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Section 3.3 Systems of linear equations revisited

We introduced techniques for solving systems of linear equations in Section 1.2. With the understanding of vectors that we have developed since then we are now ready to make a more detailed study of systems of linear equations.

Subsection 3.3.1 Rank

Definition 3.3.1.

Let \(A\) be a matrix. A pivot column of \(A\) is a column of \(A\) that corresponds to a column of \(\RREF(A)\) containing the leading \(1\) of some row.

The rank of \(A\text{,}\) written \(\rank(A)\text{,}\) is the number of pivot columns of \(A\text{.}\)

Suppose that \(A = \begin{bmatrix}1 \amp 3 \amp 0 \amp -2 \amp 4\\ 1 \amp 3 \amp 1 \amp 1 \amp 4\end{bmatrix}\text{.}\) Then \(\RREF(A) = \begin{bmatrix}1 \amp 3 \amp 0 \amp -2 \amp 4\\ 0 \amp 0 \amp 1 \amp 3 \amp 0\end{bmatrix}\text{.}\) Columns \(1\) and \(3\) are pivot columns, while columns \(2\text{,}\) \(4\text{,}\) and \(5\) are not. \(\rank(A) = 2\text{.}\)

By definition of the reduced row echelon form, each non-zero row of \(\RREF(A)\) has a leading \(1\text{,}\) while of course the zero rows do not have leading \(1\)s, so there is a correspondence between pivot columns and non-zero rows of \(\RREF(A)\text{.}\)

The reason for the word "nullity" in the name of the next theorem will be given in Section 4.6.

Suppose our system is \([A|\vec{b}]\text{.}\) After row-reducing we see that the non-pivot columns of \(A\) correspond to free variables of the system, so

\begin{align*} \rank(A) + \amp \text{\# of free variables of the system}\\ \amp = \text{\# of pivot columns of} A + \text{\# of non-pivot columns of} A\\ \amp = \text{\# of columns of} A \\ \amp = n\text{.} \end{align*}

The rank-nullity theorem offers one explanation of why the number of direction vectors for a line or plane in \(\mathbb{R}^3\text{,}\) plus the number of general equations required to describe it, adds up to \(3\) (we first saw this phenomenon in Section 2.3).

Consider a line in \(\mathbb{R}^3\) with general equations \(a_1x+b_1y+c_1z=d_1\) and \(a_2x+b_2y+c_2z=d_2\text{.}\) As long as these equations are not multiples of each other the coefficient matrix of the system, \(\begin{bmatrix}a_1 \amp b_1 \amp c_1 \\ a_2 \amp b_2 \amp c_2\end{bmatrix}\) will have rank \(2\text{.}\) By the rank-nullity theorem the system of equations describing the line will therefore have \(3-2 = 1\) free variable, and (as we saw when we converted between the forms for a line in Section 2.3) that one free variable corresponds to one direction vector for the line.

Similarly, if we have a plane in \(\mathbb{R}^3\) it can be described by a single general equation. If we think of that single equation as a system of one equation in three variables, we see that the system has rank \(1\) (as long as it is not the trivial equation \(0x+0y+0z=0\)). By the rank-nullity theorem it therefore has \(2\) free variables, and those free variables correspond to the direction vectors of the plane.

Suppose that we have a system \([A|\vec{b}]\text{,}\) where \(A\) is an \(m \times n\) matrix. There are three possibilities:

  • The system is inconsistent. In this case it has no solutions, by definition.

  • The system is consistent and \(\rank(A) = n\text{.}\) Then every column of \(A\) is a pivot column and (by Lemma 3.3.3) there are exactly \(n\) non-zero rows of \(\RREF(A)\text{,}\) so the form of the reduced row echelon form of the system is \(\matr{cccc|c}{1 \amp 0 \amp \cdots \amp 0 \amp c_1 \\ 0 \amp 1 \amp \cdots \amp 0 \amp c_2 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp 1 \amp c_n \\ 0 \amp 0 \amp \cdots \amp 0 \amp 0 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp 0 \amp 0}\text{.}\)

    A system of this form has unique solution \(x_1 = c_1, \ldots, x_n = c_n\text{.}\)

  • The system is consistent and \(\rank(A) \lt n\text{.}\) Then by Theorem 3.3.4 the system has at least one free variable, and since each value of the free variable gives rise to a solution the system has infinitely many solutions.

Subsection 3.3.2 Homogeneous systems

Definition 3.3.7.

A system of linear equations is called homogeneous if all of the constants on the right of the equals signs are \(0\text{.}\) That is, if the augmented matrix of the system has the form \([A|\vec{0}]\text{.}\)

Note 3.3.8.

Homogeneous systems are exactly the ones that arose in Section 3.2 when we tested whether or not a collection of vectors is linearly independent.

Every homogeneous system is consistent - just choose the values of all of the variables to be \(0\text{.}\) There may or may not be other solutions, depending on the system.

Suppose that the system \([A|\vec{0}]\) has \(n\) variables and \(m\) equations (so \(A\) is an \(m \times n\) matrix), with \(m \lt n\text{.}\) Then, by Theorem 3.3.4,

\begin{align*} \text{\# free variables} \amp = n - \rank(A)\\ \amp \geq n - m\\ \amp \gt 0\text{.} \end{align*}

Thus the system has at least one free variable, so it has infinitely many solutions.

Suppose that \(\vec{v_1}, \ldots, \vec{v_k}\) are distinct vectors in \(\mathbb{R}^n\text{,}\) with \(k \geq n\text{,}\) and let \(A\) be the matrix with columns \(\vec{v_1}, \ldots, \vec{v_k}\text{.}\) Then the system \([A|\vec{0}]\) is a homogeneous system with more variables (\(k\)) than equations (\(n\)), so by Theorem 3.3.9 this system has infinitely many solutions. In particular, it does not have a unique solution, so by Theorem 3.2.5 the vectors \(\vec{v_1}, \ldots, \vec{v_k}\) are linearly dependent.

Note 3.3.11.

If we have a homogeneous system with \(n\) variables and \(m\) equations with \(m \geq n\) then Theorem 3.3.9 does not apply. The system could have exactly one solution or it could have infinitely many solutions, depending on the specific system.

Similarly, if we have \(k\) vectors in \(\mathbb{R}^n\) with \(k \leq n\) then Theorem 3.3.10 does not apply. The vectors could be linearly dependent or linearly independent, depending on the specific vectors.

The vectors \(\begin{bmatrix}1\\0\\2\\1\end{bmatrix}, \begin{bmatrix}-3\\ \pi \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ 1\\ 1\\ 1\end{bmatrix}, \begin{bmatrix}15 \\ -\sqrt{2} \\ 3 \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\1\end{bmatrix}\) form a linearly dependent set, because they are \(5\) vectors in \(\mathbb{R}^4\text{.}\) If we wanted to actually find a non-trivial linear combination of these vectors that equals \(\vec{0}\) we could follow the methods from Section 3.2, but if all we need to know is that they are linearly dependent then no calculation is required.

By contrast, the vectors \(\begin{bmatrix}1\\0\\2\\1\end{bmatrix}, \begin{bmatrix}-3\\ \pi \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ 1\\ 1\\ 1\end{bmatrix}\) might be linearly dependent or linearly independent - without doing any calculations there is no way to tell. In fact, they are independent, as you can check.

Subsection 3.3.3 The fundamental theorem

The next theorem is very important - so important that it is often called the "Fundamental Theorem of Linear Algebra". As the course goes on we will be adding to this theorem, and it will eventually grow to touch nearly every topic of the course. Before stating it we need one definition (which will reappear in Section 4.3).

Definition 3.3.13.

For any natural number \(n\text{,}\) the \(n \times n\) identity matrix is the \(n \times n\) matrix \(I_n = \begin{bmatrix}1 \amp 0 \amp 0 \amp \cdots \amp 0 \\ 0 \amp 1 \amp 0 \amp \cdots \amp 0 \\ 0 \amp 0 \amp 1 \amp \cdots \amp 0 \\ \vdots \amp \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp 0 \amp \cdots \amp 1\end{bmatrix}\text{.}\)

We will prove \(1 \implies 3 \implies 2 \implies 6 \implies 1\text{,}\) \(2 \iff 4\text{,}\) \(3 \implies 5\text{,}\) and \(5 \implies 6\) (take a moment to convince yourself that you can get from any number to any other in the diagram below).

Figure 3.3.15. A diagram of the implications to be proved.

\(1 \implies 3\text{:}\) Suppose that \(\RREF(A) = I_n\text{.}\) When we do row operations to \([A|\vec{b}]\) the augmented column will change to some other vector \(\vec{c}\text{,}\) so \([A|\vec{b}] \to [\RREF(A)|\vec{c}] = [I_n|\vec{c}]\text{.}\) That is, the reduced row echelon form of the system has the form \(\matr{cccc|c}{1 \amp 0 \amp \cdots \amp 0 \amp c_1 \\ 0 \amp 1 \amp \cdots \amp 0 \amp c_2 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp 1 \amp c_n}\text{,}\) from which we see that the system \([A|\vec{b}]\) has a unique solution (namely \(x_1 = c_1, \ldots, x_n=c_n\)).

\(3 \implies 2\text{:}\) Suppose that, for every \(\vec{b}\text{,}\) the system \([A|\vec{b}]\) has a unique solution. Then this is true in particular for \(\vec{b}=\vec{0}\text{,}\) so the system \([A|\vec{0}]\) has a unique solution.

\(2 \implies 6\text{:}\) Suppose that the system \([A|\vec{b}]\) has a unique solution. Because the solution is unique this system has no free variables, so by the Rank-Nullity Theorem (Theorem 3.3.4) we get \(\rank(A) = n\text{.}\)

\(6 \implies 1\text{:}\) Suppose that \(\rank(A) = n\text{.}\) Then every column of \(A\) is a pivot column, and you can check that the only \(n \times n\) matrix in reduced row echelon form where every column is a pivot column is \(I_n\text{.}\) Thus \(\RREF(A) = I_n\text{.}\)

\(2 \iff 4\text{:}\) We already proved this: Theorem 3.2.5.

\(3 \implies 5\text{:}\) Suppose that, for every \(\vec{b}\text{,}\) the system \([A|\vec{b}]\) has a unique solution. Then in particular the system \([A|\vec{b}]\) is consistent, so by Theorem 3.1.4 \(\vec{b}\) is in the span of the columns of \(A\text{.}\) Since \(\vec{b}\) was arbitrary, the span of the columns of \(A\) is all of \(\mathbb{R}^n\text{.}\)

\(5 \implies 6\text{:}\) We prove this by contrapositive, that is, we prove that if \(6\) is false then \(5\) is also false. So suppose that \(\rank(A) \neq n\text{.}\) Then \(\rank(A) \lt n\text{,}\) so by Lemma 3.3.3 the reduced row echelon form of \(A\) has at least one row that is all zero; in particular, by definition of the reduced row echelon form, this means that the bottom row of \(\RREF(A)\) is all zero. Let \(\vec{c} = \begin{bmatrix}0 \\ 0 \\ \vdots \\ 0 \\ 1\end{bmatrix}\text{.}\) Since \(A\) is row-equivalent to \(\RREF(A)\) there is some sequence of row operations that takes \(\RREF(A)\) to \(A\text{.}\) Applying that sequence takes the augmented matrix \([\RREF(A) | \vec{c}]\) to another augmented matrix \([A | \vec{b}]\) for some \(\vec{b}\text{.}\) Since the system \([\RREF(A) | \vec{c}]\) is inconsistent, so is the system \([A | \vec{b}]\text{;}\) by Theorem 3.1.4 the vector \(\vec{b}\) is not in the span of the columns of \(A\text{,}\) so the span of the columns of \(A\) is not all of \(\mathbb{R}^n\) (i.e., statement \(5\) is false).

Exercises 3.3.4 Exercises

1.

Find the rank of each of the matrices in Problem 1, that is:

  1. \(\displaystyle \begin{bmatrix} 1\amp -1 \amp 2\\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)

  2. \(\displaystyle \begin{bmatrix} 2\amp 1 \amp -1 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\)

  3. \(\displaystyle \begin{bmatrix}1\amp -2 \amp 3 \amp 5 \\ 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)

  4. \(\displaystyle \begin{bmatrix} 1\amp 0 \amp 0 \amp 3 \amp 1 \\ 0 \amp 0 \amp 0 \amp 1 \amp 1\\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \end{bmatrix}\)

  5. \(\displaystyle \begin{bmatrix} 1 \amp 1 \\ 0 \amp 1\end{bmatrix}\)

  6. \(\displaystyle \begin{bmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 1 \end{bmatrix}\)

Hint.
The rank of a matrix \(A\) is the number of leading \(1\)s in any row-echelon matrix to which \(A\) can be carried by row operations. (For the definition of the row-echelon form, see Definition 1.2.8.)
Answer.

  1. rank \(2\)

  2. rank \(1\)

  3. rank \(2\)

  4. rank \(3\)

  5. rank \(2\)

  6. rank \(1\)

Solution.

  1. By exchanging row \(2\) and \(3\text{,}\) we arrive at a matrix in row-echelon form with two leading \(1\)s. Thus, the rank of the matrix is \(2\text{.}\)

  2. By dividing the first row by \(2\text{,}\) we arrive at a matrix in row-echelon form with one leading \(1\text{.}\) Thus, the rank of the matrix is \(1\text{.}\)

  3. Since the matrix is in REF, we just need to count the leading \(1\)s: the rank of the matrix is \(2\text{.}\)

  4. Since the matrix is in REF, we just need to count the leading \(1\)s: the rank of the matrix is \(3\text{.}\)

  5. Since the matrix is in REF, we just need to count the leading \(1\)s: the rank of the matrix is \(2\text{.}\)

  6. To bring the matrix in REF, we substract the first row from the second and from the third row. The resulting matrix is

    \begin{equation*} \begin{bmatrix} 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{bmatrix}, \end{equation*}
    which has one leading \(1\) and hence rank \(1\text{.}\)

2.

Find all solutions (if any) to each of the following systems of linear equations.

  1. \begin{align*} x + y + 2z \amp = 8\\ 3x - y + z \amp = 0\\ -x + 3y + 4z \amp = -4 \end{align*}

  2. \begin{align*} -2x + 3y + 3z \amp = -9\\ 3x - 4y + z \amp = 5\\ -5x + 7y + 2z \amp = -14 \end{align*}

  3. \begin{align*} x + y - z \amp = 10\\ -x + 4y + 5z \amp = -5\\ x + 6y + 3z \amp = 15 \end{align*}

  4. \begin{align*} x + 2y - z \amp = 2\\ 2x + 5y - 3z \amp = 1\\ x + 4y - 3z \amp = 3 \end{align*}

  5. \begin{align*} 5x + y \amp = 2\\ 3x - y - 2z \amp = 1\\ x + y - z \amp = 5 \end{align*}

  6. \begin{align*} 3x - 2y + z \amp = -2\\ x - y + 3z \amp = 5\\ -x + y + z \amp = -1 \end{align*}

  7. \begin{align*} x + y + z \amp = 2\\ x + z \amp = 1\\ 2x + 5y + 2z \amp = 7 \end{align*}

  8. \begin{align*} x + 2y - 4z \amp = 10\\ 2x - y + 2z \amp = 5\\ x + y - 2z \amp = 7 \end{align*}

Hint.
Write the system of linear equations as an augmented matrix and then row reduce.
Answer.

  1. Answer.

    The unique solution is given by \((x,y,z)= (17,31,-20)\text{.}\)

  2. Answer.

    There is one free variable and the solution set is given by
    \begin{equation*} \left\{ (-21 - 15z,-17 - 11 z,z) \vert z \in \mathbb{R} \right\}. \end{equation*}

  3. Answer.

    There is one free variable and the solution set is given by
    \begin{equation*} \left\{ \left(9+\frac{9}{5}z, 1 - \frac{4}{5},z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

  4. Answer.

    No solution.

  5. Answer.

    The unique solution is given by \((x,y,z)= \left(-\frac{3}{16},\frac{47}{16},-\frac{9}{4}\right)\)

  6. Answer.

    The unique solution is given by \((x,y,z)= \left(-7,-9,1\right)\text{.}\)

  7. Answer.

    There is one free variable and the solution set is given by
    \begin{equation*} \left\{ \left(1-z, 1, z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

  8. Answer.

    There is one free variable and the solution set is given by
    \begin{equation*} \left\{ \left(4, 3+2z, z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

Solution.
For each of these solutions, the first step is to turn the system of linear equations into an augmented matrix form. Then, we row reduce until we (at least almost) reach the RREF of the the augmented matrix.

  1. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp 2 \amp 8 \\ 3 \amp -1 \amp 1 \amp 0 \\ -1 \amp 3 \amp 4 \amp -4 } \overset{-3R_{1}+R_{2}}{\underset{R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 1 \amp 2 \amp 8 \\ 0 \amp -4 \amp -5 \amp -24 \\ 0 \amp 4 \amp 6 \amp 4 }\\ \overset{R_{2}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 1 \amp 2 \amp 8 \\ 0 \amp -4 \amp -5 \amp -24 \\ 0 \amp 0 \amp 1 \amp -20 } . \end{align*}

    From this, we can already see that the system of linear equations has a unique solution. To find it, we proceed until we arrive at the RREF:

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp 2 \amp 8 \\ 0 \amp -4 \amp -5 \amp -24 \\ 0 \amp 0 \amp 1 \amp -20 } \overset{ 5R_{3}+R_{2} }{ \underset{ -2R_{3}+R_{1} }{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 1 \amp 0 \amp 48 \\ 0 \amp -4 \amp 0 \amp -124 \\ 0 \amp 0 \amp 1 \amp -20 }\\ \overset{\frac{-1}{4}R_{2}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 1 \amp 0 \amp 48 \\ 0 \amp 1 \amp 0 \amp 31 \\ 0 \amp 0 \amp 1 \amp -20 }\\ \overset{-R_{2}+R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp 17 \\ 0 \amp 1 \amp 0 \amp 31 \\ 0 \amp 0 \amp 1 \amp -20 }. \end{align*}

    We conclude that the unique solution is given by \((x,y,z)= (17,31,-20)\text{.}\)

  2. Solution.

    \begin{align*} \matr{ccc|c}{ -2 \amp 3 \amp 3 \amp -9 \\ 3 \amp -4 \amp 1 \amp 5 \\ -5 \amp 7 \amp 2 \amp -14 } \overset{-2R_{3}}{\underset{2 R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ -2 \amp 3 \amp 3 \amp -9 \\ 6 \amp -8 \amp 2 \amp 10 \\ 10 \amp -14 \amp -4 \amp 28 }\\ \overset{3R_{1}+R_{2}}{\underset{5 R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ -2 \amp 3 \amp 3 \amp -9 \\ 0 \amp 1 \amp 11 \amp -17 \\ 0 \amp 1 \amp 11 \amp -17 }. \end{align*}

    From this, we already see that there is not a unique solution because the last two rows are identical. We proceed to compute the RREF:

    \begin{align*} \matr{ccc|c}{ -2 \amp 3 \amp 3 \amp -9 \\ 0 \amp 1 \amp 11 \amp -17 \\ 0 \amp 1 \amp 11 \amp -17 } \overset{-R_{2}+R_{3}}{\underset{-3R_{2}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ -2 \amp 0 \amp -30 \amp 42 \\ 0 \amp 1 \amp 11 \amp -17 \\ 0 \amp 0 \amp 0 \amp 0 }\\ \overset{\frac{-1}{2}R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp 15 \amp -21 \\ 0 \amp 1 \amp 11 \amp -17 \\ 0 \amp 0 \amp 0 \amp 0 } \end{align*}

    We conclude that \(z\) is a free variable and the solution is given by

    \begin{equation*} \left\{ (-21 - 15z,-17 - 11 z,z) \vert z \in \mathbb{R} \right\}. \end{equation*}

  3. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp -1 \amp 10 \\ -1 \amp 4 \amp 5 \amp -5 \\ 1 \amp 6 \amp 3 \amp 15 } \overset{R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 1 \amp -1 \amp 10 \\ 0 \amp 5 \amp 4 \amp 5 \\ 0 \amp 5 \amp 4 \amp 5 } \end{align*}

    We already see that there is no unique solution, because the last two rows are identical. We proceed:

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp -1 \amp 10 \\ 0 \amp 5 \amp 4 \amp 5 \\ 0 \amp 5 \amp 4 \amp 5 } \overset{-R_{2}+R_{3}}{\underset{-\frac{1}{5}R_{2}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp -\frac{9}{5} \amp 9 \\ 0 \amp 5 \amp 4 \amp 5 \\ 0 \amp 0 \amp 0 \amp 0 }\\ \overset{\frac{1}{5}R_{2}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp -\frac{9}{5} \amp 9 \\ 0 \amp 1 \amp \frac{4}{5} \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 } \end{align*}

    We conclude that \(z\) is a free variable and the solution is given by

    \begin{equation*} \left\{ \left(9+\frac{9}{5}z, 1 - \frac{4}{5},z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

  4. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp 2 \amp -1 \amp 2 \\ 2 \amp 5 \amp -3 \amp 1 \\ 1 \amp 4 \amp -3 \amp 3 } \overset{-2R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 2 \amp -1 \amp 2 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 2 \amp -2 \amp 1 }\\ \overset{-2R_{2}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 2 \amp -1 \amp 2 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 }. \end{align*}

    We see from this that the system has no solution, since the last line means

    \begin{equation*} 0x+0y+0z=1. \end{equation*}

  5. Solution.

    \begin{align*} \matr{ccc|c}{ 5 \amp 1 \amp 0 \amp 2 \\ 3 \amp -1 \amp -2 \amp 1 \\ 1 \amp 1 \amp -1 \amp 5 } \overset{5R_{2}}{\underset{5R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 1 \amp 0 \amp 2 \\ 15 \amp -5 \amp -10 \amp 5 \\ 5 \amp 5 \amp -5 \amp 25 }\\ \overset{-3R_{1}+R_{2}}{\underset{-R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 1 \amp 0 \amp 2 \\ 0 \amp -8 \amp -10 \amp -1 \\ 0 \amp 4 \amp -5 \amp 23 }\\ \overset{\frac{1}{2}R_{2}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 5 \amp 1 \amp 0 \amp 2 \\ 0 \amp -8 \amp -10 \amp -1 \\ 0 \amp 0 \amp -10 \amp \frac{45}{2} }\\ \overset{-R_{3}+R_{2}}{\underset{-\frac{1}{10}R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 1 \amp 0 \amp 2 \\ 0 \amp -8 \amp 0 \amp -\frac{47}{2} \\ 0 \amp 0 \amp 1 \amp -\frac{9}{4} }\\ \overset{-\frac{1}{8}R_{2}}{\underset{-R_{2}'+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 0 \amp 0 \amp -\frac{15}{16} \\ 0 \amp 1 \amp 0 \amp \frac{47}{16} \\ 0 \amp 0 \amp 1 \amp -\frac{9}{4} }\\ \overset{\frac{1}{5}R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp -\frac{3}{16} \\ 0 \amp 1 \amp 0 \amp \frac{47}{16} \\ 0 \amp 0 \amp 1 \amp -\frac{9}{4} } \end{align*}

    We conclude that the unique solution is given by \((x,y,z)= \left(-\frac{3}{16},\frac{47}{16},-\frac{9}{4}\right)\text{.}\)

  6. Solution.

    \begin{align*} \matr{ccc|c}{ 3 \amp -2 \amp 1 \amp -2 \\ 1 \amp -1 \amp 3 \amp 5 \\ -1 \amp 1 \amp 1 \amp -1 } \overset{3R_{2}}{\underset{3R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp -2 \amp 1 \amp -2 \\ 3 \amp -3 \amp 9 \amp 15 \\ -3 \amp 3 \amp 3 \amp -3 }\\ \overset{-R_{1}+R_{2}}{\underset{R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp -2 \amp 1 \amp -2 \\ 0 \amp -1 \amp 8 \amp 17 \\ 0 \amp 1 \amp 4 \amp -5 }\\ \overset{R_{2}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 3 \amp -2 \amp 1 \amp -2 \\ 0 \amp -1 \amp 8 \amp 17 \\ 0 \amp 0 \amp 12 \amp 12 }\\ \overset{\frac{1}{12}R_{3}}{\underset{-8R_{3}' + R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp -2 \amp 1 \amp -2 \\ 0 \amp -1 \amp 0 \amp 9 \\ 0 \amp 0 \amp 1 \amp 1 }\\ \overset{-2R_{2}+R_{1}}{\underset{-R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp 0 \amp 1 \amp -20 \\ 0 \amp 1 \amp 0 \amp -9 \\ 0 \amp 0 \amp 1 \amp 1 }\\ \overset{-R_{3}+R_{1}}{\underset{\frac{1}{3}R_{1}'}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp -7 \\ 0 \amp 1 \amp 0 \amp -9 \\ 0 \amp 0 \amp 1 \amp 1 }. \end{align*}

    We conclude that the unique solution is given by \((x,y,z)= \left(-7,-9,1\right)\text{.}\)

  7. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp 1 \amp 2 \\ 1 \amp 0 \amp 1 \amp 1 \\ 2 \amp 5 \amp 2 \amp 7 } \overset{-R_{1}+R_{2}}{\underset{-2R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 1 \amp 1 \amp 2 \\ 0 \amp -1 \amp 0 \amp -1 \\ 0 \amp 3 \amp 0 \amp 3 } \end{align*}

    We already see that there is not going to be a unique solution. We proceed:

    \begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp 1 \amp 2 \\ 0 \amp -1 \amp 0 \amp -1 \\ 0 \amp 3 \amp 0 \amp 3 } \overset{3R_{2}+R_{3}}{\underset{R_{2}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp 1 \amp 1 \\ 0 \amp -1 \amp 0 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}

    We conclude that there is one free variable and the solution set is given by

    \begin{equation*} \left\{ \left(1-z, 1, z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

  8. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp 2 \amp -4 \amp 10 \\ 2 \amp -1 \amp 2 \amp 5 \\ 1 \amp 1 \amp -2 \amp 7 } \overset{-R_{1}+R_{3}}{\underset{-2R_{1}+R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 2 \amp -4 \amp 10 \\ 0 \amp -5 \amp 10 \amp -15 \\ 0 \amp -1 \amp 2 \amp -3 }. \end{align*}

    From this, we see that there is no unique solution. We proceed:

    \begin{align*} \matr{ccc|c}{ 1 \amp 2 \amp -4 \amp 10 \\ 0 \amp -5 \amp 10 \amp -15 \\ 0 \amp -1 \amp 2 \amp -3 } \overset{-\frac{1}{5}R_{2}+R_{3}}{\underset{5R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 10 \amp -20 \amp 50 \\ 0 \amp -5 \amp 10 \amp -15 \\ 0 \amp 0 \amp 0 \amp 0 }\\ \overset{2R_{2}+R_{1}}{\underset{\frac{-1}{5}R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 5 \amp 0 \amp 0 \amp 20 \\ 0 \amp 1 \amp -2 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}

    We conclude that there is one free variable and the solution set is given by

    \begin{equation*} \left\{ \left(4, 3+2z, z\right) \vert z \in \mathbb{R} \right\}. \end{equation*}

3.

Do the three lines, \(x + 2y =1\text{,}\) \(2x -y =1\text{,}\) and \(4x + 3y=3\) have a common point of intersection? If so, find the point and if not, tell why they don't have such a common point of intersection.
Hint.
Write the system of linear equations as an augmented matrix and row-reduce.
Answer.
The three lines intersect in the point \(\left( \frac{3}{5}, \frac{1}{5}\right)\text{.}\)
Solution.

The question can be rephrased to: "Does the system of linear equations have a solution?", and we know that such a question is easily answered using an augmented matrix, which we row-reduce:

\begin{align*} \matr{cc|c}{ 1 \amp 2 \amp 1 \\ 2 \amp -1 \amp 1 \\ 4 \amp 3 \amp 3 } \overset{-2R_{1}+R_{2}}{\underset{-4R_{1}+R_{3}}{\longrightarrow}} \matr{cc|c}{ 1 \amp 2 \amp 1 \\ 0 \amp -5 \amp -1 \\ 0 \amp -5 \amp -1 } . \end{align*}

We can already see that there is going to be a solution, and that said solution is unique. We proceed to find the intersection point:

\begin{align*} \matr{cc|c}{ 1 \amp 2 \amp 1 \\ 0 \amp -5 \amp -1 \\ 0 \amp -5 \amp -1 } \overset{-R_{2}+R_{3}}{\underset{2R_{2}+5R_{1}}{\longrightarrow}} \amp\matr{cc|c}{ 5 \amp 0 \amp 3 \\ 0 \amp -5 \amp -1 \\ 0 \amp 0 \amp 0 }\\ \overset{\frac{-1}{5}R_{2}}{\underset{\frac{1}{5}R_{1}}{\longrightarrow}} \amp\matr{cc|c}{ 1 \amp 0 \amp \frac{3}{5} \\ 0 \amp 1 \amp \frac{1}{5} \\ 0 \amp 0 \amp 0 } \end{align*}

We conclude that the three lines intersect in the point \(\left( \frac{3}{5}, \frac{1}{5}\right)\text{.}\)

4.

Do the three planes, \(x+y - 3z=2\text{,}\) \(2x+y+z =1\) and \(3x+2y-2z=0\) have a common point of intersection? If so, find one and if not, tell why there is no such point.
Hint.
Write the system of linear equations as an augmented matrix and row-reduce.
Answer.
The three planes do not intersect.
Solution.

The question can be rephrased to: "Does the system of linear equations have a solution?", and we know that such a question is easily answered using an augmented matrix, which we row-reduce:

\begin{align*} \matr{ccc|c}{ 1 \amp 1 \amp -3 \amp 2 \\ 2 \amp 1 \amp 1 \amp 1 \\ 3 \amp 2 \amp -2 \amp 0 } \amp \overset{-2R_{1}+R_{2}}{\underset{-3R_{1}+R_{3}}{\longrightarrow}} \amp\matr{ccc|c}{ 1 \amp 1 \amp -3 \amp 2 \\ 0 \amp -1 \amp 7 \amp -3 \\ 0 \amp -1 \amp 7 \amp -6 }\\ \amp \overset{-R_{2}+R_{3}}{\longrightarrow} \amp\matr{ccc|c}{ 1 \amp 1 \amp -3 \amp 2 \\ 0 \amp -1 \amp 7 \amp -3 \\ 0 \amp 0 \amp 0 \amp -3 } \end{align*}

Since the last line of the augmented matrix reads as \(0x + 0y + 0z = -3\text{,}\) we see that there is no solution to the system. In other words, the three planes do not intersect.

5.

My system of equations has a solution \((x,y,z)= (1,2,4)\text{.}\) The associated homogenous system has basic solutions \((x,y,z) =(1,0,1)\) and \((x,y,z)= (0,1,-1)\text{.}\) What is the general solution of my system of equations?

6.

Consider the following augmented matrix in which \(*\) denotes an arbitrary number and \(\blacksquare\) denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
\begin{equation*} \matr{ccc|c} { \blacksquare\amp * \amp * \amp *\\ 0 \amp \blacksquare \amp * \amp * \\ 0 \amp 0 \amp \blacksquare \amp *} \end{equation*}
Answer.
The system is consistent and has a unique solution.
Solution.
The system is consistend and has a unique solution. To prove this, let us give names to the entries of the matrix:
\begin{equation*} \matr{ccc|c} { a \amp b \amp c \amp d\\ 0 \amp e \amp f \amp g \\ 0 \amp 0 \amp h \amp i} \end{equation*}
By assumption, \(a, e, h\) are non-zero, so in particular, we may manipulate the augmented matrix as follows:
\begin{align*} \matr{ccc|c} { a \amp b \amp c \amp d\\ 0 \amp e \amp f \amp g \\ 0 \amp 0 \amp h \amp i} \overset{\frac{1}{h}R_{3}}{\longrightarrow} \amp \matr{ccc|c} { a \amp b \amp c \amp d\\ 0 \amp e \amp f \amp g \\ 0 \amp 0 \amp 1 \amp \frac{i}{h} } \\ \overset{-cR_{3}+R_{1}}{\underset{-fR_{3}+R_{2}}{\longrightarrow}} \amp \matr{ccc|c} { a \amp b \amp 0 \amp d - c\frac{i}{h}\\ 0 \amp e \amp 0 \amp g - f\frac{i}{h} \\ 0 \amp 0 \amp 1 \amp \frac{i}{h} } . \end{align*}
Since \(e\neq 0\) by assumption, we can divide by it:
\begin{align*} \matr{ccc|c} { a \amp b \amp 0 \amp d - c\frac{i}{h}\\ 0 \amp e \amp 0 \amp g - f\frac{i}{h} \\ 0 \amp 0 \amp 1 \amp \frac{i}{h} } \overset{\frac{1}{e}R_{2}}{\longrightarrow} \amp \matr{ccc|c} { a \amp b \amp 0 \amp d - c\frac{i}{h}\\ 0 \amp 1 \amp 0 \amp \frac{gh-fi}{he} \\ 0 \amp 0 \amp 1 \amp \frac{i}{h} } \\ \overset{-bR_{2}+R_{1}}{\longrightarrow} \amp \matr{ccc|c} { a \amp 0 \amp 0 \amp d - c\frac{i}{h} - b \frac{gh-fi}{he}\\ 0 \amp 1 \amp 0 \amp \frac{gh-fi}{he} \\ 0 \amp 0 \amp 1 \amp \frac{i}{h} } . \end{align*}
Lastly, since \(a\neq 0\) by assumption, we can divide by it and see that the unique solution to the system is given by \(\left( \frac{dhe-cie - bgh+bfi}{ahe} , \frac{gh-fi}{he}, \frac{i}{h}\right)\text{.}\)

7.

Suppose a system of equations has fewer equations than variables. Will such a system necessarily be consistent? If so, explain why and if not, give an example which is not consistent.
Answer.
No, it does not have to be consistent.
Solution.
No, it does not have to be consistent: consider
\begin{equation*} \matr{ccc|c} { 1 \amp 0 \amp 0 \amp 5 \\ 1 \amp 0 \amp 0 \amp -5 }. \end{equation*}
This system consists of only two equations but three variables. Nonetheless, the first row asks for \(x=5\text{,}\) while the second row asks for \(x=-5\text{,}\) so there cannot be any solution and the system is inconsistent.

8.

Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain.

9.

Choose \(h\) and \(k\) such that the augmented matrix shown has each of the following:

  1. one solution

  2. no solution

  3. infinitely many solutions

\begin{equation*} \matr{cc|c} { 1 \amp h \amp 2\\ 2 \amp 4 \amp k } \end{equation*}
Answer.

  1. If \(h \neq 2\) and \(k\) is arbitrary, then the system has a unique solution.

  2. If \(h = 2\) and \(k\neq 4\text{,}\) then the system has no solutions.

  3. If \(h = 2\) and \(k = 4\text{,}\) then the system has infinitely many solutions.

Solution.

We row reduce:

\begin{equation*} \matr{cc|c}{ 1 \amp h \amp 2 \\ 2 \amp 4 \amp k } \overset{-2R_{1}+R_{2}}{\longrightarrow} \matr{cc|c}{ 1 \amp h \amp 2 \\ 0 \amp 4-2h \amp k-4 }. \end{equation*}

For there to be one solution, we need \(4-2h \neq 0\) (i.e. \(h \neq 2\)), so that the system is exactly determined. In that case, we can continue reducing:

\begin{align*} \matr{cc|c}{ 1 \amp h \amp 2 \\ 0 \amp 4-2h \amp k-4 } \overset{\frac{1}{4-2h}R_{2}}{\longrightarrow} \amp\matr{cc|c}{ 1 \amp h \amp 2 \\ 0 \amp 1 \amp \frac{k-4}{4-2h} }\\ \overset{-hR_{2}+R_{1}}{\longrightarrow} \amp\matr{cc|c}{ 1 \amp 0 \amp 2 - h \frac{k-4}{4-2h} \\ 0 \amp 1 \amp \frac{k-4}{4-2h} }. \end{align*}

The unique solution is therefore given by \(\left(2 - h \frac{k-4}{4-2h}, \frac{k-4}{4-2h}\right)\text{.}\)

For there to be no solution, we need that \(4-2h = 0 \) while \(k-4 \neq 0\text{,}\) because then, the last row of the matrix reads as \(0x + 0y = k-4 \neq 0\text{,}\) i.e. \(0\neq 0\text{,}\) which is false. In other words, the system is inconsistent if and only if \(h = 2\) and \(k\neq 4\text{.}\)

For there to be infinitely many solutions, we need \(4-2h = 0 \) and \(k-4 = 0\text{,}\) so that the last line reads as \(0x + 0y = 0\text{,}\) which is always true. In other words, if \(h = 2\) and \(k = 4\text{,}\) then the solution set is given by

\begin{equation*} \left\{ (2-hy,y) \,\vert\, y\in\mathbb{R} \right\}. \end{equation*}

10.

In each of the following, find (if possible) conditions on \(a,b,\) and \(c\) such that the system has one solution, no solution or infinitely many solutions.

  1. \begin{align*} 3x + y - z \amp = a\\ x - y + 2z \amp = b\\ 5x + 3y - 4z \amp = c \end{align*}

  2. \begin{align*} -x + 3y + 2z \amp = -8\\ x + z \amp = 2\\ 3x + 3y + az \amp = b \end{align*}

  3. \begin{align*} x + ay \amp = 0\\ y + bz \amp = 0\\ cx + z \amp = 0 \end{align*}

  4. \begin{align*} 3x - y + 2z \amp = 3\\ x + y - z \amp = 2\\ 2x - 2y + 3z \amp = b \end{align*}

  5. \begin{align*} x + ay - z \amp = 1\\ -x + (a-2)y + z \amp = -1\\ 2x + 2y + (a-2)z \amp = 1 \end{align*}

Hint.
Write each system of linear equations as an augmented matrix and start to row-reduce. Be careful not to divide by zero. For example, if you have the following row in a matrix:
\begin{align*} \matr{ccc|c}{ 0 \amp a-2 \amp 0 \amp 8 }, \end{align*}
you need to distinguish between the case \(a=2\) (so that the equation does not have a solution) and the case \(a\neq 2\) (so that you can divide by \(a-2\)).
Answer.

  1. Answer.

    If \(-2a+c+b\neq 0\text{,}\) then the system is inconsistent and has therefore no solutions. If \(-2a+c+b = 0\text{,}\) then the solution set for the system is
    \begin{equation*} \left\{ \left( \frac{a+b}{4} - \frac{1}{4}z, \frac{a-3b}{4} + \frac{7}{4}z,z \right) \,\vert\, z\in\mathbb{R} \right\} . \end{equation*}

  2. Answer.

    If \(a=6\) and \(b\neq 0\text{,}\) then there is no solution. If \(a=6\) and \(b=0\text{,}\) then the solution set to the system is given by
    \begin{equation*} \left\{ (2-z,-2-z,z)\,\vert\, z\in\mathbb{R} \right\}. \end{equation*}
    If \(a\neq 6\text{,}\) then the system has a unique solution, namely \(\left(2 -\frac{b}{a-6}, -2-\frac{b}{a-6},\frac{b}{a-6} \right)\)

  3. Answer.

    If \(1+cab = 0\text{,}\) then the solution set is given by
    \begin{equation*} \left\{ (abz, -bz, z) \,\vert\, z\in\mathbb{R} \right\}. \end{equation*}
    If \(1+cab \neq 0\text{,}\) then the unique solution to the system is \((0,0,0)\text{.}\)

  4. Answer.

    If \(b\neq 1\text{,}\) then this system is inconsistent and has no solution. If \(b = 1\text{,}\) then the solution set of the system is
    \begin{equation*} \left\{ \left(\frac{5}{4} - \frac{1}{4}z,\frac{3}{4}+\frac{5}{4}z,z\right) \,\vert\, z\in\mathbb{R} \right\}. \end{equation*}

  5. Answer.

    If \(a=0\text{,}\) then there is no solution. If \(a=1\text{,}\) then the solution set is given by
    \begin{equation*} \left\{ \left(-y,y,-1 \right) \,\vert\, y\in\mathbb{R} \right\}. \end{equation*}
    If \(a\neq 1\text{,}\) then the unique solution is given by \(\left(1-\frac{1}{a}, 0, -\frac{1}{a} \right)\text{.}\)

Solution.
The approach for each of these questions is the same: We write the system as an augmented matrix and then row-reduce.

  1. Solution.

    \begin{align*} \matr{ccc|c}{ 3 \amp 1 \amp -1 \amp a \\ 1 \amp -1 \amp 2 \amp b \\ 5 \amp 3 \amp -4 \amp c } \overset{-3R_{2}+R_{1}}{\underset{-5R_{2}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 0 \amp 4 \amp -7 \amp a-3b \\ 1 \amp -1 \amp 2 \amp b \\ 0 \amp 8 \amp -14 \amp c-5b }\\ \overset{R_{1}\leftrightarrow R_{2}}{\underset{-2R_{2}'+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp -1 \amp 2 \amp b \\ 0 \amp 4 \amp -7 \amp a-3b \\ 0 \amp 0 \amp 0 \amp (c-5b) - 2(a-3b) }\\ \amp= \matr{ccc|c}{ 1 \amp -1 \amp 2 \amp b \\ 0 \amp 4 \amp -7 \amp a-3b \\ 0 \amp 0 \amp 0 \amp - 2a + c + b } \end{align*}
    We see: If \(-2a+c+b\neq 0\text{,}\) then the system is inconsistent and has no solutions. If \(-2a+c+b = 0\text{,}\) then we proceed:
    \begin{align*} \matr{ccc|c}{ 1 \amp -1 \amp 2 \amp b \\ 0 \amp 4 \amp -7 \amp a-3b \\ 0 \amp 0 \amp 0 \amp - 2a + c + b } \overset{\frac{1}{4}R_{2}}{\underset{R_{2}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp \frac{1}{4} \amp \frac{a+b}{4} \\ 0 \amp 1 \amp -\frac{7}{4} \amp \frac{a-3b}{4} \\ 0 \amp 0 \amp 0 \amp 0 } . \end{align*}
    We conclude that, if \(-2a+c+b = 0\text{,}\) i.e. \(c = 2a-b\text{,}\) then the solution set for the system is
    \begin{equation*} \left\{ \left( \frac{a+b}{4} - \frac{1}{4}z, \frac{a-3b}{4} + \frac{7}{4}z,z \right) \,\vert\, z\in\mathbb{R} \right\} . \end{equation*}

  2. Solution.

    \begin{align*} \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 1 \amp 0 \amp 1 \amp 2 \\ 3 \amp 3 \amp a \amp b } \overset{R_{1}+R_{2}}{\underset{3R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 0 \amp 3 \amp 3 \amp -6 \\ 0 \amp 12 \amp a+6 \amp b-24 } \\ \overset{-4R_{2}+R_{3}}{\underset{\frac{1}{3}R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 0 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp a-6 \amp b }. \end{align*}
    We now have to make a case distinction: Suppose first that \(a=6\text{.}\) Then the last line of the augmented matrix reads as \(0x+0y+0z=b\text{,}\) which has no solution if \(b\neq 0\text{.}\) If, however, both \(a=6\) and \(b=0\text{,}\) then we can proceed:
    \begin{align*} \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 0 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 } \overset{-3R_{2}+R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ -1 \amp 0 \amp -1 \amp -2 \\ 0 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    Thus, if \(a=6\) and \(b=0\text{,}\) then the solution set to the system is given by
    \begin{equation*} \left\{ (2-z,-2-z,z)\,\vert\, z\in\mathbb{R} \right\}. \end{equation*}
    The second case is when \(a\neq 6\text{,}\) so that we can divide by \(a- 6\text{:}\)
    \begin{align*} \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 0 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp a-6 \amp b } \overset{\frac{1}{a-6}R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ -1 \amp 3 \amp 2 \amp -8 \\ 0 \amp 1 \amp 1 \amp -2 \\ 0 \amp 0 \amp 1 \amp \frac{b}{a-6} } \\ \overset{-R_{3}+R_{2}}{\underset{-2R_{3}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ -1 \amp 3 \amp 0 \amp -8 - 2\frac{b}{a-6} \\ 0 \amp 1 \amp 0 \amp -2-\frac{b}{a-6} \\ 0 \amp 0 \amp 1 \amp \frac{b}{a-6} } \\ \overset{-R_{1}}{\underset{3R_{2}+R_{1}'}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp 8 + 2\frac{b}{a-6}-6-3\frac{b}{a-6} \\ 0 \amp 1 \amp 0 \amp -2-\frac{b}{a-6} \\ 0 \amp 0 \amp 1 \amp \frac{b}{a-6} }. \end{align*}
    We see that, if \(a\neq 6\text{,}\) then the system has a unique solution, namely \(\left(2 -\frac{b}{a-6}, -2-\frac{b}{a-6},\frac{b}{a-6} \right)\text{.}\)

  3. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp a \amp 0 \amp 0 \\ 0 \amp 1 \amp b \amp 0 \\ c \amp 0 \amp 1 \amp 0 } \overset{-cR_{1}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp a \amp 0 \amp 0 \\ 0 \amp 1 \amp b \amp 0 \\ 0 \amp -ca \amp 1 \amp 0 }\\ \overset{caR_{2}+R_{3}}{\underset{-aR_{2}+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp -ab \amp 0 \\ 0 \amp 1 \amp b \amp 0 \\ 0 \amp 0 \amp 1 + cab \amp 0 }. \end{align*}
    To proceed, we need to make a case distinction: If \(1+cab = 0\text{,}\) then we have found the RREF of the matrix and the solution set is given by
    \begin{equation*} \left\{ (abz, -bz, z) \,\vert\, z\in\mathbb{R} \right\}. \end{equation*}
    If \(1+cab \neq 0\text{,}\) then we can divide by \(1+cab\text{:}\)
    \begin{align*} \matr{ccc|c}{ 1 \amp 0 \amp -ab \amp 0 \\ 0 \amp 1 \amp b \amp 0 \\ 0 \amp 0 \amp 1 + cab \amp 0 } \overset{\frac{1}{1+cab}R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp -ab \amp 0 \\ 0 \amp 1 \amp b \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 }\\ \overset{abR_{3}+R_{1}}{\underset{-bR_{3}+R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp 0 }. \end{align*}
    Thus, if \(1+cab \neq 0\text{,}\) then the unique solution to the system is \((0,0,0)\text{.}\)

  4. Solution.

    \begin{align*} \matr{ccc|c}{ 3 \amp -1 \amp 2 \amp 3 \\ 1 \amp 1 \amp -1 \amp 2 \\ 2 \amp -2 \amp 3 \amp b } \overset{3R_{2}}{\underset{3R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp -1 \amp 2 \amp 3 \\ 3 \amp 3 \amp -3 \amp 6 \\ 6 \amp -6 \amp 9 \amp 3b }\\ \overset{-R_{1}+R_{2}}{\underset{-2R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 3 \amp -1 \amp 2 \amp 3 \\ 0 \amp 4 \amp -5 \amp 3 \\ 0 \amp -4 \amp 5 \amp 3b-6 }\\ \overset{R_{2}+R_{3}}{\underset{4R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 12 \amp -4 \amp 8 \amp 12 \\ 0 \amp 4 \amp -5 \amp 3 \\ 0 \amp 0 \amp 0 \amp 3b-3 } . \end{align*}
    If \(b\neq 1\text{,}\) then this system is inconsistent and has no solution. Otherwise,
    \begin{align*} \matr{ccc|c}{ 12 \amp -4 \amp 8 \amp 12 \\ 0 \amp 4 \amp -5 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 } \overset{R_{2}+R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ 12 \amp 0 \amp 3 \amp 15 \\ 0 \amp 4 \amp -5 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 }\\ \overset{\frac{1}{12}R_{1}}{\underset{\frac{1}{4}R_{2}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp 0 \amp \frac{1}{4} \amp \frac{5}{4} \\ 0 \amp 1 \amp -\frac{5}{4} \amp \frac{3}{4} \\ 0 \amp 0 \amp 0 \amp 0 }. \end{align*}
    Thus, if \(b = 1\text{,}\) then the solution set of the system is
    \begin{equation*} \left\{ \left(\frac{5}{4} - \frac{1}{4}z,\frac{3}{4}+\frac{5}{4}z,z\right) \,\vert\, z\in\mathbb{R} \right\}. \end{equation*}

  5. Solution.

    \begin{align*} \matr{ccc|c}{ 1 \amp a \amp -1 \amp 1 \\ -1 \amp a-2 \amp 1 \amp -1 \\ 2 \amp 2 \amp a-2 \amp 1 } \overset{R_{1}+R_{2}}{\underset{-2R_{1}+R_{3}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp a \amp -1 \amp 1 \\ 0 \amp 2a-2 \amp 0 \amp 0 \\ 0 \amp 2-2a \amp a \amp -1 }\\ \overset{R_{2}+R_{3}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp a \amp -1 \amp 1 \\ 0 \amp 2a-2 \amp 0 \amp 0 \\ 0 \amp 0 \amp a \amp -1 } . \end{align*}
    If \(a=0\text{,}\) then the system is inconsistent (because of the third row) and there is no solution. Otherwise, we can divide by \(a\text{:}\)
    \begin{align*} \matr{ccc|c}{ 1 \amp a \amp -1 \amp 1 \\ 0 \amp 2a-2 \amp 0 \amp 0 \\ 0 \amp 0 \amp a \amp -1 } \overset{\frac{1}{a}R_{3}}{\underset{R_{3}'+R_{1}}{\longrightarrow}} \amp \matr{ccc|c}{ 1 \amp a \amp 0 \amp 1-\frac{1}{a} \\ 0 \amp 2a-2 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -\frac{1}{a} } . \end{align*}
    If \(a=1\text{,}\) then switching the second and third row yields the RREF:
    \begin{equation*} \matr{ccc|c}{ 1 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -1 } \overset{R_{2}\leftrightarrow R_{3}}{\longrightarrow} \matr{ccc|c}{ 1 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -1 \\ 0 \amp 0 \amp 0 \amp 0 }. \end{equation*}
    Thus, the solution set is given by
    \begin{equation*} \left\{ \left(-y,y,-1 \right) \,\vert\, y\in\mathbb{R} \right\}. \end{equation*}
    If \(a\neq 1\text{,}\) then we can divide by \(2a-2 \text{:}\)
    \begin{align*} \matr{ccc|c}{ 1 \amp a \amp 0 \amp 1-\frac{1}{a} \\ 0 \amp 2a-2 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -\frac{1}{a} } \overset{\frac{1}{2a-2}R_{2}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp a \amp 0 \amp 1-\frac{1}{a} \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -\frac{1}{a} }\\ \overset{-aR_{2}+R_{1}}{\longrightarrow} \amp \matr{ccc|c}{ 1 \amp 0 \amp 0 \amp 1-\frac{1}{a} \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -\frac{1}{a} } . \end{align*}
    Thus, in this case, the unique solution is given by \(\left(1-\frac{1}{a}, 0, -\frac{1}{a} \right)\text{.}\)