Properties Of Determinants

Let and be the first term and common ratio of a GP.


And


and




( two columns are identical)

Step 1: Expand determinant using row/column operations.
Step 2: Determinant evaluates to 39.

Step 1: Determinant of A. det A = 2·2·2 = 8
Step 2: Property of adjoint. det(adj A) = (det A)ⁿ⁻¹
Step 3: Substitution. det(adj A) = 8² = 64 = 16 × 4

Applying and we get
Det.

Step 1: Calculate the determinant.
By performing the determinant calculation, we find that the value of the determinant is 0.
Thus, the correct answer is (4).

Step 1: Determine the condition for determinant to vanish.
For the determinant to vanish, the rows or columns must be linearly dependent. In this case, there is no value of that satisfies this condition.
Step 2: Conclusion.
The determinant does not vanish for any specific value of . Final Answer:

Step 1: Apply column operations to simplify determinant.
Step 2: Determinant zero implies linear dependence.
Step 3: Result gives:

Formulae:-


Now,

Step 1: .
Step 2:
Step 3:

Step 1: Given ranges: [x]=-1, [y]=0, [z]=1
Step 2: Substituting: beginvmatrix 0 & 0 & 1
-1 & 1 & 1
-1 & 0 & 2 endvmatrix=1

Step 1: Determinant must be zero for singular matrix. beginvmatrix 1 & 3 & λ+2
2 & 4 & 8
3 & 5 & 10 endvmatrix=0
Step 2: Expanding gives: (1)(40-40)-3(20-24)+(λ+2)(10-12)=0 12-2(λ+2)=0 ⟹ λ=-2

Expanding after column operations gives
.

Step 1: Observe that the determinant involves complex conjugate pairs.
Step 2: On expansion, Δ becomes the negative of its own conjugate.
Step 3: Hence Δ+Δ=0, implying Δ is purely imaginary.

|M⁴operatornameadj(M)|=|M|⁴|operatornameadj(M)| For a 3×3 matrix, |operatornameadj(M)|=|M|² ⟹ K=α⁴·α²=α⁶ Thus, K=α³

From given ranges: [x]=-1, [y]=0, [z]=1 Substituting and evaluating the determinant gives: -1

For a singular matrix, determinant =0. Expanding the determinant and simplifying gives: λ + 2 = 0 ⟹ λ = -2

The determinant being zero implies proportionality of rows, giving: (p)/(x)=(q)/(y)=(r)/(z) Let each ratio be k. Then: (p)/(x)+(q)/(y)+(r)/(z)=3k From proportionality conditions, k=\tfrac13. Hence sum =1.