Properties of Determinants

Properties of Determinants: So far we learnt what are determinants, how are they represented and some of its applications. Let us now look at the Properties of Determinants which will help us in simplifying its evaluation by obtaining the maximum number of zeros in a row or a column. These properties are true for determinants of any order. However, we shall restrict ourselves to determinants of order 3 only.

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Properties of Determinants

Property 1

The value of the determinant remains unchanged if both rows and columns are interchanged.

Verification: Let

Expanding along the first row, we get,

= a₁ (b₂ c₃ – b₃ c₂) – a₂ (b₁ c₃ – b₃ c₁) + a₃ (b₁ c₂ – b₂ c₁)
By interchanging the rows and columns of Δ, we get the determinant

Expanding Δ₁ along first column, we get,
Δ₁ = a₁ (b₂ c₃ – c₂ b₃) – a₂ (b₁ c₃ – b₃ c₁) + a₃ (b₁ c₂ – b₂ c₁)
Hence Δ = Δ₁

Browse more Topics under Determinants

• Determinant of a Matrix
• Minors and Cofactors of Determinant
• Area of a Triangle Using Determinants
• Adjoint and Inverse of a Matrix
• Solution of System of Linear Equations using Inverse of a Matrix

Property 2:

If any two rows (or columns) of a determinant are interchanged, then the sign of determinant changes.

Verification: Let

Expanding along first row, we get,
Δ = a₁ (b₂ c₃ – b₃ c₂) – a₂ (b₁ c₃ – b₃ c₁) + a₃ (b₁ c₂ – b₂ c₁)
Interchanging first and third rows, the new determinant obtained as

Expanding along third row, we get,
Δ₁ = a₁ (c₂ b₃ – b₂ c₃) – a₂ (c₁ b₃ – c₃ b₁) + a₃ (b₂ c₁ – b₁ c₂)
= – [a₁ (b₂ c₃ – b₃ c₂) – a₂ (b₁ c₃ – b₃ c₁) + a₃ (b₁ c₂ – b₂ c₁)]
Clearly Δ₁ = – Δ
Similarly, we can verify the result by interchanging any two columns.

Property 3:

If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Proof: If we interchange the identical rows (or columns) of the determinant Δ, then Δ does not change. However, by Property 2, it follows that Δ has changed its sign, therefore Δ = – Δ or Δ = 0. So let us verify the above property by an example.

Example: Evaluate

Solution: Expanding along first row, we get,
Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0
Here both the rows R₁ and R₃ are identical.

Property 4:

If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

Verification: Let

and Δ₁ be the determinant consequently obtained by multiplying the elements of the first row by k. Then,
So now expanding along the first row, we get
Δ₁ = k a₁ (b₂ c₃ – b₃ c₂) – k b₁ (a₂ c₃ – c₂ a₃) + k c₁ (a₂ b₃ – b₂ a₃)
= k [a₁ (b₂ c₃ – b₃ c₂) – b₁ (a₂ c₃ – c₂ a₃) + c₁ (a₂ b₃ – b₂ a₃)]
= k Δ
Hence,

Property 5:

If some or all elements of a row or column of a determinant are expressed as the sum of two (or more) terms, then the determinant can be expressed as the sum of two (or more) determinants. For example,

Verification: L.H.S. =

So now expanding the determinants along the first row, we get,
Δ = (a₁ + λ₁) (b₂ c₃ – c₂ b₃) – (a₂ + λ₂) (b₁ c₃ – b₃ c₁) + (a₃ + λ₃) (b₁ c₂ – b₂ c₁)
= a₁ (b₂ c₃ – c₂ b₃) – a₂ (b₁ c₃ – b₃ c₁) + a₃ (b₁ c₂ – b₂ c₁) + λ₁ (b₂ c₃ – c₂ b₃) – λ₂ (b₁c₃ – b₃ c₁) + λ₃ (b₁ c₂ – b₂ c₁)

= R.H.S.
Similarly, we may verify Property 5 for other rows or columns.

Property 6:

If the equimultiples of corresponding elements of other rows (or columns) are added to every element of any row or column of a determinant, then the value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation R_i → R_i + k R_j or C_i → C_i + k C_j .

Verification: Let

where Δ₁ is consequently obtained by the operation R₁ → R₁ + kR₃. Here, we have multiplied the elements of the third row (R₃) by a constant k and added them to the corresponding elements of the first row (R₁). Symbolically, we write this operation as R₁ → R₁ + kR₃. Now, again

= Δ + 0 (since both R₁ and R₃ are proportional)
Hence Δ = Δ₁
Now, let’s looks at some more examples on the properties of determinants.

Solved Examples on Properties of Determinants

Question 1: $$\begin{vmatrix}  sin^{2}x & cos^{2}x & 1\\ cos^{2}x & sin^{2}x & 1  \\  -10 & 12 & 2 \end{vmatrix} $$

1. 0
2. 12 cos²x + 10 sin²x + 2
3. 12 sin²x – 10 cos²
4. 10 sin2x

Answer : $$Let A = \begin{vmatrix}  sin^{2}x & cos^{2}x & 1\\ cos^{2}x & sin^{2}x & 1  \\  -10 & 12 & 2 \end{vmatrix} $$

So, by column transformation on determinant C1 → C1 + C2

$$A = \begin{vmatrix}  1 & cos^{2}x & 1\\ 1 & sin^{2}x & 1  \\  2 & 12 & 2 \end{vmatrix} $$

C1 → C1 – C3

$$A = \begin{vmatrix} 0 & cos^{2}x & 1\\ 0 & sin^{2}x & 1  \\  0 & 12 & 1 \end{vmatrix} $$

Therefore, A = 0

Question 2: The solution of the equation $$\begin{vmatrix}  x & 2 & -1\\ 2 & 5 & x \\ -1 & 2 & x \end{vmatrix} = 0 \ are $$

1. 3, -1
2. -3, 1
3. 1, 3
4. -1, -3

Answer:

$$Since, \begin{vmatrix} x & 2 & -1\\2 & 5 & x \\ -1 & 2 & x \end{vmatrix} = 0$$

$$\begin{vmatrix} x & 2 & -1\\2 & 5 & x \\ -1 & 2 & x \end{vmatrix} = 0 R3 → R3 – R2$$

– 1(−6 + 15) − x[−3x + 6] = 0
−9 + 3x² − 6x = 0
x² −2x − 3 = 0
(x − 3) (x + 1) = 0
x = −1, 3
Thus the answer is D.

Question 3: Are determinants of positive matrix always positive?

Answer: The determinant of a positive definite matrix shall always be positive. Therefore, a positive definite matrix is the one that is always non-singular. The matrix inverse of a positive definite matrix also happens to be positive definite.

Question 4: What is meant by the value of determinant?

Answer: The determinant happens to be a scalar value that one can compute from the square matrix’s elements. Furthermore, it encodes certain properties that belong to the linear transformation as described by the matrix. The denotation of the determinant of a matrix A is as det(A), det A, or |A|.

Question 5: What does it mean when the determinant is zero?

Answer: When the determinant of a square matrix n×n A is zero, then A shall not be invertible. When the determinant of a matrix is zero, the equations system in association with it is linearly dependent. This means that if the determinant of a matrix is zero, a minimum of one row of that matrix is a scalar multiple of another.

Question 6:  Can determinants ever be negative?

Answer: Yes, it is possible for a determinant to be a negative number.