### 3 Properties of determinants

Often, especially with determinants of large order, we can simplify the evaluation rules. In this Section we quote some useful properties of determinants in general.

1. If two rows (or two columns) of a determinant are interchanged then the value of the determinant is multiplied by $\left(-1\right)$ .

For example $\left|\begin{array}{cc}\hfill 4\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=8-3=5$ but (interchanging columns) $\left|\begin{array}{cc}\hfill 3\hfill & \hfill 4\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=3-8=-5$ and (interchanging rows) $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 3\hfill \end{array}\right|=3-8=-5.$

2. The determinant of a matrix $A$ and the determinant of its transpose ${A}^{T}$ are equal.

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right|=4-6=-2$

3. If two rows (or two columns) of a matrix $A$ are equal then it has zero determinant.

For example, the following determinant has two identical rows:

$\begin{array}{rcll}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 5\hfill & \hfill 6\hfill \end{array}\right|& =& 1×\left(\left|\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 5\hfill & \hfill 6\hfill \end{array}\right|\right)+2×\left(-\left|\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 6\hfill \end{array}\right|\right)+3×\left(\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 5\hfill \end{array}\right|\right)& \text{}\\ & =& -3+2×\left(6\right)+3×\left(-3\right)=0.\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$
4. If the elements of one row (or one column) of a determinant are multiplied by $k$ , then the resulting determinant is $k$ times the given determinant:

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 8\hfill & \hfill 6\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill \end{array}\right|=2\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 3\hfill \\ \hfill 7\hfill & \hfill 8\hfill & \hfill 9\hfill \end{array}\right|.$

Note that if one row (or column) of a determinant is a multiple of another row (or column) then the value of the determinant is zero. (This follows from properties 3 and 4.)

For example:

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill 2& \hfill 4& \hfill -1\\ \hfill 4& \hfill 2& \hfill 1\\ \hfill -4& \hfill -8& \hfill 2\end{array}\right|=2×\left|\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -8& \hfill 2\end{array}\right|+4×\left(-\left|\begin{array}{cc}\hfill 4& \hfill 1\\ \hfill -4& \hfill 2\end{array}\right|\right)-1×\left|\begin{array}{cc}\hfill 4& \hfill 2\\ \hfill -4& \hfill -8\end{array}\right|$

$\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=2\left(12\right)+4\left(-12\right)-\left(-24\right)=0$

This is predictable as the 3rd row is $\left(-2\right)$ times the first row.

5. If we add (or subtract) a multiple of one row (or column) to another, the value of the determinant is unchanged.

Given $\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 5\hfill \end{array}\right|$ , add (2 $×$ row 1) to (row 2) gives

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4+2×1\hfill & \hfill 5+2×2\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 6\hfill & \hfill 9\hfill \end{array}\right|=9-12=-3=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 5\hfill \end{array}\right|$

6. The determinant of a lower triangular matrix, an upper triangular matrix or a diagonal matrix is the product of the elements on the leading diagonal.

As an example, it is easily confirmed that each of the following determinants has the same value $1×4×6=24$ .

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \end{array}\right|,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 5\hfill & \hfill 6\hfill \end{array}\right|,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 6\hfill \end{array}\right|$

This task is in four parts. Consider

$\phantom{\rule{2em}{0ex}}\Delta =\left|\begin{array}{cccc}\hfill 1& \hfill 4\phantom{\rule{1em}{0ex}}& \hfill 8& \hfill 2\\ \hfill 2& \hfill -1\phantom{\rule{1em}{0ex}}& \hfill 1& \hfill -3\\ \hfill 0& \hfill 2\phantom{\rule{1em}{0ex}}& \hfill 4& \hfill 2\\ \hfill 0& \hfill 3\phantom{\rule{1em}{0ex}}& \hfill 6& \hfill 3\end{array}\right|$

1. Use property 2 to find another matrix whose determinant is equal to $\Delta$ :

$\Delta =\left|\begin{array}{cccc}\hfill 1& \hfill 2& \hfill 0& \hfill 0\\ \hfill 4& \hfill -1& \hfill 2& \hfill 3\\ \hfill 8& \hfill 1& \hfill 4& \hfill 6\\ \hfill 2& \hfill -3& \hfill 2& \hfill 3\end{array}\right|$ , by transposing the matrix.

2. Now expand along the top row to express $\Delta$ as the sum of two products, each of a number and a $3×3$ determinant:

$\Delta =1×\left|\begin{array}{ccc}\hfill -1& \hfill 2& \hfill 3\\ \hfill 1& \hfill 4& \hfill 6\\ \hfill -3& \hfill 2& \hfill 3\end{array}\right|-2×\left|\begin{array}{ccc}\hfill 4\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 8\hfill & \hfill 4\hfill & \hfill 6\hfill \\ \hfill 2\hfill & \hfill 2\hfill & \hfill 3\hfill \end{array}\right|$

3. Use the statement after property 4 to show that the second of the $3×3$ determinants is zero:

In the second $3×3$ determinant, row $2=2×$ row 1 hence the determinant has value zero.

4. Use the statement after property 4 to evaluate the first determinant:

In the first $3×3$ determinant column 3 $=\frac{3}{2}\phantom{\rule{1em}{0ex}}×$ column 2. Hence this determinant is also zero. Therefore $\Delta =0.$

##### Exercises
1. Use Laplace expansion along the 1st row to determine

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill 3& \hfill 1& \hfill -4\\ \hfill 6& \hfill 9& \hfill -2\\ \hfill -1& \hfill 2& \hfill 1\end{array}\right|$

Show that the same value is obtained if you choose any other row or column for your expansion.

2. Using any of the properties of determinants to minimise the arithmetic, evaluate

$\phantom{\rule{2em}{0ex}}\text{(a)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|\begin{array}{ccc}\hfill 12& \hfill 27& \hfill 12\\ \hfill 28& \hfill 18& \hfill 24\\ \hfill 70& \hfill 15& \hfill 40\end{array}\right|\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{(b)}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\left|\begin{array}{cccc}\hfill 2& \hfill 4& \hfill 6& \hfill 4\\ \hfill 0& \hfill 4& \hfill 6& \hfill 9\\ \hfill 2& \hfill 1& \hfill 4& \hfill 0\\ \hfill 1& \hfill 2& \hfill 3& \hfill 2\end{array}\right|$

3. Find the cofactors of $x,y,z$ in the determinant

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill 1& \hfill 1& \hfill 1\\ \hfill 2& \hfill 3& \hfill 4\\ \hfill x& \hfill y& \hfill z\end{array}\right|$

4. Prove that, no matter what the values of $x,y,z,$ are

$\phantom{\rule{2em}{0ex}}\left|\begin{array}{ccc}\hfill y+z\hfill & \hfill z+x\hfill & \hfill x+y\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|=0$

1. $3\left|\begin{array}{cc}\hfill 9& \hfill -2\\ \hfill 2& \hfill 1\end{array}\right|-1\left|\begin{array}{cc}\hfill 6& \hfill -2\\ \hfill -1& \hfill 1\end{array}\right|-4\left|\begin{array}{cc}\hfill 6& \hfill 9\\ \hfill -1& \hfill 2\end{array}\right|\begin{array}{c}=3\left(9+4\right)-1\left(6-2\right)-4\left(12+9\right)=-49\hfill \end{array}$
1. Take out common factors in rows and columns

$720\left|\begin{array}{ccc}\hfill 2\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill 7\hfill & \hfill 3\hfill & \hfill 3\hfill \\ \hfill 7\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right|=720\left|\begin{array}{ccc}\hfill 0& \hfill 0& \hfill 1\\ \hfill 1& \hfill -6& \hfill 3\\ \hfill 3& \hfill -5& \hfill 2\end{array}\right|$   using $\left(-2{C}_{3}+{C}_{1}\right)$ then $\left(-3{C}_{3}+{C}_{2}\right)$ .

The value of the determinant (expand along top row) is then easily found

as $720×13=9360$ .

2. Zero since (row 1) is $2$   $×$ (row 4).
2. Cofactors of $x,y,z$ are $1,-2,1$ respectively.