It is very easy to calculate the determinant of a 3x3 matrix.

Generally, the determinant of a 3x3 matrix is computed as follows: if,

$$\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & i \end{bmatrix}$$

Then, the determinant of the 3x3 can be calculated as follows:

$$ a(ei-hf)-d(bi-hc)+g(bf-ec)$$

In words, the value of the determinant 3x3 matrix is the sum of the products of the elements and their cofactors of any rows or any column of the corresponding 3x3 matrices.

## How To Solve The Determinant Of The 3x3 Matrix.

1. Given a matrix.

$$\begin{bmatrix} a & d & g \\ b & e & h \\ c & f & I \end{bmatrix}$$

2. Multiply $a$ by the determinant of the 2x2 matrix which is not in a's row or column.

$$a\left(det\begin{bmatrix} e & h \\ f & i \end{bmatrix}\right)$$

If you are familiar with the determinant of a 2x2 matrix, you should be aware by now that the determinant of a 2x2 matrix is the difference between the product of top left and bottom right entries, and the product of top right and bottom left entries

Hence, $a\left(ei-hf\right)$

3. Multiply $d$ by the determinant of the 2x2 matrix which is not in d's row or column.

$$d\left(det\begin{bmatrix} b & h \\ c & i \end{bmatrix}\right)$$

$$d(bi-hc)$$

4. Do the same for $g$.

$$g\left(det\begin{bmatrix} b & e \\ c & f \end{bmatrix}\right)$$

$$d(bf-ec)$$

5. Finally, add the result of multiplying $g$ by the determinant 2x2 matrix after subtracting the result of multiplying $d$ by the determinant of 2x2 matrix from the result of multiplying $a$ by the determinant of 2x2 matrix.

$$a(ei-hf)-d(bi-hc)+g(bf-ec)$$.

__Example 1__

__Example 1__

Given that $A=\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9\end{bmatrix}$. Find |A|

**Solution:**

Recall $ a(ei-fh)-d(bi-ch)+g(bf-ce)$

$1(5×9-6×8)-2(4×9-7×6)-3(4×8-5×7)$

$1(45-48)-2(36-42)+3(32-35)$

$1(-3)-2(-6)+3(-3)$

$-3+12-9=0$

__Example 2__

What is the determinant of $\begin{bmatrix} 2 & -5 & -6 \\ -1 & -2 & 3 \\ 0 & 6 & -1 \end{bmatrix}$?

**Solution:**

$2(-2×-1-3×6)-(-5)(-1×-1-3×0)+(-6)(-1×6-0×-2)$

$2(2-18)+5(1-0)-6(-6-0)$

$2(-16)+5(1)-6(-6)$

$-32+5+36=9$

Hence, the determinant is 9.

READ ALSO: **MULTIPLICATION OF 2X2, 3X3 MATRICES **

__Example 3__

Given that the determinant of

$\begin{bmatrix} \frac{2}{3} & \frac{-3}{2} & \frac{-3}{2} \\ 0 & 0 & 0 \\ 4 & \frac{2}{6} & \frac{34}{24} \end{bmatrix}$

**Solution:**

The determinant of this matrix is 0 because the second column is a zero-row.

To recall, a property of the determinant says that, if in a matrix, any row or column consists entirely of zeros, then, the determinant of such matrix is zero.

To summarize, to obtain the determinant of a 3x3 determinant, the result of multiplying the top-left entry to the determinant of the middle-centre, middle-right, bottom-centre, and bottom-right entries must be subtracted the result of multiplying top-center entry to the determinant of middle left, middle right, bottom left and bottom right entries before adding the result of multiplying the top-right entry to the determinant of the middle-left, middle-centre, bottom-left, and bottom-centre entries.

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