Advanced Higher Maths
Systems of Equations

We form the augmented matrix and then use elementary row operations (EROs) to reduce its left hand side to upper triangular form. When speaking of the entire augmented matrix, this is called row echelon form. In either case, the entries below the main diagonal should all be zero.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 2 & 1 & 1 & 2 \cr 1 & -3 & -1 & 5 \cr 1 & 1 & 2 & 3 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_3\!-\!R_2\normalsize\cr \small R_3\rightarrow 2R_3\!-\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 2 & 1 & 1 & 2 \cr 0 & 4 & 3 & -2 \cr 0 & 1 & 3 & 4 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow 4R_3\!-\!R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 2 & 1 & 1 & 2 \cr 0 & 4 & 3 & -2 \cr 0 & 0 & 9 & 18 \end{array} \right) \\[10pt] \end{eqnarray} $$

Now we perform back substitution, obtaining \(\raise 0.2pt{z\small,}\) \(\raise 0.2pt{y}\) and \(\raise 0.2pt{x\small,}\) in that order, as we read up the rows of the matrix.

From row 3:

From row 2:

From row 1:

The general equation of a parabola is:

So we substitute each coordinate point \(\left(x,y\right)\) to obtain a system of three equations in \(\raise 0.2pt{a\small,}\) \(b\) and \(\raise 0.2pt{c:}\)

$$ \begin{gather} -4 = a(1^2)+b(1)+c \\[4pt] -2 = a(2^2)+b(2)+c \\[4pt] 10 = a(3^2)+b(3)+c \end{gather} $$

These simplify to:

$$ \begin{gather} a+b+c=-4 \\[4pt] 4a+2b+c=-2 \\[4pt] 9a+3b+c=10 \end{gather} $$

We then form the augmented matrix and use elementary row operations to reduce it to row echelon form.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & -4 \cr 4 & 2 & 1 & -2 \cr 9 & 3 & 1 & 10 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_2\!-\!4R_1\normalsize\cr \small R_3\rightarrow R_3\!-9\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & -4 \cr 0 & -2 & -3 & 14 \cr 0 & -6 & -8 & 46 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow R_3\!-\!3R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & -4 \cr 0 & -2 & -3 & 14 \cr 0 & 0 & 1 & 4 \end{array} \right) \\[10pt] \end{eqnarray} $$

Now we perform back substitution, obtaining \(\raise 0.2pt{c\small,}\) \(b\) and \(\raise 0.2pt{a\small,}\) in that order, as we read up the rows of the matrix.

From row 3:

From row 2:

From row 1:

So the parabola has equation: \(y=5x^2-13x+4.\)

We start by using EROs to reduce the augmented matrix to row echelon form.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & 4 \cr 3 & -1 & -2 & 13 \cr 2 & -2 & 1 & 9 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_2\!-\!3R_1\normalsize\cr \small R_3\rightarrow R_3\!-\!2R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & 4 \cr 0 & -4 & -1 & 1 \cr 0 & -4 & -1 & 1 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow R_2\!-\!R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 1 & 1 & 4 \cr 0 & -4 & -1 & 1 \cr 0 & 0 & 0 & 0 \end{array} \right) \\[10pt] \end{eqnarray} $$

Rows 2 and 3 being the same led to a row of zeros, so we have redundancy. As a consequence, there is no unique solution and the best we can do is to obtain equations connecting \(\raise 0.2pt{x\small,}\) \(\raise 0.2pt{y}\) and \(\raise 0.2pt{z}\) parametrically.

Start by letting \(\raise 0.2pt{z\!=\!\lambda\small,}\) say. Then from row 2:

Then from row 1:

So a parametric solution is:

Note: Interpreted geometrically, our three original equations represent three 2D planes. However, because of the redundancy, the three planes do not intersect at a point, but instead along a straight line. Our parametric solution gives us the parametric equations of this line.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 3 & -1 & 1 & 2 \cr 1 & 2 & 2 & 6 \cr 1 & -5 & k & -10 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow 3R_2\!-\!R_1\normalsize\cr \small R_3\rightarrow 3R_3\!-\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 3 & -1 & 1 & 2 \cr 0 & 7 & 5 & 16 \cr 0 & -14 & 3k\!-\!1 & -32 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow R_3\!+\!2R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 3 & -1 & 1 & 2 \cr 0 & 7 & 5 & 16 \cr 0 & 0 & 3k\!+\!9 & 0 \end{array} \right) \\[10pt] \end{eqnarray} $$

For redundancy, \(3k+9=0\) so this occurs when \(k=-3\small.\)

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 6 & 5 \cr 1 & -4 & -2 & 1 \cr 1 & -1 & \lambda & -3 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_1\!-\!R_2\normalsize\cr \small R_3\rightarrow R_1\!-\!R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 6 & 5 \cr 0 & 6 & 8 & 4 \cr 0 & 3 & 6\!-\!\!\lambda & 8 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow 2R_3\!-\!R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 6 & 5 \cr 0 & 6 & 8 & 4 \cr 0 & 0 & 4\!-\!2\lambda & 12 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 3:

The system is inconsistent when \(4\!-\!2\lambda=0\small,\) which is when \(\lambda\!=\!2\small.\)

Note: A system is ill-conditioned if changing an entry in the augmented matrix by a small amount results in very different solutions.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 10 & 9 & 5 \cr 9 & 8 & 4 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow 9R_1\!-10\!R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 10 & 9 & 5 \cr 0 & 1 & 5 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 2, \(y\!=\!5\small,\) and from row 1:

Now we make a very small change to one of the coefficients and repeat this process to find out whether or not the new solutions are close to \(x\!=\!\!-\!4\small,\normalsize\ y\!=\!5\small.\)

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 10 & 9 & 5 \cr 9 & 8.001 & 4 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow 9R_1\!-10\!R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 10 & 9 & 5 \cr 0 & 0.99 & 5 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 2, \(0.99y\!=\!5\) so \(y=\large\frac{500}{99}\normalsize\approx 5.06\small.\)

From row 1:

The values of \(\raise 0.2pt{x}\) and \(y\) have changed very little. This system of equations is well-conditioned (that is, not ill-conditioned).

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 300 & -1 & 1 \cr 299 & -1 & -2 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow 299R_1\!-300\!R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 300 & -1 & -1 \cr 0 & 1 & 301 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 2, \(y\!=\!301\small,\) and from row 1:

Now we make a very small change to one of the coefficients and repeat this process to find out whether or not the new solutions are close to \(x\!=\!1\small,\normalsize\ y\!=\!301\small.\)

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 300 & -1 & -1 \cr 299 & -1.01 & -2 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow 299R_1\!-300\!R_2\normalsize \end{array} && \left( \begin{array}{cc:c} 300 & -1 & -1 \cr 0 & 4 & 301 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 2, \(4y\!=\!301\) so \(y=75.25\small.\)

From row 1:

The values of \(\raise 0.2pt{x}\) and \(y\) have changed considerably. This system of equations is therefore ill-conditioned.

We create the augmented matrix and use EROs to reduce it to row echelon form. For redundancy, all entries in the third row will be zero.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 3 & 3 \cr 2 & -1 & 4 & 5 \cr 1 & -3 & 2\lambda & 2 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_2\!-\!2R_1\normalsize\cr \small R_3\rightarrow R_3\!-\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 3 & 3 \cr 0 & -5 & -2 & -1 \cr 0 & -5 & 2k\!-\!3 & -1 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow R_3\!-\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & 2 & 3 & 3 \cr 0 & -5 & -2 & -1 \cr 0 & 0 & 2\lambda\!-\!1 & 0 \end{array} \right) \\[10pt] \end{eqnarray} $$

For redundancy, \(2\lambda-1=0\small.\) This occurs when \(\lambda=\frac12\small.\)

First we create the augmented matrix. Then we use Gaussian elimination to reduce it to row echelon form.

$$ \begin{eqnarray} \begin{array}{r} \small R_1\normalsize\cr \small R_2\normalsize\cr \small R_3\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -3 & 1 & -1 \cr 3 & -2 & 4 & 11 \cr 1 & 4 & 2 & 15 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small R_2\rightarrow R_2\!-\!3R_1\normalsize\cr \small R_3\rightarrow R_3\!-\!R_1\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -3 & 1 & -1 \cr 0 & 7 & 1 & 14 \cr 0 & 7 & 1 & 16 \end{array} \right) \\[10pt] \begin{array}{r} \small\textsf{Frozen}\normalsize\cr \small\textsf{Frozen}\normalsize\cr \small R_3\rightarrow R_3\!-\!R_2\normalsize \end{array} && \left( \begin{array}{ccc:c} 1 & -3 & 1 & -1 \cr 0 & 7 & 1 & 14 \cr 0 & 0 & 0 & 2 \end{array} \right) \\[10pt] \end{eqnarray} $$

From row 3, we can see that the system shows inconsistency, because \(0\neq 2\small.\)

Note: We could also have concluded that the system is inconsistent at the previous stage, because \(14\neq 16\small.\)