## Understanding Properties of Determinants

There are many properties of determinants. Listed here are some properties that may be helpful in calculating the determinant of a matrix.

### A General Note: Properties of Determinants

1. If the matrix is in upper triangular form, the determinant equals the product of entries down the main diagonal.
2. When two rows are interchanged, the determinant changes sign.
3. If either two rows or two columns are identical, the determinant equals zero.
4. If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero.
5. The determinant of an inverse matrix ${A}^{-1}$ is the reciprocal of the determinant of the matrix $A$.
6. If any row or column is multiplied by a constant, the determinant is multiplied by the same factor.

### Example 7: Illustrating Properties of Determinants

Illustrate each of the properties of determinants.

### Solution

Property 1 states that if the matrix is in upper triangular form, the determinant is the product of the entries down the main diagonal.

$A=\left[\begin{array}{rrr}\hfill 1& \hfill 2& \hfill 3\\ \hfill 0& \hfill 2& \hfill 1\\ \hfill 0& \hfill 0& \hfill -1\end{array}\right]$

Augment $A$ with the first two columns.

$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 1\\ 0& 0& -1\end{array}|\begin{array}{c}1\\ 0\\ 0\end{array}\begin{array}{c}2\\ 2\\ 0\end{array}\right]$

Then

$\begin{array}{l}\mathrm{det}\left(A\right)=1\left(2\right)\left(-1\right)+2\left(1\right)\left(0\right)+3\left(0\right)\left(0\right)-0\left(2\right)\left(3\right)-0\left(1\right)\left(1\right)+1\left(0\right)\left(2\right)\hfill \\ =-2\hfill \end{array}$

Property 2 states that interchanging rows changes the sign. Given

$\begin{array}{l}\begin{array}{l}\\ A=\left[\begin{array}{cc}-1& 5\\ 4& -3\end{array}\right],\mathrm{det}\left(A\right)=\left(-1\right)\left(-3\right)-\left(4\right)\left(5\right)=3 - 20=-17\end{array}\hfill \\ \hfill \\ B=\left[\begin{array}{cc}4& -3\\ -1& 5\end{array}\right],\mathrm{det}\left(B\right)=\left(4\right)\left(5\right)-\left(-1\right)\left(-3\right)=20 - 3=17\hfill \end{array}$

Property 3 states that if two rows or two columns are identical, the determinant equals zero.

$\begin{array}{l}A=\left[\begin{array}{ccc}1& 2& 2\\ 2& 2& 2\\ -1& 2& 2\end{array}\text{ }|\text{ }\begin{array}{c}1\\ 2\\ -1\end{array} \begin{array}{c}2\\ 2\\ 2\end{array}\right]\hfill \\ \hfill \\ \mathrm{det}\left(A\right)=1\left(2\right)\left(2\right)+2\left(2\right)\left(-1\right)+2\left(2\right)\left(2\right)+1\left(2\right)\left(2\right)-2\left(2\right)\left(1\right)-2\left(2\right)\left(2\right)\hfill \\ =4 - 4+8+4 - 4-8=0\hfill \end{array}$

Property 4 states that if a row or column equals zero, the determinant equals zero. Thus,

$A=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right],\mathrm{det}\left(A\right)=1\left(0\right)-2\left(0\right)=0$

Property 5 states that the determinant of an inverse matrix ${A}^{-1}$ is the reciprocal of the determinant $A$. Thus,

$\begin{array}{l}A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],\mathrm{det}\left(A\right)=1\left(4\right)-3\left(2\right)=-2\hfill \\ \hfill \\ {A}^{-1}=\left[\begin{array}{cc}-2& 1\\ \frac{3}{2}& -\frac{1}{2}\end{array}\right],\mathrm{det}\left({A}^{-1}\right)=-2\left(-\frac{1}{2}\right)-\left(\frac{3}{2}\right)\left(1\right)=-\frac{1}{2}\hfill \end{array}$

Property 6 states that if any row or column of a matrix is multiplied by a constant, the determinant is multiplied by the same factor. Thus,

$\begin{array}{l}A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],\mathrm{det}\left(A\right)=1\left(4\right)-2\left(3\right)=-2\hfill \\ \hfill \\ B=\left[\begin{array}{cc}2\left(1\right)& 2\left(2\right)\\ 3& 4\end{array}\right],\mathrm{det}\left(B\right)=2\left(4\right)-3\left(4\right)=-4\hfill \end{array}$

### Example 8: Using Cramer’s Rule and Determinant Properties to Solve a System

Find the solution to the given 3 × 3 system.

$\begin{array}{ll}2x+4y+4z=2\hfill & \left(1\right)\hfill \\ 3x+7y+7z=-5\hfill & \left(2\right)\hfill \\ \text{ }x+2y+2z=4\hfill & \left(3\right)\hfill \end{array}$

### Solution

Using Cramer’s Rule, we have

$D=|\begin{array}{ccc}2& 4& 4\\ 3& 7& 7\\ 1& 2& 2\end{array}|$

Notice that the second and third columns are identical. According to Property 3, the determinant will be zero, so there is either no solution or an infinite number of solutions. We have to perform elimination to find out.

1. Multiply equation (3) by –2 and add the result to equation (1).
$\frac{\begin{array}{l}-2x - 4y - 4x=-8\hfill \\ \text{ }2x+4y+4z=2\hfill \end{array}}{0=-6}$

Obtaining a statement that is a contradiction means that the system has no solution.