Properties of Determinants.

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 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 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 + k R3. Now, again

= Δ + 0 (since both R1 and R3 are proportional)
Hence Δ = Δ1Now, let’s looks at some more examples on the properties of determinants.