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} \cdots 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 $A$ be a 2 x 2 matrix.

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

Different ways to represent determinant of matrix $A$.

\begin{aligned} &det(A) \hspace{1cm}(1)\\\\ &|A| \hspace{1cm}(2)\\\\ &\begin{vmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{vmatrix} \hspace{1cm}(3) \end{aligned}

### 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, 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 $A$ be a square matrix of order $1 \times 1$.

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

Then the determinant of $A$ is,

|A| = |2| = 2

**Determinant of $ 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 $A$ be 2 x 2 square matrix.

\begin{aligned} &A = \begin{bmatrix}2 & 3\\1 & 5\end{bmatrix}\\\\ &|A| = a \times d - b \times c\\\\ &|A| = 2 \times 5 - 3 \times 1\\\\ &|A| = 10 - 3 = 7 \end{aligned}

**Example #3**

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

**Solution:**

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

\begin{aligned} &B = \begin{bmatrix}5 & -1\\4 & -3\end{bmatrix}\\\\ &|B| = a \times d - b \times c\\\\ &|B| = 5 \times (-3) - (-1) \times 4\\\\ &|B| = (-15) - (-4)\\\\ &|B| = (-15) + 4 = -11 \end{aligned}

### Determinant Of $3 \times 3$ 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 $a$. then row $i = 1$ and column $j = 1$. Then the sign of first element

\begin{aligned} &= (-1)^{i+j}a(ei - fh)\\\\ &= (-1)^{1 + 1}a(ei - fh)\\\\ &= (-1)^{2})a(ei - fh)\\\\ &= +a(ei - fh)\\\\ \end{aligned}

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

= -b(di - fg)

Third element from top row( a b c).

= +c(dh - eg)

**Step 3: **Write down the determinant function obtained previously.

The determinant of $3 \times 3$ matrix.

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

Let us see few examples of determinant of 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 $C$, the determinant is following function.

\begin{aligned} &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}) \end{aligned}

Therefore,

\begin{aligned} &|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(-39) - 1(-9) + 4(24)\\\\ &|C| = 2(-39) - 1(-9) + 4(24) = -78 +9 + 96\\\\ &|C| = 27 \end{aligned}

**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 $3 \times 3$.

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})

Therefore,

\begin{aligned} &|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(-52)+ 1 + (-6)(21) = 156 + 1 -126\\\\ &|A| = 31 \end{aligned}

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.