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1.1 Systems of Linear Equations

Why linear algebra?

  • Find the solution set of a system of linear eqs.
  • How many solutions, if any?

Examples:

\[\begin{aligned} x_1 + 2x_2 &= 6 \\ 4x_1 + 7x_2 &= 26 \end{aligned}\]

\((x_1, x_2) = ?\)

\(\Rightarrow\) intersection of two lines

Two intersecting lines crossing at a single point.

one solution

Two parallel lines that never intersect.

none

A single line representing two overlapping lines.

infinitely-many

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\[\begin{aligned} x_1 - 2x_2 + x_3 &= 0 \\ 2x_2 - 8x_3 &= 8 \\ 5x_1 \quad - 5x_3 &= 10 \end{aligned}\]

intersection of 3 planes

Back to

\[\begin{aligned} x_1 + 2x_2 &= 6 \\ 4x_1 + 7x_2 &= 26 \end{aligned}\]

In matrix notation the coefficient matrix is

\[\begin{bmatrix} 1 & 2 \\ 4 & 7 \end{bmatrix}\]
\(\rightarrow 2 \times 2\) matrix (2 rows, 2 columns)

The augmented matrix of the system is

\[\left[ \begin{array}{cc|c} 1 & 2 & 6 \\ 4 & 7 & 26 \end{array} \right]\]
\(\rightarrow 2 \times 3\) matrix
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Solving Systems of Linear Equations

Substitution Method

Solution of:

\[\begin{cases} x_1 + 2x_2 = 6 \\ 4x_1 + 7x_2 = 26 \end{cases}\]

From the first equation, we can express \(x_1\) in terms of \(x_2\):

\[x_1 = 6 - 2x_2\]

Substituting this into the second equation:

\[4(6 - 2x_2) + 7x_2 = 26\]\[24 - 8x_2 + 7x_2 = 26 \rightarrow x_2 = -2\]

Now, solve for \(x_1\):

\[x_1 = 6 - 2(-2) = 10\]

Unique solution: \((10, -2)\)

Matrix Method

Matrix way:

\[\begin{bmatrix} 1 & 2 & 6 \\ 4 & 7 & 26 \end{bmatrix}\]

Eliminate \(x_1\) (the first column) from the second row. Multiply row 1 by \(-4\):

\[\begin{bmatrix} -4 & -8 & -24 \\ 4 & 7 & 26 \end{bmatrix}\]
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Row Operations and Triangular Form

Add row 1 to row 2:

\[\begin{bmatrix} -4 & -8 & -24 \\ 0 & -1 & +2 \end{bmatrix}\]

"triangular form"

Eliminate \(x_2\) from row 1. Add to row 1 \(-8\) times row 2:

\[\begin{bmatrix} -4 & 0 & -40 \\ 0 & -1 & 2 \end{bmatrix}\]

Multiply row 1 by \(-1/4\) and multiply row 2 by \(-1\):

\[\begin{bmatrix} 1 & 0 & 10 \\ 0 & 1 & -2 \end{bmatrix}\]

Final Solution

  • Row 1: \(1 \cdot x_1 + 0 \cdot x_2 = 10 \rightarrow x_1 = 10\)
  • Row 2: \(0 \cdot x_1 + 1 \cdot x_2 = -2 \rightarrow x_2 = -2\)
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Elementary Row Operations (ERO's)

The things we did to the matrix are called Elementary Row Operations (ERO's).

  1. Add multiple of one row to another
  2. Swap/interchange two rows
  3. Multiply a row by any nonzero constant

ERO's do NOT change the solution set of the system.

ERO's are reversible.

Two matrices are row equivalent if one can be transformed into another by ERO's.

So

\[ \begin{bmatrix} 1 & 2 & 6 \\ 4 & 7 & 26 \end{bmatrix} \text{ is row equivalent to } \begin{bmatrix} 1 & 0 & 10 \\ 0 & 1 & -2 \end{bmatrix} \]
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careful! two systems are equivalent if they have the same solution set.

Bigger system examples

example:

\[ \begin{aligned} x_2 + 4x_3 &= -3 \\ x_1 + 3x_2 + 6x_3 &= 4 \\ 2x_1 + 5x_2 + 8x_3 &= 5 \end{aligned} \]

Augmented matrix:

\[ \begin{bmatrix} 0 & 1 & 4 & -3 \\ 1 & 3 & 6 & 4 \\ 2 & 5 & 8 & 5 \end{bmatrix} \]

try to keep \( x_1 \) in row 1, get rid of it in other rows

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Row Operations and Inconsistent Systems

Swap row 1 and row 2

\[\begin{bmatrix} 1 & 3 & 6 & 4 \\ 0 & 1 & 4 & -3 \\ 2 & 5 & 8 & 5 \end{bmatrix}\]

\(-2 \cdot R_1 + R_3\)

\[\begin{bmatrix} 1 & 3 & 6 & 4 \\ 0 & 1 & 4 & -3 \\ 0 & -1 & -4 & -3 \end{bmatrix}\]

Keep \(x_2\) in \(R_2\), get rid of it from others

\(R_2 + R_3\)
\[\begin{bmatrix} 1 & 3 & 6 & 4 \\ 0 & 1 & 4 & -3 \\ 0 & 0 & 0 & -6 \end{bmatrix}\]

row 3: \(0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 = -6\)

\(0 = -6\)

this means the system is inconsistent \(\rightarrow\) no solution

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Example: the augmented matrix of a system is

\[\begin{bmatrix} 1 & -1 & 1 & 7 \\ 3 & 2 & -12 & 11 \\ 4 & 1 & -11 & 18 \end{bmatrix}\]

\(-3 \cdot R_1 + R_2\) AND \(-4 \cdot R_1 + R_3\)

\[\begin{bmatrix} 1 & -1 & 1 & 7 \\ 0 & 5 & -15 & -10 \\ 0 & 5 & -15 & -10 \end{bmatrix}\]

\(-1 \cdot R_2 + R_3\)

\[\begin{bmatrix} 1 & -1 & 1 & 7 \\ 0 & 5 & -15 & -10 \\ 0 & 0 & 0 & 0 \end{bmatrix}\]

Zero row \(\rightarrow\) arbitrary solution in one or more variables

row 3: \(0 = 0\)

\(0 \cdot x_1 + 0 \cdot x_2 + 0 \cdot x_3 = 0 \rightarrow\) at least one of them is arbitrary