DETERMINANT OF 3X3 MATRIX

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

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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