Matrix Algebra

Product Rule

Properties of Row Operations. By the effect of transpose in determinant (see Laplace expansions) properties 1–3 [also 4–5] of the column operations preserve for row vectors. Thus, we obtain the properties involving row operations.

$\det B = \det A$ , if replaces by adding a multiple of the th row to the th row of .
$\det B = - \det A$ , if interchanges the th and th rows of .
$\det B = c\cdot\det A$ , if is produced by multiplying the th row of by .

Further Discussion of Properties. Recall that the row operations “replacement,” “interchange,” and “scaling” correspond to elementary matrices, say $E_1$ , $E_2$ , and $E_3$ , and recall how we can construct $E_1$ , $E_2$ , and $E_3$ from the identity matrix . Since $\det I = 1$ , we can complete the following calculations.

$\det E_1 = \det I = 1$ ;
$\det E_2 = -\det I = -1$ ;
$\det E_3 = c\cdot\det I = c$ .

Then the properties of determinant over row operations can be simply expressed as

$\displaystyle \det E A = (\det E)(\det A)$ for every elementary matrix

. $\displaystyle$

Determining Invertibility. If is invertible, there is a series of elementary matrices $E_1, E_2, \ldots, E_p$ so that $E_p \cdots E_2 E_1 A = I$ . By applying the above equation repeatedly, we obtain $(\det E_p)\cdots (\det E_2)(\det E_1)(\det A) = 1$ . Since $\det E_i \neq 0$ for all , we found ${\det A \neq 0}$ . Similarly if is not invertible, we obtain ${\det A = 0}$ (why?). Together we conclude that

$\displaystyle \det A \neq 0$ if and only if is invertible. $\displaystyle$

Product Rule. If is invertible, we can express $A = E_1^{-1} \cdots E_p^{-1}$ (why?). Since $E_i^{-1}$ 's are also elementary matrices, we obtain

$\begin{displaymath} \begin{array}{ll} \det AB &= \det E_1^{-1} \cdots E_p^... ... \cdots E_p^{-1}) (\det B) = (\det A) (\det B) \end{array} \end{displaymath}$

Thus, we have shown that

$\displaystyle \det AB = (\det A) (\det B)$

is not invertible, neither is

(why?); thus, ${\det AB = (\det A) (\det B) = 0}$ .

EXAMPLES 3. Find an LU factorization of , and then compute $\det {A}$ using the upper triangular matrix in each of the following.

$A = \begin{bmatrix} 1 &-4 & 2 \\ -2 & 8 &-9 \\ -1 & 7 & 0 \end{bmatrix}$
$A = \begin{bmatrix} 2 &-8 & 6 & 8 \\ 3 &-9 & 5 &10 \\ -3 & 0 & 1 &-2 \\ 1 &-4 & 0 & 6 \end{bmatrix}$