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3.1 + 3.2 Determinants

"HW 13" + "HW 14" due together

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

What about \( 3 \times 3 \) and beyond?

\[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \sim \dots \sim \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{11}a_{22} - a_{12}a_{21} & a_{11}a_{23} - a_{13}a_{21} \\ 0 & 0 & a_{11} \Delta \end{bmatrix} \]

where

\[ \Delta = a_{11}a_{22}a_{33} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} \]

if \( \Delta \neq 0 \), then \( A^{-1} \) exists, just like \( 2 \times 2 \) case when \( \det(A) \neq 0 \).

\( \Delta \) is the determinant of \( 3 \times 3 \) \( A \).

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rewrite \( \Delta \)

\[ \Delta = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

determinant of

\[ \begin{bmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{bmatrix} \]

\( A \) w/ first row and first col covered

det of

\[ \begin{bmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{bmatrix} \]

\( A \) w/ 2nd col and 1st row covered

det of

\[ \begin{bmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \]

\( A \) w/ 3rd col and 1st row covered

example

\( A = \begin{bmatrix} 2 & 0 & 4 \\ 3 & 4 & 2 \\ 0 & 4 & -2 \end{bmatrix} \)\( \det(A) = ? \)(\(\det A\))
\[ \det(A) = \begin{vmatrix} 2 & 0 & 4 \\ 3 & 4 & 2 \\ 0 & 4 & -2 \end{vmatrix} \]
sign change at every even col or even row
\[ = (2) \begin{vmatrix} 4 & 2 \\ 4 & -2 \end{vmatrix} - (0) \begin{vmatrix} 3 & 2 \\ 0 & -2 \end{vmatrix} + (4) \begin{vmatrix} 3 & 4 \\ 0 & 4 \end{vmatrix} \]
\[ = (2)(-8-8) - (0)(-6-0) + (4)(12-0) = 16 \]
"cofactor expansion"
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Cofactor Expansion and Rule of Sarrus

We can do cofactor expansion along any row or column.

Last example: along first row

Try along 2nd column

\[ A = \begin{bmatrix} 2^+ & 0^- & 4^+ \\ 3^- & 4^+ & 2^- \\ 0^+ & 4^- & -2^+ \end{bmatrix} \]
\[ \det(A) = -(0) \begin{vmatrix} 3 & 2 \\ 0 & -2 \end{vmatrix} + (4) \begin{vmatrix} 2 & 4 \\ 0 & -2 \end{vmatrix} - (4) \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} \]
\[ = (4)(-4) - (4)(-8) = 16 \]

Another way for \( 3 \times 3 \): Rule of Sarrus

Copies of 1st and 2nd columns are appended to the right of the matrix to facilitate diagonal multiplication.

Rule of Sarrus diagram showing a 3x3 matrix with its first two columns repeated and arrows indicating diagonal products.
\[ \det(A) = (2)(4)(-2) + (0)(2)(0) + (4)(3)(4) \]\[ - (0)(4)(4) - (4)(2)(2) - (-2)(3)(0) = 16 \]
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Determinants of Larger Matrices

\( 4 \times 4 \) and beyond are just series of \( 3 \times 3 \)'s.

Example

\[ \begin{vmatrix} 5^+ & 0^- & 0^+ & 4^- \\ 2^+ & 7^+ & 3^- & -8^+ \\ 2^+ & 0^- & 0^+ & 0^- \\ 8^- & 3^+ & 1^- & 9^+ \end{vmatrix} \]

Cofactor expansion along ANY column or row, but row/col with lots of zeros are best.

Expanding along the third row (which contains three zeros):

\[ = (2) \begin{vmatrix} 0^+ & 0^- & 4^+ \\ 7 & 3 & -8 \\ 3 & 1 & 9 \end{vmatrix} + (0) \begin{vmatrix} \text{I don't} \\ \text{care} \end{vmatrix} + (0) |IDC| + (0) |IDC| \]

Now expanding the resulting \( 3 \times 3 \) matrix along its first row:

\[ = (2) \left\{ (0) |IDC| - (0) |IDC| + (4) \begin{vmatrix} 7 & 3 \\ 3 & 1 \end{vmatrix} \right\} \]
\[ = (2)(4)(7 - 9) = -16 \]
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Determinants of Triangular Matrices

If \( A \) is triangular then \( \det(A) \) is the product of main diagonal elements.

\[ \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} = 3 \]
\[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} = (1) \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = (1)(4)(6) = 24 \]

How row operations affect determinants

  1. Each time two rows are interchanged, the determinant changes sign.
  2. If one row of \( A \) is multiplied by \( k \) to produce \( B \), then \( \det(B) = k \det(A) \).
  3. If multiples of one row is added to another, the determinant does not change.
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Example: Row Interchange

\( A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \quad \det(A) = 3 \)

Interchange rows: \( \begin{bmatrix} 0 & 3 \\ 1 & 2 \end{bmatrix} \quad \det = -3 \)

Example: Row Scaling

\( A = \begin{bmatrix} \frac{1}{10} & \frac{1}{5} \\ 3 & 4 \end{bmatrix} \quad \det(A) = -\frac{1}{5} \)

Factor out \( \frac{1}{10} \) out of first row: \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)

\[ \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} = -2 \]

\( (\frac{1}{10})(-2) = -\frac{1}{5} \)

Just as in \( 2 \times 2 \), any square matrix \( A \) is invertible if and only if \( \det(A) \neq 0 \).

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Properties of Determinants

\[ \det(A^T) = \det(A) \]

This means we can do elementary column ops.

\[ \det(AB) = \det(A) \det(B) \]

But \( \det(A+B) \neq \det(A) + \det(B) \) in general.

\[ \det(A^n) = [\det(A)]^n \]

Linearity Property

If one col of \( A \) is multiplied by some \( c \), then determinant is also multiplied by \( c \).

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \det(A) = -2 \]
\[ B = \begin{bmatrix} 2 & 2 \\ 6 & 4 \end{bmatrix} \quad \det(B) = -4 \]

col 1 of A times 2

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If one col \( A \) is linear combo of column vectors, then determinant is the linear combo of determinants of matrices w/ those vectors.

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \det(A) = -2 \]
\[ B = \begin{bmatrix} 2 & 2 \\ 4 & 4 \end{bmatrix} \quad \det(B) = 0 \]
\[ \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
\[ \det\left(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\right) + \det\left(\begin{bmatrix} 1 & 2 \\ 1 & 4 \end{bmatrix}\right) \]
\[ = -2 + 2 = 0 \]
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What is the determinant?

\[ \vec{u} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \quad \vec{v} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} \]

Consider the vectors \( \vec{u} \) and \( \vec{v} \) plotted on a coordinate system. We want to find the area of the parallelogram formed by these vectors.

Geometric Interpretation

The area of the parallelogram can be calculated by transforming it into a rectangle with the same base and height.

A coordinate graph showing vectors u=(1,2) and v=(3,0) forming a parallelogram in the first quadrant.

By shifting the triangular part of the parallelogram, we obtain a rectangle with a base of 3 and a height of 2.

Area = 6
Diagram showing a parallelogram with base 3 and height 2 being rearranged into a 3 by 2 rectangle.

Algebraic Calculation

We can also find this area using the determinant of the matrix formed by the vectors:

\[ A = \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix} \] \[ |\det(A)| = |0 - 6| = 6 \]

Note: The absolute value of the determinant corresponds exactly to the area calculated geometrically.