Systems of linear equations

Solution 2.7.1

(See Exercise 2.6.) If \(a \ne 0\) or \(b \ne 0\), then the equation \(ax+by = c\) has infinitely many solutions. Indeed, if, say \(a \ne 0\), we can subtract \(by\) and divide by \(a\), which gives \(x = \frac{c-by}a\). Thus, for any \(y \in {\bf R}\), the pair \((x=\frac{c-by}a, y)\) is a solution. A similar analysis works if \(b \ne 0\). It remains to consider the case in which \(a=0\) and \(b=0\). In this case the solution set of the equation depends on \(c\):

  • If \(c = 0\), then any pair \((x,y)\) is a solution. Indeed: \(0 x + 0y= 0\) holds true then. Thus, if \(a=b=c=0\), there are infinitely many solutions.

  • If \(c \ne 0\), the equation \(0x+0y = c\) has no solution, since the left hand side is always 0, while the right hand side is nonzero. So, in the case \(a=b=0\) but \(c \ne 0\), there is no solution.

Solution 2.7.2

(See Exercise 2.10.) The matrix associated to the system is as follows, and we bring it to reduced row echelon form:

\[ \begin{align*} \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ -2 & -3 & 1 & 1 \\ 0 & 1 & -1 & 1 \end{array} \right ) \leadsto & \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 1 & -1 & 1 \\ 0 & 1 & -1 & 1 \end{array} \right ) \\ \leadsto & \left ( \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right ) \leadsto \left ( \begin{array}{ccc|c} \underline 1 & 0 & +1 & -2 \\ 0 & \underline 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right ) \end{align*} \]

We have two leading ones (underlined), so the third unknown \(x_3\) is a free variable and \(x_1\) and \(x_2\) are non-free, and we have \(x_2 = 1+x_3\) and \(x_1 = -2 -x_3\). Thus, the solution set is

\[ \{(-2-x_3, 1+x_3, x_3) \ | \ x_3 \in {\bf R} \}. \]

Solution 2.7.3

(See Exercise 2.12.) We apply Method 2.31. The matrix associated to the system is

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 1 & -1 & 0 & 1 & -3 \\ 1 & -1 & 3 & -2 & 0 \end{array} \right ). \]

We compute the reduced row-echelon form of that matrix using Gaussian eliminiation (Method 2.29): we subtract the first row from the third, which gives

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & -1 & 1 & -1 \\ 1 & -1 & 3 & -2 & 0 \end{array} \right ). \]

We then subtract the first row from the fourth:

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & -1 & 1 & -1 \\ 0 & 0 & 2 & -2 & 2 \end{array} \right ). \]

We add the second line to the third:

\[ \left ( \begin{array}{cccc|c} 1 & -1 & 1 & 0 & -2 \\ 0 & 0 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 2 & -2 & 2 \end{array} \right ). \]

We then add \((-2)\) times the second line to the fourth (equivalently, subtract \(2\) times the second line from the fourth):

\[ \left ( \begin{array}{cccc|c} {\underline 1} & -1 & 1 & 0 & -2 \\ 0 & 0 & {\underline 1} & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

This matrix is in row-echelon form, with the leading 1’s being underlined above. We finally bring it into reduced row-echelon form by subtracting the second from the first line, which gives

\[ \left ( \begin{array}{cccc|c} {\underline 1} & -1 & 0 & 1 & -3 \\ 0 & 0 & {\underline 1} & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right ). \]

The matrix has no entry of the form \(0 \ \dots \ 0 \ 1\), so the system does have a solution. The first column of the matrix corresponds to the variable \(x_1\) etc., so that the free variables are \(x_2\) and \(x_4\). We let \(x_2 = \alpha\), \(x_4 = \beta\), where \(\alpha\) and \(\beta\) are arbitrary real numbers. The non-free variables \(x_1\) and \(x_3\) are uniquely determined by \(\alpha\) and \(\beta\). To compute them, we use the equations obtained by the matrix

\[ \begin{align*} x_3 - \beta & = 1 \\ x_1 -\alpha + \beta & = -3 \end{align*} \]

which we solve as \(x_3 = 1 + \beta\) and \(x_1 = \alpha-\beta-3\). Thus, the solution set is

\[ \{(\alpha - \beta - 3, \alpha, 1+\beta, \beta) \ | \ \alpha, \beta \in {\bf R}\}. \]

Solution 2.7.4

(See Exercise 2.14.) Hint: we will apply Gaussian elimination, but it simplifies the calculations to do a certain change of rows first. (Why is that allowed?)

Solution 2.7.5

(See Exercise 2.17.) Suppose \(x_1 = 1-t\), \(x_2 = 2+3t\) and \(x_3 = 4t\). We have to determine whether there is some \(t\in {\bf R}\) such that for these choices of \(x_1, x_2, x_3\), we have a solution of the given system, i.e., whether

\[ \begin{align*} x_1 + x_2 + x_3 & = (1-t)+(2+3t)+4t & = 1 \\ x_1 - x_3 & = 1-t-4t & = 0. \end{align*} \]

Simplifying these equations gives the system

\[ \begin{align*} 6t + 3 & = 0 \\ 1-5t & = 0. \end{align*} \]

This system has no solutions, so there is no \(t \in {\bf R}\) such that the vector \((1-t, 2+3t, 4t)\) is a solution to the original system.

Solution 2.7.6

(See Exercise 2.19.) We substitute \(x_1 = 1+t\), \(x_2 = t+q\) and \(x_3 = -t+2q+1\) into the given equation and get the equation

\[ 3(1+t)+2(t+q)-(-t+2q+1)=5. \]

This simplifies to

\[ 6t+2=5 \]

which has the solution \(t = -\frac 12\). Since the variable \(q\) does not appear in that equation it is a free variable. Thus, for all \(q \in {\bf R}\), the vector

\[ (x_1 = 1-\frac 12, x_2 = -\frac 12 + q, x_3 = \frac 12 + 2q+1) = (\frac 12, -\frac 12+q, \frac 32 + 2q) \]

satisfies the requested conditions. Note that these are infinitely many solutions.

Solution 2.7.7

(See Exercise 2.20.) We have to find \(a_0, \dots, a_3\), so these are the unknowns. The conditions amount to the linear (!) system

\[ \begin{align*} p(1) & = a_0 + a_1 + a_2 + a_3 & = 0 \\ p(2) & = a_0 + a_1 \cdot 2+ a_2 \cdot 2^2 + a_3 \cdot 2^3 & = 3. \end{align*} \]

This can be rewritten as

\[ \begin{align*} a_0 + a_1 + a_2 + a_3 & = 0 \\ a_0 + 2 a_1 + 4 a_2 + 8 a_3 & = 3. \end{align*} \]

Using Gaussian elimination to solve this: the associated matrix is

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 1 & 2 & 4 & 8 & 3 \end{array} \right ). \]

Subtracting the first from the second row gives

\[ \left ( \begin{array}{cccc|c} 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 3 & 7 & 3 \end{array} \right ). \]

Subtracting the second from the first yields a reduced row echelon matrix:

\[ \left ( \begin{array}{cccc|c} 1 & 0 & -2 & -6 & -3 \\ 0 & 1 & 3 & 7 & 3 \end{array} \right ). \]

The variables \(a_0\) and \(a_1\) correspond to the leading 1’s, the variables \(a_2\) and \(a_3\) are therefore free variables. Thus, there are infintely many solutions. One solution, for \(a_2 = a_3 = 0\) is

\[ a_0 = -3, \ a_1 = 3, \]

so that

\[ p(x) = -3 + 3x \]

is a solution to the problem. Another solution would be \(a_2 = a_3 = 1\), which gives \(a_1 = -7\) and \(a_0 = 5\), i.e.,

\[ p(x) = 5 - 7 x + x^2 + x^3. \]

Solution 2.7.8

(See Exercise 2.25.) We solve, for each \(\lambda\in\mathbf R\), the system by Gaussian elimination (Method 2.31):

\[ \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & \lambda & 1+\lambda & 1\\ \lambda & 1 & 2 & 3 \end{array} \right ) \xrightarrow{R_3\leftarrow R_3-R_1} \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & \lambda & 1+\lambda & 1\\ 0 & 1 & 2 & 3 \end{array} \right ). \]

We would like to divide the first row by \(\lambda\), which however is only a legitimate row operation (Definition 2.28) if \(\lambda \ne 0\). So we distinguish two cases:

  • Case \(\lambda=0\). The matrix becomes
\[ \left ( \begin{array}{ccc|c} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 1 & 2 & 3 \end{array} \right ), \]

so \(x_3=1\), \(x_2=1\), and \(x_1\) is free. Hence the solution set is \(\{(t, 1, 1) \mid t \in \mathbf R\}\).

  • Case \(\lambda\neq0\). Divide row 2 by \(\lambda\) and eliminate the \(1\) in row 3, column 2:
\[ \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & 1 & \frac{1+\lambda}{\lambda} & \frac1\lambda\\ 0 & 1 & 2 & 3 \end{array} \right ) \xrightarrow{R_3\leftarrow R_3-R_2} \left ( \begin{array}{ccc|c} \lambda & 0 & 0 & 0\\ 0 & 1 & \frac{1+\lambda}{\lambda} & \frac1\lambda\\ 0 & 0 & \frac{\lambda-1}{\lambda} & \frac{3\lambda-1}{\lambda} \end{array} \right ). \]

The next possible division is by \(\lambda-1\), so we split again:

  • If \(\lambda=1\), the last row is \(0\;0\;0\mid2\), so there is no solution.

  • If \(\lambda\neq1\), then

\[ x_1=0,\qquad x_3=\frac{3\lambda-1}{\lambda-1}=\frac{1-3\lambda}{1-\lambda},\qquad x_2=3-2x_3. \]
So the solution is unique.

Summary:

  • If \(\lambda=0\): infinitely many solutions, namely \((t,1,1)\) with \(t\in\mathbf R\).

  • If \(\lambda=1\): no solution.

  • If \(\lambda\neq0,1\): unique solution \(\left(0,\,3-2\frac{1-3\lambda}{1-\lambda},\,\frac{1-3\lambda}{1-\lambda}\right)\).