10 Math Problems: Determinant of a Matrix

Determinant of a Matrix

Determinants originally appeared in the study of the linear equations. We shall, however, associate the notion of the determinant with matrices. For this purpose, we have to consider the square matrices of order 1, 2, 3 etc.

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If A is a square matrix, its determinant is denoted by |A| or det.(A). And, the value of determinant is calculated differently for different ordered square matrices. Here is mentioned about the calculation of determinants of square matrices of order 1, 2 and 3.

Determinant of a 1×1 Matrix

Let A = (a₁₁) be a square matrix of order 1×1. Then, its determinant |A| is given simply by the value of its single element a₁₁.

i.e. |A| = |a₁₁| = a₁₁

For example: Let, A = (2), then

Determinant of A = |A| = |2| = 2

Determinant of a 2×2 Matrix

Let A = (■(a_11&a_12@a_21&a_22 )) be a square matrix of order 2×2. Then, its determinant |A| is given by the value of a11a22 – a12a21. i.e. |A| = |■(a_11&a_12@a_21&a_22 )| = a11a22 – a12a21 For example: Let, A = (■(2&3@1&4)), then Determinant of A = |A| = |■(2&3@1&4)| = 2×4 - 3×1 = 8 – 3 = 5

Determinant of a 3×3 Matrix

Let A = (■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )) be a square matrix of order 3×3. Then its determinant |A| is given by the value of a11A11 + a12A12 + a13A13. i.e. |A| = |■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )| = a11A11 + a12A12 + a13A13 Where, A11, A12 and A13 are called the cofactors of the elements a11, a12 and a13 respectively and are given by the following calculations. A11 = (-1)1+1. |■(a_22&a_23@a_32&a_33 )| = a22a33 – a23a32 A12 = (-1)1+2. |■(a_21&a_23@a_31&a_33 )| = - (a21a33 – a23a31) A13 = (-1)1+3. |■(a_21&a_22@a_31&a_32 )| = a21a32 – a22a31 Here, the 2×2 determinant |■(a_22&a_23@a_32&a_33 )| is called minor of the element a11, which is obtained by omitting the 1st row and 1st column of 3×3 matrix A. Similarly, |■(a_21&a_23@a_31&a_33 )| is the minor of a12, and is obtained by omitting 1st row and 2nd column of A. And, |■(a_21&a_22@a_31&a_32 )| is the minor of a13, obtained by omitting 1st row and 3rd column of A.

Here is an easy pattern to find the minors of any element of a 3×3 matrix:

Note: Determinant of 3×3 matrix A can also be calculated by taking the products of the elements of any row (or column) and the corresponding cofactors. i.e.

|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

Or,

|A| = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃

Or,

|A| = a₃₁A₃₁ + a₃₂A₃₂ + a₃₃A₃₃

Or,

|A| = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁

Or,

|A| = a₁₂A₁₂ + a₂₂A₂₂ + a₃₂A₃₂

Or,

|A| = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃

All will give the same value for the determinant of A.

For example: Let, A = (■(1&2&0@2&3&1@5&4&2)) Now, A11 = (-1)1+1. |■(3&1@4&2)| = 6 – 4 = 2 A12 = (-1)1+2. |■(2&1@5&2)| = -(4 – 5) = 1 A13 = (-1)1+3. |■(2&3@5&4)| = 8 – 15 = -7 A21 = (-1)2+1. |■(2&0@4&2)| = -(4 – 0) = - 4 A22 = (-1)2+2. |■(1&0@5&2)| = 2 – 0 = 2 A23 = (-1)2+3. |■(1&2@5&4)| = -(4 – 10) = 6 A31 = (-1)3+1. |■(2&0@3&1)| = 2 – 0 = 2 A32 = (-1)3+2. |■(1&0@2&1)| = -(1 – 0) = - 1 A33 = (-1)3+3. |■(1&2@2&3)| = 3 – 4 = - 1

Now,

|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = 1×2 + 2×1 + 0×-7 = 4

|A| = a₂₁A₂₁ + a₂₂A₂₂ + a₂₃A₂₃ = 2×-4 + 3×2 + 1×6 = 4

|A| = a₃₁A₃₁ + a₃₂A₃₂ + a₃₃A₃₃ = 5×2 + 4×-1 + 2×-1 = 4

|A| = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁ = 1×2 + 2×-4 + 5×2 = 4

|A| = a₁₂A₁₂ + a₂₂A₂₂ + a₃₂A₃₂ = 2×1 + 3×2 + 4×-1 = 4

|A| = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃ = 0×-7 + 1×6 + 2×-1 = 4

Here, we got |A| = 4 in every calculations, so we can use any one of the above methods. Generally, we use the first method to calculate the determinant of a 3×3 matrix A. i.e.

|A| = a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃

Mechanical Rule (Rule of Sarrus)

A mechanical rule for finding the value of a third order determinant is as indicated below:-

Let |A| = |■(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )| Add, to the three columns of a determinant, the first two columns respectively as the fourth and the fifth columns. |A| = a11 a12 a13 a11 a12 + a21 a22 a23 a21 a22 a31 a32 a33 a31 a32 – = (a11a22a33 + a12a23a31 + a13a21a32) – (a31a22a13 + a32a23a11 + a33a21a12)

From the algebraic sum of the products of the elements of the diagonals pointing downwards, subtract the algebraic sum of the products of the elements of the diagonals pointing upwards.

This rule of expansion of a third-order determinant is known as The Rule of Sarrus.

For example: Find the determinant of |A| = |■(1&2&-3@2&0&4@3&2&1)| by the rule of Sarrus. Solution: |A| = 1 2 -3 1 2 2 0 4 2 0 3 2 1 3 2 = (0 + 24 – 12) – (0 + 8 + 4) = 12 – 12 = 0

Note: This rule does not work for the determinants of the order greater than 3.

Properties of 3×3 Determinant

We have seen how to evaluate a 3×3 determinant. There are some properties of 3×3 determinant. Look at them and their verification.

Prop. 1: The value of the determinant is unaltered by interchanging its rows and columns.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )| = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 ……… (i) Now, |At|= |■(a_1&a_2&a_3@b_1&b_2&b_3@c_1&c_2&c_3 )| = a1|■(b_2&b_3@c_2&c_3 )| - b1|■(a_2&a_3@c_2&c_3 )| + c1|■(a_2&a_3@b_2&b_3 )| = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 …… (ii) From (i) and (ii), we get,|A| = |At| Hence verified.

Prop. 2: Interchanging any two adjacent rows (or columns) changes the sign of the determinant.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )| = a1(b2c3 – b3c2) – b1(a2c3 – a3c2) + c1(a2b3 – a3b2) = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 Let |B| be the determinant obtained by interchanging 1st and 2nd rows of |A|. ∴ |B|= |■(a_2&b_2&c_2@a_1&b_1&c_1@a_3&b_3&c_3 )| = a2|■(b_1&c_1@b_3&c_3 )| – b2|■(a_1&c_1@a_3&c_3 )| + c2|■(a_1&b_1@a_3&b_3 )| = a2(b1c3 – b3c1) – b2(a1c3 – a3c1) + c2(a1b3 – a3b1) = a2b1c3 – a2b3c1 – a1b2c3 + a3b2c1 + a1b3c2 – a3b1c2 = - (a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1) = - |A| Hence verified.

Prop. 3: If any two rows (or columns) of a determinant are identical, then the value of the determinant is zero.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_1&b_1&c_1@a_3&b_3&c_3 )| In this determinant first two rows are identical. Let |B| be the determinant obtained by interchanging first two rows in |A|. Then the determinant itself remains unchanged, but by prop. 2. its value is - |A| i.e. |B| = - |A| Hence, |A| = -|A| Or, |A| + |A| = 0 Or, 2|A| = 0 Or, |A| = 0 Hence verified.

Prop. 4: If all the elements of any row (or column) are multiplied by a constant k, then the value of the determinant is multiplied by k.

Verification:

Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| = a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1 (as obtained before) And, let |B| be the determinant obtained from A by multiplying the elements of the 1st column by a constant k (say). Then, |B| = |■(〖ka〗_1&b_1&c_1@〖ka〗_2&b_2&c_2@〖ka〗_3&b_3&c_3 )| = ka1|■(b_2&c_2@b_3&c_3 )| - b1|■(〖ka〗_2&c_2@〖ka〗_3&c_3 )| + c1|■(〖ka〗_2&b_2@〖ka〗_3&b_3 )| = ka1(b2c3 – b3c2) – b1(ka2c3 – ka3c2) + c1(ka2b3 – ka3b2) = ka1b2c3 – ka1b3c2 – ka2b1c3 – ka3b1c2 + ka2b3c1 – ka3b2c1 = k(a1b2c3 – a1b3c2 – a2b1c3 – a3b1c2 + a2b3c1 – a3b2c1) = k|A| Hence verified.

Prop. 5: If each element of any row (or column) of a determinant is expressed as the sum of two terms, then the determinant can be expressed as the sum of determinants.

Verification:

Let |A| = |■(a_1+α&b_1&c_1@a_2+β&b_2&c_2@a_3+γ&b_3&c_3 )|, |B| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| and |C| = |■(α&b_1&c_1@β&b_2&c_2@γ&b_3&c_3 )|. Then we have to show that |A| = |B| + |C|. Here, |A| = |■(a_1+α&b_1&c_1@a_2+β&b_2&c_2@a_3+γ&b_3&c_3 )| = (a1+α)|■(b_2&c_2@b_3&c_3 )| - (b1+β)|■(a_2&c_2@a_3&c_3 )| + (c1+γ)|■(a_2&b_2@a_3&b_3 )| = a1|■(b_2&c_2@b_3&c_3 )| - b1|■(a_2&c_2@a_3&c_3 )| + c1|■(a_2&b_2@a_3&b_3 )| + α|■(b_2&c_2@b_3&c_3 )| - β|■(a_2&c_2@a_3&c_3 )| + γ|■(a_2&b_2@a_3&b_3 )| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| + |■(α&b_1&c_1@β&b_2&c_2@γ&b_3&c_3 )| = |B| + |C| Hence verified.

Prop. 6: If to the elements of any row (or column) a multiple of any other row (or column) is added, the value of the determinant remains unaltered.

Verification: