How to Find the determinant of a 3×3 . matrix

The determinant of a matrix is commonly used in calculus, linear algebra, and advanced geometry. Outside the academic world, engineers and computer graphics programmers also have to resort to matrices and their determinants. With this article, the wikiHow will show you how to find the determinant of a 3×3 matrix.

Table of Contents

Find the determinant

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Write your 3×3 matrix. We will start with a matrix A of size 3×3 and try to find the determinant |A| its. This is the general matrix notation we will be using, and the matrix in the example here is:

$USA$ $=$ $($ $a$ $11$ $a$ $twelfth$ $a$ $13$ $a$ $21$ $a$ $22$ $a$ $23$ $a$ $thirty first$ $a$ $32$ $a$ $33$ $)$ $=$ $($ $first$ $5$ $3$ $2$ $4$ $7$ $4$ $6$ $2$ $)$ ${displaystyle M={begin{pmatrix}a_{11}&a_{12}&a_{13}a_{21}&a_{22}&a_{23}a_{31}&a_{32}&a_{33}end{pmatrix }}={begin{pmatrix}1&5&32&4&74&6&2end{pmatrix}}}$ $M={begin{pmatrix}a_{{11}}&a_{{12}}&a_{{13}}a_{{21}}&a_{{22}}&a_{{23}}a_{{31} }&a_{{32}}&a_{{33}}end{pmatrix}}={begin{pmatrix}1&5&32&4&74&6&2end{pmatrix}}$

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Select a row or a column. That will be your reference row or column. Either way, the final answer is the same. For now, just pick the first row. Then we’ll have some advice on how to choose the easiest option to calculate.

Select the first row of matrix A in our example. Circle row 1 5 3. Or in general terms, circle a ₁₁ a ₁₂ a ₁₃ .

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Cross out rows and columns past your first element. Look at the row or column you’ve circled and select the first element. Draw a line over its row and column. At this point, you have only four numbers left. We will treat them as a 2×2 matrix.

In this example, our reference row is 1 5 3. The first element is in row 1 and column 1. Cross out all rows 1 and column 1. Write the remaining elements as a 2×2 matrix. :
~~1 5 3~~
2 4 7
4 6 2

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Find the determinant of a 2×2 matrix. Remember, the determinant of the matrix

(

a

b

c

d

)

{displaystyle {begin{pmatrix}a&bc&dend{pmatrix}}}

{begin{pmatrix}a&bc&dend{pmatrix}}

is ad – bc . ^{[1] X Research Source} You can remember this formula by drawing an X going from side to side of the 2×2 matrix. Multiply the two numbers connected by the diagonal of X. Next, subtract the product of the two numbers joined by the diagonal /. Use this formula to calculate the determinant of the matrix you just found.

In the above example, the determinant of the matrix $($ $4$ $7$ $6$ $2$ $)$ ${displaystyle {begin{pmatrix}4&76&2end{pmatrix}}}$ ${begin{pmatrix}4&76&2end{pmatrix}}$ = 4 * 2 – 7 * 6 = -34 .
This determinant is called a child determinant of the element selected from the original matrix. ^{[2] X Research Source} In this case, we have just found the subdeterminance of a ₁₁ .

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Multiply the determinant you just found by the element you selected. Remember that when specifying rows and columns to cross out, you selected the element from the row (or column) reference. Multiply this element by the determinant you just calculated from the 2×2 matrix.

In the above example, we have chosen a ₁₁ , the element has the value 1. Multiply it by -34 (the determinant of the 2×2 matrix), we get 1*-34 = -34 .

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Define your result mark. You will then multiply this result by 1 or -1 to get the algebraic subsection of the selected element. Using 1 or -1 depends on the element’s position in the 3×3 matrix. Memorize this simple sign table to determine the sign of each element:

+ – +
– + –
+ – +
Since we have chosen a ₁₁ , the element marked a +, we will multiply the result by +1 (in other words, do nothing with the result). That would still be -34 .
Or, you can specify the sign with the formula (-1) ^i+j , where i and j are the row and column of the element, respectively. ^{[3] X Research Sources}

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Repeat this process for the second element on your reference row or column. Go back to the original 3×3 matrix and the row or column you circled earlier. Repeat the above process for this element:

Cross out the row and column of that element. In our case, the selected element is a ₁₂ (which has a value of 5). Cross out row one (1 5 3) and column two $($ $5$ $4$ $6$ $)$ ${displaystyle {begin{pmatrix}546end{pmatrix}}}$ ${begin{pmatrix}546end{pmatrix}}$ .
View the remaining elements as a 2×2 matrix. Here, it’s the matrix $($ $2$ $7$ $4$ $2$ $)$ ${displaystyle {begin{pmatrix}2&74&2end{pmatrix}}}$ ${begin{pmatrix}2&74&2end{pmatrix}}$
Find the determinant of this 2×2 matrix. Use the formula ad – bc (2*2 – 7*4 = -24).
Multiply by the element selected from the 3×3 matrix. -24 * 5 = -120
Determines whether it is necessary to multiply it by -1. Use the sign table or the formula (-1) ^i+j . Here, the selected element is a ₁₂ , which carries the – sign in the sign table. We have to change the sign of the result: (-1)*(-120) = 120 .

Simplify the math

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Select the reference row or column with the most zero elements. Remember that you can select any row or column to reference. The answer obtained is unchanged. When selecting rows or columns with multiple zero elements, you only need to calculate the algebraic complement of the non-zero elements. Because:

Suppose you select row 2, with the elements a ₂₁ , a ₂₂ , and a ₂₃ . To solve the problem, we have three different 2×2 matrices. Call them A ₂₁ , A ₂₂ , and A ₂₃ respectively.
The determinant of a 3×3 matrix is a ₂₁ |A ₂₁ | – a ₂₂ |A ₂₂ | + a ₂₃ |A ₂₃ |.
If the elements a ₂₂ and a ₂₃ are both 0, that would be a ₂₁ |A ₂₁ | – 0*|A ₂₂ | + 0*|A ₂₃ | = a ₂₁ |A ₂₁ | – 0 + 0 = a ₂₁ |A ₂₁ |. Now we only have to calculate the algebraic complement of an element.

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Add rows to make the matrix easier. If you take the values of one row and add them to another row, the determinant of the matrix will not change. Similarly, the determinant of the matrix will also remain the same when we do the same with the column. You can manipulate multiple times or multiply the values of the elements of a row or column by a constant before adding them to another row or column to get the maximum number of zeros in the matrix. This saves a lot of computation time.

For example, let’s say you have a 3×3 matrix: $($ $9$ $-$ $first$ $2$ $3$ $first$ $0$ $7$ $5$ $-$ $2$ $)$ ${displaystyle {begin{pmatrix}9&-1&23&1&07&5&-2end{pmatrix}}}$ ${begin{pmatrix}9&-1&23&1&07&5&-2end{pmatrix}}$ .
To remove the 9 at position a ₁₁ , we can multiply the second row by -3 and add the resulting value to the top row. The new row is [9 -1 2] + [-9 -3 0] = [0 -4 2].
The new matrix is $($ $0$ $-$ $4$ $2$ $3$ $first$ $0$ $7$ $5$ $-$ $2$ $)$ ${displaystyle {begin{pmatrix}0&-4&23&1&07&5&-2end{pmatrix}}}$ ${begin{pmatrix}0&-4&23&1&07&5&-2end{pmatrix}}$ . Try the same trick with columns to get a ₁₂ to 0.

Advice

This method can be extended to all square matrices. For example, if we use it for a 4×4 matrix, the “cross out” step gives us a 3×3 matrix and we can find the determinant of this 3×3 matrix with the steps described above. Be forewarned, though, that hand calculations can get pretty tedious!
When every element of a row or a column is 0, the matrix has a determinant of 0.

Steps

Find the determinant

Simplify the math

Advice

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