13 3 23, 30 7 53,39 9 70,=1 using properties of determinants
Answers
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
Verification: L.H.S. =
Properties of Determinants
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)
Properties of Determinants
= 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
Properties of Determinants
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
Properties of Determinants
= 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.
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