# Determinant of a Matrix of Order 3 × 3

#### description

• 1st, 2nd and 3rd Row
• 1st, 2nd and 3rd Columns
• Expansion along the first Row (R1)
• Expansion along the second row (R2)
• Expansion along the first Column (C1)

#### notes

Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order 3 corresponding to each of three rows (R_1, R_2 and R_3) and three columns (C_1, C_2 and C_3) giving the same value as shown below. Consider the determinant of square matrix A = [a_(ij)]_(3 × 3)
i.e. |A| = |(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)|

Expansion along first Row (R_1)
Step 1: Multiply first element a_11 of R_1 by (–1)^((1 + 1)) [(–1)^("sum of suffixes in"  a_11)] and with the second order determinant obtained by deleting the elements of first row (R_1) and first column (C_1) of
| A | as a_11 lies in R_1 and C_1,

i.e., (-1)^(1+1) a_11|(a_22,a_23),(a_32,a_33)|

Step 2:  Multiply 2nd element a_12 of R_1 by (–1)^(1 + 2) [(–1)^("sum of suffixes in"  a_12)] and the second order determinant obtained by deleting elements of first row (R_1) and 2nd column (C_2) of | A | as
a_12 lies in R_1 and C_2,
i.e., (-1)^(1+2)   a_12|(a_21,a_23),(a_31,a_33)|

Step 3: Multiply third element a_13 of R_1 by (–1)^(1 + 3) [(–1)^("sum of suffixes in"  a_13)] and the second order determinant obtained by deleting elements of first row (R_1) and third column(C_3) of | A | as a_13 lies in R_1 and C_3,
i.e., (-1)^(1+3)  a_13 |(a_21,a_22),(a_31,a_33)|

Step 4:  Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by

det A = |A| = (–1)^(1 + 1) a_11 |(a_22,a_23),(a_32,a_33)| + (-1)^(1+2)  a_12 |(a_21,a_23),(a_31,a_33)| + (-1)^(1+3)  a_13 |(a_21,a_22)(a_31,a_32)|

|A|
= a_11 (a_22 a_33 – a_32 a_23) – a_12 (a_21 a_33 – a_31 a_23)
+ a_13 (a_21 a_32 – a_31 a_22)

= a_11 a_22 a_33 – a_11 a_32 a_23 – a_12 a_21 a_33 + a_12 a_31 a_23 + a_13 a_21 a_32 – a_13 a_31 a_22                    ... (1)

Expansion along second row (R_2):
|A| = |(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)|
Expanding along R2, we get

|A| = (-1)^(2+1) a_21 |(a_12,a_13),(a_32,a_33)| +(-1)^(2+2) a_22 |(a_11,a_13),(a_31,a_33)|  + (-1)^(2+3) a_23 |(a_11,a_12),(a_31,a_32)|

= – a_21 (a_12 a_33 – a_32 a_13) + a_22 (a_11 a_33 – a_31 a_13) – a_23 (a_11 a_32 – a_31 a_12)

| A | = – a_21 a_12 a_33 + a_21 a_32 a_13 + a_22 a_11 a_33 – a_22 a_31 a_13 – a_23 a_11 a_32 + a_23 a_31 a_12

= a_11 a_22 a_33 – a_11 a_23 a_32 – a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 – a_13 a_31 a-22                      ... (2)

Expansion along first Column (C_1):
|A| = |(a_11,a_12,a_13),(a_21,a_22,a_23),(a_31,a_32,a_33)|
By expanding along C1, we get

|A| = a_11 (-1)^(1+1) |(a_22,a_23),(a_32,a_33)| + a_21 (-1)^(2+1) |(a_12,a_13)(a_32,a_33)| + a_31(-1)^(3+1) |(a_12,a_13)(a_22,a_23)|
= a_11 (a_22 a_33 – a_23 a_32) – a_21 (a_12 a_33 – a_13 a_32) + a_31 (a_12 a_23 – a_13 a_22)

| A | = a_11 a_22 a_33 – a_11 a_23 a_32 – a_21 a_12 a_33 + a_21 a_13 a_32 + a_31 a_12 a_23 – a_31 a_13 a_22
= a_11 a_22 a_33 – a_11 a_23 a_32 – a_12 a_21 a_33 + a_12 a_23 a_31 + a_13 a_21 a_32 – a_13 a_31 a_22    ... (3)

Clearly, values of |A| in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along R_3, C_2 and C_3 are equal to the value of |A| obtained in (1), (2) or (3).
Hence, expanding a determinant along any row or column gives same value.

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Determinants part 4 (Expanding determinant along 2nd 3rd row) [00:07:58]
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