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:
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$|I|=1$
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Row exchange reverses the sign of the determinant:
Permutation Matrices are derived by row exchange of Identity Matrix. Hence, $|P|= \pm1$.
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(a) For any square matrix $A$, $\begin{vmatrix} ta & tb \\ c & d \end{vmatrix}=t\begin{vmatrix} a & b \\ c & d \end{vmatrix}$
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(b) For any square matrix $A$, $|A|$ behaves like a linear function of a row if all the other rows are keep fixed. $\begin{vmatrix} a+a^{’} & b+b^{’} \\ c & d \end{vmatrix}=\begin{vmatrix} a & b \\ c & d \end{vmatrix}+\begin{vmatrix} a^{’} & b^{’} \\ c & d \end{vmatrix}$
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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. $|A| = -|A| \implies |A|=0$.
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Subtracting a multiple of one row from another, doesn’t change the determinant:
$$\begin{align} |A| = \begin{vmatrix} a & b \\ c-la & d-lb \end{vmatrix}= \begin{vmatrix} a & b \\ c & d \end{vmatrix}+\begin{vmatrix} a & b \\ -la & -lb \end{vmatrix} [\text{From 3(b)}] \end{align}$$
$$\begin{align} =\begin{vmatrix} a & b \\ c & d \end{vmatrix}+(-l)\begin{vmatrix} a & b \\ a & b \end{vmatrix} [\text{From 3(a)}]=\begin{vmatrix} a & b \\ c & d \end{vmatrix}+(-l)0 [\text{From 4}] = |A| \end{align}$$
- 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
$$\begin{align} t|A| = t\begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix}= \begin{vmatrix} 0 & 0 \\ c & d \end{vmatrix}=|A| [\text{From 3(b)}] \implies t|A| = |A| \implies |A| = 0 \end{align}$$
- 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
$$\begin{align} |U|= |D| = \begin{vmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{vmatrix}=d_1d_2d_3\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\text{[From 3(a)]}=d_1d_2d_3|I|=d_1d_2d_3 \end{align}$$
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$|A|=0$ 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.
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$|AB| = |A||B|$: 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 $A$).
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$|A^T| = |A|$: 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.