Solution. The system is not linear in \(x,y,z\), but it is linear in the reciprocals. Put \[u=\frac 1x,\qquad v=\frac 1y,\qquad w=\frac 1z,\] which is legitimate because none of \(x,y,z\) may be zero for the equations to make sense. The system becomes \[u+2v-4w=1,\qquad 2u+3v+8w=0,\qquad -u+9v+10w=5,\] and row reducing the augmented matrix gives \[u=-\frac {7}{13},\qquad v=\frac {54}{91},\qquad w=-\frac {8}{91}.\] Inverting, \[x=-\frac {13}{7},\qquad y=\frac {91}{54},\qquad z=-\frac {91}{8}.\] Substituting back into the first equation as a check: \(-\frac {7}{13}+2\cdot \frac {54}{91}-4\cdot \left (-\frac {8}{91}\right ) =-\frac {49}{91}+\frac {108}{91}+\frac {32}{91}=\frac {91}{91}=1.\)

The lesson is that a substitution can make a non-linear system linear; what must be checked is that the substitution is reversible on the domain in question, which here means \(x,y,z\neq 0\).

Solution. Divide the second equation by \(2\) and the third by \(3\): \[x_1+x_2+x_3=a,\qquad x_1+x_3=\tfrac {b}{2},\qquad x_2+x_3=\tfrac {c}{3}.\] Subtracting the second from the first gives \(x_2=a-\frac {b}{2}\). Substituting into the third gives \[x_3=\frac {c}{3}-x_2=\frac {c}{3}-a+\frac {b}{2},\] and then the second equation gives \[x_1=\frac {b}{2}-x_3=\frac {b}{2}-\frac {c}{3}+a-\frac {b}{2}=a-\frac {c}{3}.\] So the solution is \[x_1=a-\frac {c}{3},\qquad x_2=a-\frac {b}{2},\qquad x_3=\frac {b}{2}+\frac {c}{3}-a,\] and it exists and is unique for every \(a,b,c\): the coefficient matrix has rank \(3\), so by Theorem 3.2.3 there is never a leading element in the augmented column. A square system with an invertible coefficient matrix is consistent whatever the right-hand side.

Solution. Row reduce the augmented matrix, using the first row to clear the first column: \[\begin {pmatrix}1&2&-3&|&4\\3&-1&5&|&2\\4&1&a^2-14&|&a+2\end {pmatrix} \xrightarrow [\ r_3\rightarrow r_3-4r_1\ ]{\ r_2\rightarrow r_2-3r_1\ } \begin {pmatrix}1&2&-3&|&4\\0&-7&14&|&-10\\0&-7&a^2-2&|&a-14\end {pmatrix}.\] Now \(r_3\rightarrow r_3-r_2\) gives \[\begin {pmatrix}1&2&-3&|&4\\0&-7&14&|&-10\\0&0&a^2-16&|&a-4\end {pmatrix}.\] Everything is decided by the last row, whose coefficient is \(a^2-16=(a-4)(a+4)\).

(ii)

If \(a\neq 4\) and \(a\neq -4\) the coefficient is non-zero, the third variable is determined, and there is a unique solution.

(iii)

If \(a=4\) the last row is \(0=0\), which imposes no condition. There are two leading elements and three variables, so one free variable remains and there are infinitely many solutions.

(i)

If \(a=-4\) the last row reads \(0=-8\), which is impossible, so there is no solution.

Note that \(a=4\) and \(a=-4\) both make the determinant vanish, but they behave differently. A zero determinant tells you the solution is not unique; it does not tell you which of the other two cases you are in — for that the augmented column must be looked at.

Solution. The determinant of the coefficient matrix is \[\det \begin {pmatrix}2&3&1\\3&-1&\lambda \\1&7&-6\end {pmatrix}=-11(\lambda -8).\]

(i)

If \(\lambda \neq 8\) the determinant is non-zero, the coefficient matrix is invertible, and there is a unique solution for every \(\mu \).

For \(\lambda =8\) the determinant vanishes and \(\mu \) decides. Eliminate \(x\) using the third equation: \[r_1-2r_3:\ -11y+13z=5-2\mu ,\qquad r_2-3r_3:\ -22y+26z=2-3\mu .\] The left-hand side of the second is exactly twice that of the first, so the two are consistent only if \(2-3\mu =2(5-2\mu )\), that is \(\mu =8\).

(iii)

If \(\lambda =8\) and \(\mu =8\) the second equation is a consequence of the first, leaving two independent equations in three unknowns: infinitely many solutions.

(ii)

If \(\lambda =8\) and \(\mu \neq 8\) the two equations contradict each other: no solution.

Solution. Every one of these lines passes through the origin, since \((0,0)\) satisfies each equation — so the system is always consistent, and the question is only whether the three lines meet anywhere else.

(i)

Only the trivial solution. The three lines have exactly one common point, the origin. So at least two of them are genuinely different lines: they cross at the origin and nowhere else.

(ii)

Non-trivial solutions. If \((x_0,y_0)\neq (0,0)\) satisfies all three equations, then so does \(t(x_0,y_0)\) for every scalar \(t\), since the equations are homogeneous. The common solutions therefore form a whole line through the origin, and all three lines must be that line — they coincide.

The dividing line is rank: the solution set is a point when the coefficient matrix has rank \(2\), and a line when it has rank \(1\). (Rank \(0\) would mean all six coefficients are zero, in which case the “lines” are not lines at all and every point of the plane is a solution.)

Solution.

(a)

Always false — the reduced echelon form of a matrix is unique. Different routes may take different numbers of steps, but they arrive at the same matrix. This is exactly why it can be spoken of as “the reduced echelon form”.

(b)

Always true. Echelon form, unlike reduced echelon form, is not unique, because nothing forces the entries above a leading element to be cleared or the leading entries to be \(1\). For example \[\begin {pmatrix}1&2\\2&5\end {pmatrix}\longrightarrow \begin {pmatrix}1&2\\0&1\end {pmatrix} \qquad \text {and}\qquad \begin {pmatrix}1&2\\2&5\end {pmatrix}\longrightarrow \begin {pmatrix}2&5\\0&-1/2\end {pmatrix}\] are both echelon forms of the same matrix.

(c)

Sometimes false. A row of zeros means one equation carried no information, but it says nothing on its own about the number of solutions. The system \[x=1,\qquad y=2,\qquad 0=0\] has augmented matrix with reduced echelon form \(\begin {pmatrix}1&0&|&1\\0&1&|&2\\0&0&|&0\end {pmatrix}\) — a row of zeros, and exactly one solution. What produces infinitely many solutions is a free variable, that is, fewer leading elements than variables; a zero row only guarantees fewer leading elements than equations.

Solution.

(a)

Sometimes false. Having more unknowns than equations guarantees that there is no unique solution, but not that a solution exists. The system \[x_1+x_2+x_3+x_4+x_5=1,\qquad x_1+x_2+x_3+x_4+x_5=2\] (padded with a third equation \(0=0\)) is inconsistent. The correct statement is the homogeneous one: a homogeneous system with more unknowns than equations always has a non-trivial solution, which is Theorem 3.3.3(1).

(b)

Sometimes false. Extra equations may simply repeat information. Take \(x=1\), \(y=1\), \(z=1\) and then repeat two of them: five equations in three unknowns, perfectly consistent with the unique solution \((1,1,1)\).

(c)

Sometimes false, and the subtlety is which matrix the leading ones are counted in. The augmented matrix of a system of \(n\) equations in \(n\) unknowns has \(n+1\) columns. If one of the \(n\) leading ones falls in the last column, that row reads \(0=1\) and the system is inconsistent. For example \[\begin {pmatrix}1&0&|&0\\0&0&|&1\end {pmatrix}\] has two leading ones for a system of two equations in two unknowns and no solution at all. The statement becomes true if the \(n\) leading ones are required to lie in the coefficient columns.