Determinant

15.1 Determinant

Every square matrix has a number assosiated with it can we call this number as determinant, often denoted as \(det(A) = |A|\) for matrix \(A\).

Prperties of determinant is as follows:

1. \(|I|=1\)

2. Row exchange reverses the sign of the determinant:

Permutation Matrices are derived by row exchange of Identity Matrix. Hence, \(|P|= \pm1\).

3. (a) For any square matrix \(A\), \(\begin{vmatrix} ta & tb \\ c & d \end{vmatrix}=t\begin{vmatrix} a & b \\ c & d \end{vmatrix}\)

4. (b) For any square matrix \(A\), \(|A|\) behaves like a linear function of a row if all the other rows are keep fixed. \(\begin{vmatrix} a+a^{'} & b+b^{'} \\ c & d \end{vmatrix}=\begin{vmatrix} a & b \\ c & d \end{vmatrix}+\begin{vmatrix} a^{'} & b^{'} \\ c & d \end{vmatrix}\)

5. If two rows of a square matrix \(A\) are equal, \(|A| = 0\): This can be proved using property 2. Exchanging the rows changes the sign of the determinant, but for a matrix \(A\) which has two equal rows, the matrix obtained by exchanging these equal rows is same as \(A\), i.e. \(|A| = -|A| \implies |A|=0\).

6. Subtracting a multiple of one row from another, doesn’t change the determinant:

\[\begin{align} |A| = \begin{vmatrix} a & b \\ c-la & d-lb \end{vmatrix}= \begin{vmatrix} a & b \\ c & d \end{vmatrix}+\begin{vmatrix} a & b \\ -la & -lb \end{vmatrix} [\text{From 3(b)}] \end{align}\]\[\begin{align} =\begin{vmatrix} a & b \\ c & d \end{vmatrix}+(-l)\begin{vmatrix} a & b \\ a & b \end{vmatrix} [\text{From 3(a)}]=\begin{vmatrix} a & b \\ c & d \end{vmatrix}+(-l)0 [\text{From 4}] = |A| \end{align}\]

6. For a square matrix \(A\), row of zeroes lead to \(|A|=0\): Let \(t\) be a multiple other than one for one of the rows of \(A\), then

\[\begin{align} t|A| = t\begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix}= \begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix}=|A| [\text{From 3(b)}] \implies t|A| = |A| \implies |A| = 0 \end{align}\]

7. For an upper traingular matrix \(U\) with \(d_is\) are the diagonal elements, \(|U| = \prod_{i=1}^{n}d_i\): Any upper triangular matrix \(U\) can be converted to a diagonal matrix \(D\) just by row elimination steps. Hence, \(|U| = |D|\). For a \(3 \times 3\) matrix, we can say that

\[\begin{align} |U|= |D| = \begin{vmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{vmatrix}=d_1d_2d_3\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\text{[From 3(a)]}=d_1d_2d_3|I|=d_1d_2d_3 \end{align}\]

8. \(|A|=0\) exactly when \(A\) is singular and \(|A| \neq 0\) when \(A\) is invertible: If the matrix is singular, by elimination steps we can get a row of all zeroes and hence \(0\) determinant.

9. \(|AB| = |A||B|\): Using this property, \(A^{-1}A = I \implies |A^{-1}A| = |I| \implies |A^{-1}||A| = |I| \implies |A^{-1}| = \frac{1}{|A|}\). Similarly, \(|A^2| = |A|^2\) and \(|2A| = 2^n|A|\) (if multiplying a matrix by 2 means doubling all the entries of \(A\)).

10. \(|A^T| = |A|\): This means that all the properties about the rows hold good for columns as well. Proof: Any matrix \(A\) can be factored as \(LU\). Using this,

\[\begin{align} |A^T| = |A| \implies |U^TL^T| = |LU| \end{align}\]

Now, \(L\) is a matrix where all the diagonal elements are \(1\) with the elements in upper half \(0\), i.e. \(|L| = 1\). \(L^T\) is an upper triangular matrix where all the diagonal elements are \(1\) and hence \(|L^T| = 1\). Similarly, \(|U^T| = |U|\). Hence the equality holds.