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PAGE 1

3.6 Determinants (part 1)

Scalar: \( a^{-1} \) exists if \( a \neq 0 \), \( a^{-1} = \frac{1}{a} \) (\( a \neq 0 \))

Matrix: \( A^{-1} \) exists if determinant of \( A \) is not zero

if \( A \) is \( 2 \times 2 \), its determinant, \( \det A = \det(A) = |A| \)

\( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) then \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \)

\( A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)

\( \det A = ad - bc \)

for \( 3 \times 3 \) and beyond, no formula like \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \) exists

however, using cofactor expansion we can find the determinant of \( 3 \times 3 \) as sum of several \( 2 \times 2 \)

(\( 4 \times 4 \) as sum of several \( 3 \times 3 \) which are sum of \( 2 \times 2 \))

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Cofactor expansion

example

\( A = \begin{bmatrix} 2 & 0 & 4 \\ 3 & 4 & 2 \\ 0 & 4 & -2 \end{bmatrix} \) find \( \det A \)

choose ANY one row/column to expand

here, let's use column 1

\( \det A = (2)(-1)^{1+1} \begin{vmatrix} 4 & 2 \\ 4 & -2 \end{vmatrix} + (3)(-1)^{2+1} \begin{vmatrix} 0 & 4 \\ 4 & -2 \end{vmatrix} + (0)(-1)^{3+1} \begin{vmatrix} 0 & 4 \\ 4 & 2 \end{vmatrix} \)

Note: \( a_{11} = 2 \) is the 1st # in col 1. The first term uses row 1, col 1. The determinant is of the submatrix after blocking out row 1, col 1.

Note: \( a_{21} = 3 \) and \( a_{31} = 0 \) are the subsequent elements in column 1.

\( = (2)(1)(-8 - 8) + (3)(-1)(-16) + (0) \)

\( = 16 \)

\( (-1)^{i+j} \) distributes signs like this:

\( \begin{bmatrix} 2^+ & 0^- & 4^+ \\ 3^- & 4^+ & 2^- \\ 0^+ & 4^- & -2^+ \end{bmatrix} \)

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Determinant Calculation: Row Expansion

Try another row/col

this time row 2

\[ A = \begin{bmatrix} 2^+ & 0^- & 4^+ \\ 3^- & 4^+ & 2^- \\ 0^+ & 4^- & -2^+ \end{bmatrix} \]
\[ \det A = -(3) \begin{vmatrix} 0 & 4 \\ 4 & -2 \end{vmatrix} + (4) \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} - (2) \begin{vmatrix} 2 & 0 \\ 0 & 4 \end{vmatrix} \]
\[ = (-3)(-16) + (4)(-4) - (2)(8) = \boxed{16} \]
pick row/col w/ many zeros to make it easier
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Example: Efficient Determinant Expansion

\[ A = \begin{bmatrix} 8 & 0 & 0 & 5 \\ 5 & 8 & 3 & -7 \\ 2 & 0 & 0 & 0 \\ 7 & 2 & 1 & 7 \end{bmatrix} \]
\[ \text{cofactor signs: } \begin{bmatrix} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{bmatrix} \]

"good" choices of row/col: row 3 (has 3 zeros)

expand along row 3

\[ \det A = (2) \begin{vmatrix} 0 & 0 & 5 \\ 8 & 3 & -7 \\ 2 & 1 & 7 \end{vmatrix} + (0) \begin{vmatrix} I.D.C \end{vmatrix} + (0) \begin{vmatrix} I.D.C. 2 \end{vmatrix} + (0) \begin{vmatrix} I.D.C. 3 \end{vmatrix} \]

expand along row 1

\[ = (2) \left( (5) \begin{vmatrix} 8 & 3 \\ 2 & 1 \end{vmatrix} \right) = (10)(8-6) = \boxed{20} \]
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Cramer's Rule

Another way to solve system (other ways: row reduction, find inverse)

\( a_{11}x_1 + a_{12}x_2 = b_1 \)
\( a_{21}x_1 + a_{22}x_2 = b_2 \)
\( \rightarrow \)
\[ \underbrace{\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}}_{A} \underbrace{\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}}_{\vec{x}} = \underbrace{\begin{bmatrix} b_1 \\ b_2 \end{bmatrix}}_{\vec{b}} \]
\[ x_1 = \frac{\begin{vmatrix} b_1 & a_{12} \\ b_2 & a_{22} \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}} \]

\( \leftarrow \) det of \( A \) w/ 1st col replaced by \( \vec{b} \)

\( \leftarrow \) det \( A \)

\[ x_2 = \frac{\begin{vmatrix} a_{11} & b_1 \\ a_{21} & b_2 \end{vmatrix}}{\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix}} \]

\( \leftarrow \) det of \( A \) w/ 2nd col replaced by \( \vec{b} \)

\( \leftarrow \) det \( A \)

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\( x_1 + 3x_2 = 9 \)
\( 2x_1 + x_2 = 8 \)
seen this before. \( x_1 = 3, x_2 = 2 \)
\[ A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix} \]
\[ x_1 = \frac{\begin{vmatrix} 9 & 3 \\ 8 & 1 \end{vmatrix}}{\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}} = \frac{-15}{-5} = 3 \]
\[ x_2 = \frac{\begin{vmatrix} 1 & 9 \\ 2 & 8 \end{vmatrix}}{\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}} = \frac{-10}{-5} = 2 \]
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Cramer's Rule for Higher Dimensions

Same idea for \(3 \times 3\) and beyond.

Example

Consider the system of equations:

\[\begin{aligned} x_1 + x_2 &= 4 \\ -4x_1 + 3x_3 &= 0 \\ x_2 - 3x_3 &= 3 \end{aligned}\]

This can be written in matrix form \(A\vec{x} = \vec{b}\) where:

\[A = \begin{bmatrix} 1 & 1 & 0 \\ -4 & 0 & 3 \\ 0 & 1 & -3 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 4 \\ 0 \\ 3 \end{bmatrix}\]

Find \(x_1\)

Using Cramer's Rule, we replace the first column of \(A\) with the vector \(\vec{b}\) to find the numerator determinant:

\[x_1 = \frac{\begin{vmatrix} 4 & 1 & 0 \\ 0 & 0 & 3 \\ 3 & 1 & -1 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 0 \\ -4 & 0 & 3 \\ 0 & 1 & -3 \end{vmatrix}} = \dots = -\frac{1}{5}\]
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Matrix Operations

Transpose of a Matrix

Definition: Interchange rows and columns.

Example 1: \(2 \times 2\) Matrix

\[A = \underset{2 \times 2}{\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}}\]\[A^T = \underset{2 \times 2}{\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}}\]

Example 2: \(2 \times 3\) Matrix

\[B = \underset{2 \times 3}{\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}}\]\[B^T = \underset{3 \times 2}{\begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}}\]