## Minor Matrix $$A_{ij}$$

Let’s first define what is a minor matrix.

A minor matrix $$A_{ij}$$ is the matrix $$A$$ with its $$i$$th row and $$j$$th coloumn crossed out.

With minor matrix notation, we can write the third-order determinant as following:

$a_{11} |A_{11}| - a_{12}|A_{12}| + a_{13}|A_{13}|$

The numbers $$a\prime_{11} = |A_{11}|$$, $$a\prime_{12} = |A_{12}|$$, and $$a\prime_{13} = |A_{13}|$$ are called the cofactors of $$a_{11}$$, $$a_{12}$$, $$a_{13}$$

We can now define the cofactor of $$a_{ij}$$ in matrix $$A$$ is

$a\prime_{ij} = (-1)^{i+j} det(A_{ij})$

Then the determinant is simply

$a_{11}a\prime{11} + a_{12}a\prime{12} + … + a_{1n}a\prime{1n}$

This is called expansion by minors on the first row.

#### General Expansion by Minors

Let $$A$$ be an $$n \times n$$ matrix, and let $$r$$ and $$s$$ be any selections from the list of numbers $$1,2,…,n$$. Then

$det(A) = a_{r1}a\prime_{r1} + a_{r2}a\prime_{r1} + … + a_{rn}a\prime_{rn}$

also

$det(A) = a_{1s}a\prime_{1s} + a_{2s}a\prime_{2s} + … + a_{ns}a\prime_{ns}$

## Determinant Properties

#### The Transponse Property

Therom: For any square matrix $$A$$, we have $$det(A) = det(A^T)$$

Proof:

Order 1 determinants:

$$a_{11} = a_{11}$$, ╮(╯_╰)╭

Order 2 determinants:

$$b_1c_2 - b_2c_1 = b_1c_2 - c_1b_2$$

Let $$n > 2$$, assume property holds for square matrices of size smaller than $$n \times n$$.

$det(A) = a_{11}|A_{11}| - a_{12}|A_{12}| + … + (-1)^{n+1}a_{1n}|A_{1n}|$

Let $$B = A^T$$, we have

$det(B) = b_{11}|B_{11}| - b_{21}|B_{21}| + … + (-1)^{n+1}b_{n1}|B_{n1}|$

However, we know $$a_{1j} = b_{j1}$$ and $$B_{j1} = {A_{1j}}^T$$. Applying our induction hypothesis to $$(n-1)$$st-order determinant $$|A_{1j}|$$, we have $$|A_{1j}| = |B_{j1}|$$.

We conclude $$det(A) = det(B) = det(A^T)$$.

$$Q.E.D.$$

#### The Row-Interchange Property

Theorem: If two different rows of a square matrix $$A$$ are interchanged, the determinant of the resulting matrix is $$-det(A)$$

Proof:

The proof for $$n = 2$$ is trivial. Assume $$n > 2$$, and this holds for all matrices smaller than $$n \times n$$.

Let $$B$$ be matrix obtained from $$A$$ by interchanging $$i$$th and $$r$$th rows. Because $$n > 2$$, we can choose another $$k$$th row for expansion. Consider these two cofactors

$(-1)^{k+j}|A_{kj}| \ \text{and} \ (-1)^{k+j}|B_{kj}|$

These numbers must have opposite signs by induction, since $$A_{kj}$$ and $$B_{kj}$$ have size of $$(n-1) \times (n-1)$$. Therefore we have $$|B_{kj}| = -|A_{kj}|$$.

Expanding by minors on $$k$$th row will give us $$det(A) = -det(B)$$

$$Q.E.D.$$

#### The Equal-Rows Property

Theorem: If two rows of a square matrix $$A$$ are equal, then $$det(A) = 0$$.

Proof:

Let $$B$$ be the matrix obtained by interchanging two equal rows of $$A$$.

We have $$det(B) = -det(A)$$, but $$B = A$$, thus $$det(A) = 0$$. _(┐「ε:)_

$$Q.E.D.$$

#### The Scalar-Multiplication Property

Theorem: If a single row of a square matrix $$A$$ is multiplied by a scalar $$r$$, the determinant of the resulting matrix is $$r \cdot det(A)$$.

Proof:

Let $$r$$ be any scalar, and let $$B$$ be the matrix obtained from $$A$$ by replacing the $$k$$th row of $$A$$ by scalar multiple $$r$$ of it.

Since all rows are equal except $$k$$, we know all cofactors are equal. Thus we can conclude $$det(B) = r \cdot det(A)$$.

$$Q.E.D.$$

Theorem: If the product of one row of a square matrix $$A$$ by a scalar is added to a different row of $$A$$, the determinant remain the same.
Same as above, we can see all cofactors are equal since other than $$k$$th row, everything else remains the same.
By expanding by minors on $$k$$th row, we have
$r \cdot det(C) + det(A)$, and note $$C$$ is just matrix $$A$$ with two same rows. Thus the final answer is still $$det(A)$$.
$$Q.E.D.$$