Question

explain deeply properties of determinant ​

Answer 1

interchange rows or columns

row multiplication

column multiplication, etc

Answer 2

Properties of Determinants

Property 1

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

Verification: Let

Properties of Determinants

Expanding along the first row, we get,

Properties of Determinants

= 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

Properties of Determinants

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 sign of determinant changes.

Verification: Let

Properties of Determinants

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

Properties of Determinants

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

Properties of Determinants

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

Properties of Determinants

and Δ1 be the determinant consequently obtained by multiplying the elements of the first row by k. Then,Properties of Determinants

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,

Properties of Determinants

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,

Properties of Determinants