Evaluation of Determinants MCQ Quiz - Objective Question with Answer for Evaluation of Determinants - Download Free PDF

Calculation:

Statement I

⇒ Statement I is correct.

Statement II

For a non-singular matrix, , where is the identity matrix.

⇒ Statement II is incorrect unless .

Statement III

The determinant of a matrix is equal to the determinant of its transpose:

⇒ Statement III is correct.

Out of the three statements, two are correct: I and III.

Calculation:

Given,

Let ω be a non-real cube root of unity, so and .

Consider the determinant

Step 1 — Column operation:  Replace the first column by :

Step 2 — Expansion along the third row:

which simplifies to

Step 3 — Equate to zero:

∴ The root of the equation is  .

Hence, the correct answer is Option 1.

Concept:

When A² + B² + C² = 0, it implies A = B = C = 0 (since the squares of real numbers are non-negative).

Substitute the values of A, B, and C for determinant calculation into the matrix

Calculation:

Since, Cos0 =1

Thus Matrix becomes 

Now determinant = 1[(1×1 - 1×1)] - 1[(1×1 - 1×1)] + 1[(1×1 - 1×1)]

= = 1(0) - 1(0) + 1(0) = 0

∴ The value of the determinant is 0.

Hence, the correct answer is Option 2.

Calculation:

Determinant Δ = 

Now, For our matrix, 

calculate the subdeterminants

⇒ 

⇒ 

⇒ 

⇒ Δ = 

⇒ Δ = 

⇒ 

Since we are given that  comparing the real and imaginary parts, we find:

A  = -6 and B = 0

Thus A + B = -6 + 0 = - 6

Hence, the Correct answer is Option 2.

Calculation:

Given,

Δ  = 

Simplify the determinant by the column operation  :

Expanding along the third row,

Thus  .

Sign analysis

• 0\): if , Δ ">1\), Δ > 0  ⇒ Statement I is false.
• :    ⇒ Statement II is true.
• :    ⇒ Statement III is false.

∴ Only Statement II is correct  ⇒  exactly one statement is true.

Hence, the correct answer is Option 2.

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

Apply R3 → R3 - R2

= 

As we can see that the first and the third row of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

 = 0

Concept:

i² = -1 , i³ = - i, i⁴ = 1, i⁶ = - 1 , i⁸ = 1 , i⁹ = i, i ¹² = 1, and i¹⁵ = - i

Calculations: 

Given determinant is 

Since, we have, 

i2 = -1 , i3 = - i, i4 = 1, i6 = - 1 , i8 = 1 , i9 = i, i 12 = 1, and i15 = - i

=

=i(i - 1) + 1(-i - i) - i (1 + i)

= i² - i - 2i - i - i²

= - 4i

Concept:

If  then determinant of A is given by:

|A| = a11 × a22 – a21 × a12

|An| = |A|n 

Calculation:

Given that,

 and |A2| = 64

⇒ |A| = x2 - 8           .... (1)

Given |A2| = 64

⇒ |A|2 = 64          [∵ |An| = |A|n]

⇒ |A| = (64)1/2 = 8       ....(2)

From equation 1 and 2

⇒ x2 - 8 = 8

⇒ x2 = 16

⇒ x = ± 4

Concept:

Property of determinants:

If A and B are two square matrices then |AB| = |A||B|

Calculation:

Given: A =  and B = 

Now,

|A| = 2 × 1 - 5 × 2 = 2 - 10 = -8

|B| = 4 × 5 - (-3 × 1) = 20 + 3 = 23

As we know that, |AB| = |A||B|

= -8 × 23 = -184

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

= 

Now, Expanding along C₃

= 1 (0 – ab) – 0 + 0 = -ab

Concept:

Elementary row or column transformations do not change the value of the determinant of a matrix.

Calculation:

Applying R₂ → R₂ – R₁, R₃ → R₃ – R₁, we get

Now, Expanding along C₁

= 1 (xy – 0) – 0 + 0 = xy

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

Apply C2 → 5C2 + C₁, we get

= 

As we can see that the second and the third column of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

∴ = 0

CONCEPT:

• If  is a square matrix of order 3, then determinant of A is given by |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• If A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

CALCULATION:

Given:  and |2A| = k

⇒ |A| = 3 × (30 - 63) - 4 × (55 - 56) + 9 × (99 - 48)

⇒ |A| = - 99 + 4 + 459 = 364

As we know that, if A is a matrix of order n, then |k ⋅ A| = kn ⋅ |A|, where k ∈ R.

⇒ |2A| = 2³ ⋅ 364 = 2912

Hence, the correct option is 3.

Calculation:

Given  = 0

⇒ x(x² - 2) - 2(x - 3) + 3(2 - 3x) = 0

Now, x³ - 2x - 2x + 6 + 6 - 9x = 0

⇒ x³ - 13x + 12 = 0

∵ x = 3 is a root of the equation, ∴ (x - 3) = 0

If we divide (x3 - 13x + 12) from (x - 3) we will get (x2 + 3x - 4)

⇒ (x - 3)(x² + 3x - 4) = 0

⇒ x2 + 3x - 4 = 0

⇒ (x + 4)(x - 1) = 0

⇒ x = 1, -4

Concept:

If  then determinant of A is given by: |A| = (a­₁₁ × a₂₂) – (a₁₂ – a₂₁).

Calculation:

Let, 

⇒ |A| = (2 + i) (i – 1) – (2 – i) (1 + i)

= 2i + i² – 2 – i – (2 – i + 2i – i²)

= i – 1 – 2 – (2 + i + 1)                   (∵ i² = -1)

= i – 3 – 2 – i – 1

= -6

∴ |A| is real number.