=-2x_5,\quad x_2=2x_3+x_4+3x_5=2x_3+x_5,\quad x_1=-x_2-x_3-x_4-2x_5=-3x_3-x_5[/tex]
For [tex]x_3=1[/tex], [tex]x_5=0[/tex] we get [tex]\left(-3 , 2 , 1 , 0 , 0\right)[/tex], and for [tex]x_3=0[/tex], [tex]x_5=1[/tex] we get [tex]\left(-1 , 1 , 0 , -2 , 1\right)[/tex]. These two solutions are linearly independent (look at the third and fifth coordinates), and every solution equals
[tex]x_3\left(-3 , 2 , 1 , 0 , 0\right)+x_5\left(-1 , 1 , 0 , -2 , 1\right)[/tex]
so they form a fundamental system of solutions (check of the second vector in [tex]E_1[/tex]: [tex]-3+4+0-4+3=0[/tex]). Answer:
[tex]\left(-3 , 2 , 1 , 0 , 0\right),\ \left(-1 , 1 , 0 , -2 , 1\right)[/tex]

Solution:
Three vectors are linearly independent if and only if the matrix with these vectors as rows has rank [tex]3[/tex]. We replace the second row by [tex]\vec{x}_2-\vec{x}_1=\left(0 , a , 1 , 3-b\right)[/tex] and the third row by [tex]\vec{x}_3+(\vec{x}_2-\vec{x}_1)=\left(0 , 0 , 3 , 4-b\right)[/tex]:
[tex]\begin{bmatrix} 1 & 0 & a & b \\ 1 & a & 1+a & 3 \\ 0 & -a & 2 & 1 \end{bmatrix}\sim\begin{bmatrix} 1 & 0 & a & b \\ 0 & a & 1 & 3-b \\ 0 & 0 & 3 & 4-b \end{bmatrix}[/tex]
If [tex]a\ne 0[/tex], the leading entries [tex]1[/tex], [tex]a[/tex], [tex]3[/tex] stand in the first three columns, so the rank is [tex]3[/tex] and the vectors are independent for every [tex]b[/tex].
If [tex]a=0[/tex], the rows are [tex]\left(1 , 0 , 0 , b\right)[/tex], [tex]\left(0 , 0 , 1 , 3-b\right)[/tex], [tex]\left(0 , 0 , 3 , 4-b\right)[/tex]; subtracting three times the second row from the third gives
[tex]\left(0 , 0 , 0 , 4-b-3(3-b)\right)=\left(0 , 0 , 0 , 2b-5\right)[/tex]
so the rank is [tex]3[/tex] for [tex]b\ne\frac{5}{2}[/tex] and [tex]2[/tex] for [tex]b=\frac{5}{2}[/tex]. Hence the vectors are linearly dependent only for [tex]a=0[/tex], [tex]b=\frac{5}{2}[/tex], and they are linearly independent exactly when
[tex]\left(a , b\right)\ne\left(0 , \frac{5}{2}\right)[/tex]

Solution:
The rank of a matrix equals the dimension of the linear span of its columns (we denote the linear span by [tex]\ell[/tex]). Let [tex]\vec{a}_1,\vec{a}_2,\vec{a}_3[/tex] be the columns of [tex]A[/tex] and [tex]\vec{b}_1,\vec{b}_2,\vec{b}_3[/tex] the columns of [tex]B[/tex]. Since [tex]\operatorname{rank}A=\operatorname{rank}B=1[/tex], the columns of [tex]A[/tex] span a one-dimensional space [tex]\ell(\vec{u})[/tex] and the columns of [tex]B[/tex] span a one-dimensional space [tex]\ell(\vec{v})[/tex] for some vectors [tex]\vec{u},\vec{v}[/tex], i.e.
[tex]\vec{a}_j=\alpha_j\vec{u},\qquad \vec{b}_j=\beta_j\vec{v},\qquad j=1,2,3[/tex]
Hence the [tex]j[/tex]-th column of [tex]A+B[/tex] is
[tex]\vec{a}_j+\vec{b}_j=\alpha_j\vec{u}+\beta_j\vec{v}\in\ell(\vec{u},\vec{v})[/tex]
The space [tex]\ell(\vec{u},\vec{v})[/tex] has dimension at most [tex]2[/tex], and any three vectors of a space of dimension at most [tex]2[/tex] are linearly dependent. Therefore the three columns of [tex]A+B[/tex] are linearly dependent and
[tex]\operatorname{rank}(A+B)\le 2[/tex]
Remark: the same argument proves the general inequality [tex]\operatorname{rank}(A+B)\le\operatorname{rank}A+\operatorname{rank}B[/tex].