#### description

- 1st, 2nd and 3rd Row
- 1st, 2nd and 3rd Columns
- Expansion along first Row (R1), Expansion along second row (R2),Expansion along 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.

Video link : https://youtu.be/21LWuY8i6Hw