# Finding Determinant By Cross Multiplication

In the previous article, we have seen how determinant decides whether a system of equation (read square matrix) has inverse, or it has a solution, only when the determinant is not zero. The determinant is obtained from the equation given below.

$determinant = \sum \pm a_{1\alpha}a_{2\beta}...a_{nv}$

To know more about finding determinant in this way , read previous article. Here we will discuss about finding determinant by cross multiplication but before that let us understand the different notations used to represent determinants.

### Notation For Determinants

There are several notation for determinants given by earlier mathematicians. Suppose $A$ represents a augmented matrix from a system of linear equations, then determinant of $A$ is given below.

Let the matrix $A$$A$ be a 2 x 2 matrix.

$A = \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}$$A = \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}$

Different ways to represent determinant of matrix $A$$A$

$det(A)$$det(A)$  -- (1)

$|A|$$|A|$ -- (2)

$\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$$\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix}$ -- (3)

### Determinant as a Function

Imagine determinant to a function that take a square matrix as input and give a single value as output. For example,$f(x) = x^3$ be a function where $x$ could be any real number. Similarly, $det(A)$ is a function that matrix as input and give a determinant value $d$. The determinant value is always integer because it is linear combination of integers, that is, all values are integers in the matrix.

Determinant of $1 \times 1$ Matrix

If $A$ is a matrix with just one element, then its determinant is the same element.

Example #1

Let $A$$A$ be a square matrix of order $1 \times 1$$1 \times 1$

$A = \begin{bmatrix}2\end{bmatrix}$$A = \begin{bmatrix}2\end{bmatrix}$

Then the determinant of $A$$A$ is

$|A| = |2| = 2$$|A| = |2| = 2$

### Determinant of $2 \times 2$$2 \times 2$ Matrix

The determinant of a $2 \times 2$ matrix is obtained by performing cross multiplication. See the following figure.

Example #2

Let $A$ be a $2 \times 2$ square matrix. Find the determinant of the matrix $A$.

Solution:

Let the $A$$A$ be 2 x 2 square matrix.

$A = \begin{bmatrix}2 & 3\\1 & 5\end{bmatrix}$$A = \begin{bmatrix}2 & 3\\1 & 5\end{bmatrix}$

$|A| = a \times d - b \times c$$|A| = a \times d - b \times c$

$|A| = 2 \times 5 - 3 \times 1$$|A| = 2 \times 5 - 3 \times 1$

$|A| = 10 - 3 = 7$$|A| = 10 - 3 = 7$

Example #3

Let $B$ be a square matrix of order $2 \times 2$. Find the determinant of the matrix $B$.

Solution:

Let $B$$B$ be a square matrix of order $2 \times 2$$2 \times 2$.

$B = \begin{bmatrix}5 & -1\\4 & -3\end{bmatrix}$$B = \begin{bmatrix}5 & -1\\4 & -3\end{bmatrix}$

$|B| = a \times d - b \times c$$|B| = a \times d - b \times c$

$|B| = 5 \times (-3) - (-1) \times 4$$|B| = 5 \times (-3) - (-1) \times 4$

$|B| = (-15) - (-4)$$|B| = (-15) - (-4)$

$|B| = (-15) + 4 = -11$$|B| = (-15) + 4 = -11$

### Determinant Of $3 \times 3$$3 \times 3$ Matrix

The determinant of a $3 \times 3$ matrix is also possible through cross multiplication; Since we have a larger matrix we need to convert the larger matrix into smaller matrix to compute determinant. See figure below.

Step 1: Select the first row, first element and strike out rest of the elements from first row and first column. Then use to remaining element to create 2 x 2 matrix and find its determinant. See image below.

Each selected element must be multiplied with the respective 2 x 2 determinant obtained by eliminating 1st row and the respective column of selected element.

Step 2: Add all terms together and assign a negative or positive sign to each term.

If the selected element from top row (a b c) is $a$$a$. then row $i = 1$$i = 1$ and column $j = 1$$j = 1$

Then the sign of first element

$= (-1)^{i+j}a(ei - fh)$$= (-1)^{i+j}a(ei - fh)$

$= (-1)^{1 + 1}a(ei - fh)$$= (-1)^{1 + 1}a(ei - fh)$

$= (-1)^{2})a(ei - fh)$$= (-1)^{2})a(ei - fh)$

$= +a(ei - fh)$$= +a(ei - fh)$

Similarly, second element from top row( a b c).

$= -b(di - fg)$$= -b(di - fg)$

Third element from top row( a b c).

$= +c(dh - eg)$$= +c(dh - eg)$

Step 3: Write down the determinant function obtained previously.

The determinant of $3 \times 3$$3 \times 3$ matrix.

$|A| = +a(ei - fh) - b(di - fg) + c(dh - eg)$$|A| = +a(ei - fh) - b(di - fg) + c(dh - eg)$

Let us see few examples of determinant of $3 \times 3$ matrices.

Example #4

Find the determinant of the following $3 \times 3$ matrix.

$C = \begin{bmatrix}2 & 1 & 4\\5 & 3 & 7\\2 & 6 & 1\end{bmatrix}$.

Solution:

Given matrix $C$$C$, the determinant is following function.

$C = \begin{bmatrix}2 & 1 & 4\\5 & 3 & 7\\2 & 6 & 1\end{bmatrix}$$C = \begin{bmatrix}2 & 1 & 4\\5 & 3 & 7\\2 & 6 & 1\end{bmatrix}$

$|C| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$|C| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Therefore,

$|C| = 2(3 \times 1 - 7 \times 6) - 1(5 \times 1 - 7 \times 2) + 4(5 \times 6 - 3 \times 2)$$|C| = 2(3 \times 1 - 7 \times 6) - 1(5 \times 1 - 7 \times 2) + 4(5 \times 6 - 3 \times 2)$

$|C| = 2(3 - 42) - 1(5 - 14) + 4(30 - 6)$$|C| = 2(3 - 42) - 1(5 - 14) + 4(30 - 6)$

$|C| = 2(-39) - 1(-9) + 4(24)$$|C| = 2(-39) - 1(-9) + 4(24)$

$|C| = 2(-39) - 1(-9) + 4(24) = -78 +9 + 96$$|C| = 2(-39) - 1(-9) + 4(24) = -78 +9 + 96$

$|C| = 27$$|C| = 27$

Example #5

Find the determinant of the following $3 \times 3$ matrix below.

$A = \begin{bmatrix}-3 & 1 & -6\\2 &-1 & 5\\3 & 9 & 7\end{bmatrix}$.

Solution:

Given matrix $3 \times 3$$3 \times 3$

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

The determinant is given by following function,

$|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$

Therefore,

$|A| = -3((-1)\times 7 - 5 \times 9)- 1(2 \times 7 - 5 \times 3)+ (-6)(2 \times 9 - (-1)\times 3)$$|A| = -3((-1)\times 7 - 5 \times 9)- 1(2 \times 7 - 5 \times 3)+ (-6)(2 \times 9 - (-1)\times 3)$

$|A| = -3((-7) - 45)- 1(14 - 15)+ (-6)(18 - (-3))$$|A| = -3((-7) - 45)- 1(14 - 15)+ (-6)(18 - (-3))$

$|A| = -3(-52)+ 1 + (-6)(21) = 156 + 1 -126$$|A| = -3(-52)+ 1 + (-6)(21) = 156 + 1 -126$

$|A| = 31$$|A| = 31$

Note that the bigger the matrix , we must find more smaller matrices of size 2 x 2 and its determinant and then add all to get the determinant of original matrix. The process of finding the determinant of smaller matrices and making a matrix of determinants in called matrix of minors.

We shall discuss more about it in the next article.