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"Homework 4" and "Homework 5" are due Tomorrow

1.4 The Matrix Equation \( A\vec{x} = \vec{b} \)

A system of linear equations:

\[ \begin{aligned} x_1 + 2x_2 + 3x_3 &= 4 \\ 5x_1 + 6x_2 + 7x_3 &= 8 \\ 9x_1 + 10x_2 + 11x_3 &= 12 \end{aligned} \]

can be written as a vector equation:

\[ x_1 \begin{bmatrix} 1 \\ 5 \\ 9 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 6 \\ 10 \end{bmatrix} + x_3 \begin{bmatrix} 3 \\ 7 \\ 11 \end{bmatrix} = \begin{bmatrix} 4 \\ 8 \\ 12 \end{bmatrix} \]

Matrix Equation Representation

This can be expressed as a matrix equation:

\[ \underbrace{\begin{bmatrix} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 9 & 10 & 11 \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}}_{\vec{x}} = \underbrace{\begin{bmatrix} 4 \\ 8 \\ 12 \end{bmatrix}}_{\vec{b}} \]

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In general, if \( \vec{a}_1, \vec{a}_2, \vec{a}_3, \dots, \vec{a}_n \) are columns of \( A \) and \( x_1, x_2, \dots, x_n \) are elements of column vector \( \vec{x} \), then:

\[ \underbrace{[ \vec{a}_1 \ \vec{a}_2 \ \dots \ \vec{a}_n ]}_{A} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} = x_1\vec{a}_1 + x_2\vec{a}_2 + \dots + x_n\vec{a}_n \]

This gives us a way to multiply a matrix by a vector.

Example

\[ \begin{bmatrix} 6 & 5 \\ -4 & -3 \\ 7 & 6 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} = 2 \begin{bmatrix} 6 \\ -4 \\ 7 \end{bmatrix} + 3 \begin{bmatrix} 5 \\ -3 \\ 6 \end{bmatrix} = \begin{bmatrix} 27 \\ -17 \\ 32 \end{bmatrix} \]

Dimensions: \( 3 \times \mathbf{2} \) and \( \mathbf{2} \times 1 \)

These MUST match

"Row-Vector Rule"

Multiply first row of \( A \) by first column of \( \vec{x} \), then 2nd row of \( A \) by first column of \( \vec{x} \), and so on.

\[ = \begin{bmatrix} 6 \cdot 2 + 5 \cdot 3 \\ -4 \cdot 2 + -3 \cdot 3 \\ 7 \cdot 2 + 6 \cdot 3 \end{bmatrix} = \begin{bmatrix} 27 \\ -17 \\ 32 \end{bmatrix} \]

Each element is a sum of products.

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Matrix Multiplication and Linear Systems

Example: Matrix-Vector Product

\[\begin{bmatrix} 7 & -3 \\ 2 & 1 \\ 9 & -6 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ -5 \end{bmatrix} = \begin{bmatrix} 1 \\ -9 \\ 12 \\ -4 \end{bmatrix}\]

The dimensions of the matrices are: \(4 \times 2\) multiplied by \(2 \times 1\) results in a \(4 \times 1\) vector.

Equation \(A\vec{x} = \vec{b}\)

\[\begin{bmatrix} 1 & 2 & 4 \\ 0 & 1 & 5 \\ -2 & -4 & -3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ 9 \end{bmatrix}\]

Solution

We use the augmented matrix and row reduction:

\[\begin{bmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 5 & 2 \\ -2 & -4 & -3 & 9 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 \end{bmatrix}\]

So \(x_1 = 0\), \(x_2 = -3\), \(x_3 = 1\).

\[\vec{x} = \begin{bmatrix} 0 \\ -3 \\ 1 \end{bmatrix}\]

So this means:

\[0 \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} + (-3) \begin{bmatrix} 2 \\ 1 \\ -4 \end{bmatrix} + (1) \begin{bmatrix} 4 \\ 5 \\ -3 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ 9 \end{bmatrix}\]

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Therefore, for a solution \(\vec{x}\) to exist, the vector \(\vec{b}\) MUST be a linear combination of columns of \(A\).

\(\Rightarrow \vec{b}\) must be in \(\text{span} \{ \vec{a}_1, \vec{a}_2, \vec{a}_3 \}\)

If we know \(\vec{b}\), we know how to find \(\vec{x}\).

Question

For an \(A\) that is \(3 \times 3\), can we form every possible vector in \(\mathbb{R}^3\)?

In other words, is \(\text{span} \{ \vec{a}_1, \vec{a}_2, \vec{a}_3 \} = \mathbb{R}^3\)?

Suppose \(A = \begin{bmatrix} 1 & 2 & 4 \\ 0 & 1 & 5 \\ -2 & -4 & -3 \end{bmatrix}\).

Is there a solution \(\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) such that:

\[x_1 \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 1 \\ -4 \end{bmatrix} + x_3 \begin{bmatrix} 4 \\ 5 \\ -3 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}\]

for ANY \(b_1, b_2, b_3\)?

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\[ \begin{bmatrix} 1 & 2 & 4 & b_1 \\ 0 & 1 & 5 & b_2 \\ -2 & -4 & -3 & b_3 \end{bmatrix} \sim \dots \sim \begin{bmatrix} 1 & 0 & 0 & \frac{17}{5}b_1 - 2b_2 + \frac{6}{5}b_3 \\ 0 & 1 & 0 & b_2 - 2b_1 - b_3 \\ 0 & 0 & 1 & \frac{2}{5}b_1 + \frac{1}{5}b_3 \end{bmatrix} \]

There is a pivot in each column, so a solution always exists for ANY \(b_1, b_2, b_3\).

Therefore, this matrix's columns span \(\mathbb{R}^3\).

What if we got

\[ \begin{bmatrix} 1 & 0 & 0 & b_1 - 2b_2 + b_3 \\ 0 & 1 & 0 & b_2 - 3b_1 + b_3 \\ 0 & 0 & 0 & b_1 + b_2 + b_3 \end{bmatrix} \]

So if \(b_1 + b_2 + b_3 \neq 0\), then there is no solution.

So \(\text{span} \{ \vec{a_1}, \vec{a_2}, \vec{a_3} \} \neq \mathbb{R}^3\).

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The columns of an \(m \times n\) matrix \(A\) span \(\mathbb{R}^m\) if the reduced row echelon form of \(A\) has a pivot position in every column.

Then, the following four statements are logically equivalent:

a. For each \(\vec{b}\) in \(\mathbb{R}^m\), \(A\vec{x} = \vec{b}\) has a solution.
b. Each \(\vec{b}\) in \(\mathbb{R}^m\) is a linear combo of columns of \(A\).
c. Columns of \(A\) span \(\mathbb{R}^m\).
d. \(A\) has pivot position in every row.

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1.5 Solution sets of Linear Systems

The equation \(A\vec{x} = \vec{b}\) is called homogeneous if \(\vec{b} = \vec{0} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix}\).

\(A\vec{x} = \vec{0}\) always has the trivial solution \(\vec{x} = \vec{0}\).

\(A\vec{x} = \vec{0}\) has nontrivial solutions if \(A\) does not have pivot position in every row.

example

\[A = \begin{bmatrix} 3 & 3 & 6 \\ -9 & -9 & -18 \\ 0 & -4 & 12 \end{bmatrix}\]

\[A\vec{x} = \vec{0} \rightarrow \begin{bmatrix} 3 & 3 & 6 & 0 \\ -9 & -9 & -18 & 0 \\ 0 & -4 & 12 & 0 \end{bmatrix}\]

\[\sim \dots \sim \begin{bmatrix} 1 & 1 & 2 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}\]

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\(x_3\) free
\(x_2 = 3x_3\)
\(x_1 = -x_2 - 2x_3 = -3x_3 - 2x_3 = -5x_3\)

so \(\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -5x_3 \\ 3x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix}\)

line in \(\mathbb{R}^3\)

Figure: 3D coordinate system with x, y, z axes showing a vector line passing through the origin.

parametric vector form (\(x_3\) is the parameter)

nonhomogeneous system

same \(A\) :

\[A = \begin{bmatrix} 3 & 3 & 6 \\ -9 & -9 & -18 \\ 0 & -4 & 12 \end{bmatrix}\]

\[\vec{b} = \begin{bmatrix} 12 \\ -36 \\ 8 \end{bmatrix}\]

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Solving Linear Systems: Augmented Matrices and Parametric Solutions

Solve the following system represented by an augmented matrix:

\[ \begin{bmatrix} 3 & 3 & 6 & 12 \\ -9 & -9 & -18 & -36 \\ 0 & -4 & 12 & 8 \end{bmatrix} \]

Performing row reduction to reach row-echelon form:

\[ \sim \begin{bmatrix} 1 & 1 & 2 & 4 \\ 0 & 1 & -3 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

Parametric Solution

From the reduced matrix, we identify the variables:

\( x_3 \) is a free variable.
\( x_2 = -2 + 3x_3 \)
\( x_1 = -x_2 - 2x_3 + 4 = 6 - 5x_3 \)

The general solution vector \( \vec{x} \) is:

\[ \text{solution: } \vec{x} = \begin{bmatrix} 6 - 5x_3 \\ -2 + 3x_3 \\ x_3 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix} \]

Note: The term \( x_3 \begin{bmatrix} -5 \\ 3 \\ 1 \end{bmatrix} \) represents the solution of the homogeneous system \( A\vec{x} = \vec{0} \).

Geometric Interpretation

The solution of a non-homogeneous system \( A\vec{x} = \vec{b} \) is a shifted version of the solution of the homogeneous system \( A\vec{x} = \vec{0} \), provided that \( A\vec{x} = \vec{b} \) is consistent. Geometrically, these solutions represent parallel lines or planes (having the same slopes).

Figure: A 3D coordinate system with x, y, and z axes. Two parallel lines are drawn, one labeled as the solution of Ax=0 and the other as the solution of b != 0.

Key Concept:

Solution of \( A\vec{x} = \vec{b} \) is a shifted version of solution of \( A\vec{x} = \vec{0} \) if \( A\vec{x} = \vec{b} \) is consistent.