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3.6 Determinants (part 2)

Transpose of A

transpose of A:

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

rows \(\leftrightarrow\) cols

note \( \det A = \det A^T \)

True for 3x3 and beyond, too

\[ A = \begin{bmatrix} 3 & 0 & 3 \\ 3 & 4 & 3 \\ 0 & 5 & -2 \end{bmatrix} \quad A^T = \begin{bmatrix} 3 & 3 & 0 \\ 0 & 4 & 5 \\ 3 & 3 & -2 \end{bmatrix} \]

choose col 1

\[ \det A = (3) \begin{vmatrix} 4 & 3 \\ 5 & -2 \end{vmatrix} - (3) \begin{vmatrix} 0 & 3 \\ 5 & -2 \end{vmatrix} = (3)(-23) - (3)(-15) \]\[ = -69 + 45 = -24 \]

choose row 1

\[ \det A^T = \text{same work as above} = \dots = -24 \]

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Triangular Matrix

\[ A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \]

Upper Triangular

(all #s below main diagonal are 0)

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

Lower Triangular

(all #s above main diagonal are 0)

determinant of triangular matrix is the product of the main diagonal numbers

\[ A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 3 & 0 \\ 4 & 5 & 6 \end{bmatrix} \quad \xrightarrow{\text{row 1}} \quad \det A = (1) \begin{vmatrix} 3 & 0 \\ 5 & 6 \end{vmatrix} = (1)(3)(6) = 18 \]

Zeros are good for determinants

Can we do row operations to introduce zeros (or make matrix triangular) and then find determinant?

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Determinant Properties and Row Operations

yes, but ...

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

Row Operation of Swapping Rows

\[ \xrightarrow{\text{swap}(R_1, R_2)} \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} \quad \text{determinant is } 2 \]

each swap of two rows switches the sign of the determinant

\[ A = \begin{bmatrix} \frac{1}{10} & \frac{2}{10} \\ 3 & 4 \end{bmatrix} \quad \det A = \frac{4}{10} - \frac{6}{10} = -\frac{2}{10} = -\frac{1}{5} \]

\[ B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \det B = -2 = 10 \cdot -\frac{1}{5} = 10 \det A \]

(row 1 of A times 10)

multiplying one row by \( k \) (\( k \neq 0 \)) changes the determinant by a factor of \( k \)

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

\[ \xrightarrow{(-3)R_1 + R_2} \begin{bmatrix} 1 & 2 \\ 0 & -2 \end{bmatrix} \quad \det \text{ is } (1)(-2) = -2 \]

upper triangular

same

multiplying one row and add to another does NOT affect the determinant

this is good because we use the operation to introduce zeros

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Example: Calculating Determinants by Introducing Zeros

Consider the matrix:

\[ A = \begin{bmatrix} 2 & -2 & 3 \\ 3 & -3 & 2 \\ 5 & 1 & 9 \end{bmatrix} \]

Let's try to introduce zeros or make it triangular. Notice column 2 has the easiest numbers to use to introduce zeros (column 1 and 3 involve fractions).

These do NOT change the determinant:

\((2)R_3 + R_1\)
\((3)R_3 + R_2\)

\[ \begin{bmatrix} 12^+ & 0^- & 21^+ \\ 18^- & 0^+ & 29^- \\ 5^+ & 1^- & 9^+ \end{bmatrix} \]

Now column 2 is good to expand on:

\[ \det A = -(1) \begin{vmatrix} 12 & 21 \\ 18 & 29 \end{vmatrix} = 30 \]

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Example: 4x4 Matrix Determinant

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

Column 1 has nice numbers, so there is room to introduce more zeros.

Makes upper a 1 which is a good pivot:

Swap \((R_1, R_3)\)

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

One swap \(\rightarrow\) det flips sign

These do NOT change det:

\((-1)R_1 + R_2\)
\((-2)R_1 + R_3\)

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

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\((-2)R_2 + R_3\)

\((-2)R_2 + R_4\)

\(\rightarrow\)

\[\begin{bmatrix} 1 & 1 & 2 & 4 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & -12 & -2 \\ 0 & 0 & -12 & 3 \end{bmatrix}\]

\((-1)R_3 + R_4\)

\(\rightarrow\)

\[\begin{bmatrix} 1 & 1 & 2 & 4 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & -12 & -2 \\ 0 & 0 & 0 & 5 \end{bmatrix}\]

Note: The matrix is now in upper triangular form. The diagonal elements are 1, 1, -12, and 5.

\[\det A = -(1)(1)(-12)(5) = 60\]

\(\uparrow\) from swap earlier

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Determinant Properties: Transpose and Operations

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

col operations \(\longleftrightarrow\) row operations
row ops \(\longleftrightarrow\) col ops

Normally, we do NOT do column operations (they change solutions to the system).

But, for the purpose of finding det, we can do column ops.

\[A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\]

\((-2)C_1 + C_2\)

\(\rightarrow\)

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

\(\det A = -2\)

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

\((-2)R_1 + R_2\)

\(\rightarrow\)

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

\(\det A^T = -2\)