 # 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.

### Suggested Videos        ## 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

### 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 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 Ri → Ri + k Rj or Ci → Ci + k Cj .

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

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

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

– 1(15− x[360
3x2 − 60
x2 2− 0
(− 3(10
13

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.

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jannat

Is determinant available just for square matrix? Guest
Stephen

Yes, it is only defined for square matrices. Guest
J. A. Zahálka

It actually is. It is equal to zero for all non-square matrices. Guest
Anonymous

Yes Guest
John

you mean if the determinant os non zero