Theorems on determinants | Improve Society

The value of a determinant remains the same if rows and columns are interchanged. In symbols, $\det(A) = \det(A^T)$ .

If all elements of any row [or column] are zero except for one element, then the value of the determinant is equal to the product of that element by its cofactor. In particular, if all elements of a row [or column] are zero the determinant is zero.

An interchange of any two rows [or columns] changes the sign of the determinant.

If all elements in any row [or column] are multiplied by a number, the determinant is also multiplied by this number.

If any two rows [or columns] are the same or proportional, the determinant is zero.

If we express the elements of each row [or column] as the sum of two terms, then the determinant can be expressed as the sum of two determinants having the same order.

If we multiply the elements of any row [or column] by a given number and add to corresponding elements of any other row [or column], then the value of the determinant remains the same.

If $A$ and $B$ are square matrices of the same order, then
$\det(AB) = \det(A)\det(B)\cdots(11)$

The sum of the products of the elements of any row [or column] by the cofactors of another row [or column] is zero. In symbols,
$\displaystyle \sum^n_{k=1}a_{qk}A_{pk} = 0$ or $\displaystyle \sum^n_{k=1}a_{kq}A_{kp} = 0$ if $p \ne q\cdots(12)$
If $p = q$ , the sum is $\det(A)$ by (10).

Let $v_1,\ v_2,\ \cdots,\ v_n$ represent row vectors [or column vectors] of a square matrix $A$ of order n. Then $\det(A) = 0$ if and only if there exist constants [scalars] $\lambda_1,\ \lambda_2,\ \cdots,\ \lambda_n$ not all zero such that
$\lambda_1v_1 + \lambda_2v_2 + \cdots + \lambda_nv_n = O \cdots(13)$
where O is the null or zero row matrix. If condition (13) is satisfied we say that the vectors $v_1,\ v_2,\ \cdots,\ v_n$ are linearly dependent. A matrix $A$ such that $\det(A) = 0$ is called a singular matrix. If $\det(A) \ne 0$ , then $A$ is a non-singular matrix.

In practice we evaluate a determinant of order n by using Theorem 7 successively to replace all but one of the elements in a row or column by zeros and then using Theorem 2 to obtain a new determinant of order n – 1. We continue in this manner, arriving ultimately at determinants of order 2 or 3 which are easily evaluated.