Determinant

For any square matrix of order 2, we have found a necessary and sufficient condition for invertibility. Indeed, consider the matrix

$\begin{displaymath}A = \left(\begin{array}{cc} a&b\\ c&d\\ \end{array}\right).\end{displaymath}$

The matrix A is invertible if and only if $ad - bc \neq 0$ . We called this number the determinant of A. It is clear from this, that we would like to have a similar result for bigger matrices (meaning higher orders). So is there a similar notion of determinant for any square matrix, which determines whether a square matrix is invertible or not?

In order to generalize such notion to higher orders, we will need to study the determinant and see what kind of properties it satisfies. First let us use the following notation for the determinant

$\begin{displaymath}\mbox{determinant of $\left(\begin{array}{cc} a&b\\ c&d\\ \... ...begin{array}{cc} a&b\\ c&d\\ \end{array}\right\vert = ad -bc.\end{displaymath}$

Properties of the Determinant

1.

Any matrix A and its transpose have the same determinant, meaning

$\begin{displaymath}\det A = \det A^T\end{displaymath}$

This is interesting since it implies that whenever we use rows, a similar behavior will result if we use columns. In particular we will see how row elementary operations are helpful in finding the determinant. Therefore, we have similar conclusions for elementary column operations.

2.

The determinant of a triangular matrix is the product of the entries on the diagonal, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a&b\\ 0&d\\ \end{array}\right\v... ...rt\begin{array}{cc} a&0\\ b&d\\ \end{array}\right\vert = ad .\end{displaymath}$

3.

If we interchange two rows, the determinant of the new matrix is the opposite of the old one, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a&b\\ c&d\\ \end{array}\right\v... ...left\vert\begin{array}{cc} c&d\\ a&b\\ \end{array}\right\vert\end{displaymath}$

4.

If we multiply one row with a constant, the determinant of the new matrix is the determinant of the old one multiplied by the constant, that is

$\begin{displaymath}\left\vert\begin{array}{cc} \lambda a&\lambda b\\ c&d\\ \en... ...rray}{cc} a&b\\ \lambda c&\lambda d\\ \end{array}\right\vert.\end{displaymath}$

In particular, if all the entries in one row are zero, then the determinant is zero.

5.

If we add one row to another one multiplied by a constant, the determinant of the new matrix is the same as the old one, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a + \lambda c&b + \lambda d\\ c&... ...} a&b\\ c + \lambda a&d + \lambda b\\ \end{array}\right\vert.\end{displaymath}$

Note that whenever you want to replace a row by something (through elementary operations), do not multiply the row itself by a constant. Otherwise, you will easily make errors (due to Property 4).

6.

We have

$\begin{displaymath}\det(AB) = \det(A) \det(B).\end{displaymath}$

In particular, if A is invertible (which happens if and only if $\det(A) \neq 0$ ), then

$\begin{displaymath}\det(A^{-1}) = \frac{1}{\det(A)}.\end{displaymath}$

If A and B are similar, then $\det(A) = \det(B)$ .

Let us look at an example, to see how these properties work.

Example. Evaluate

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ -1&3\\ \end{array}\right\vert.\end{displaymath}$

Let us transform this matrix into a triangular one through elementary operations. We will keep the first row and add to the second one the first multiplied by $\displaystyle \frac{1}{2}$ . We get

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ -1&3\\ \end{array}\right\... ...} 2&1\\ 0&\displaystyle \frac{7}{2}\\ \end{array}\right\vert.\end{displaymath}$

Using the Property 2, we get

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ 0&\displaystyle \frac{7}{2}\\ \end{array}\right\vert = 2 \cdot \frac{7}{2} = 7.\end{displaymath}$

Therefore, we have

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ -1&3\\ \end{array}\right\vert= 7\end{displaymath}$

which one may check easily.

Determinant

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