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

Given:

M = 

N = 

Calculation:

First, calculate 2M:

2M =

Now, calculate 2M + N :

2M + N =  + 

∴ Option 4 is the correct answer.

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.

Concept:

If  then determinant of A is given by:

|A| = (a­11 × a22) - (a12 × a21)

Calculation:

Given: 

To find: 2f(x) – f(2x) =?

So,                  (put x = 2x)

Hence, option (4) is correct.

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

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

⇒ AX = B

⇒ X = A^-1 B = 

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

⇒ For the given equations to have no solution, |A| = 0

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2. 

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:

1. Determinant of a 3 × 3 matrix:

• Let A be a 3 × 3 matrix given by:

then the value of |A| also written as det(A) is:

det (A) = a (ei - dh) – b (fi - dg) + c (fh - eg)

2. Property of determinant of a matrix:

• Let A be a matrix of order n × n and det(A) = k. Then for a scaler c, the following property holds:

          det(cA) = cⁿ det(A)

Calculation:

First evaluate the determinant of the given matrix:

det(A) = 4(15 - 0) – 7(-5 + 4) + 1(0 + 6)

= 4(15) -7(-1) + 1(6)

= 60 + 7 + 6

= 73

Now using the property the value of det(3A) is:

det(3A) = 3³ det(A)

= 27 × 73

= 1971

Concept:

Area of a triangle with vertices (x1, y1) , (x2, y2), (x3, y3) is given by

​Area = 

Calculations:

Given that, vertices of triangle are (-3, 0), (3, 0) and (0, k)

By using the above formula,

​⇒ Area = 

⇒ Area = [-3(0 - k) - 0 + 1(3k)]

⇒ Area = 3k

According to the question, area of triangle is 9 square unit,

⇒ 3k = 9

⇒ k = 3

∴ Required value of k is 3 unit.

Concept:

Area of equilateral triangle = ×a2

Calculation:

Given: The co-ordinates of its vertices are (x1, y1); (x2, y2): (x3, y3) then the square of the determinant 

⇒ (△ ABC ) =   = ( ) ×a²

On squaring both side, 

⇒    = 

⇒  = 

Concept:

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

If  then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) - (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) - (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) - (a₂₂ × a₃₁)}

Calculation:

⇒ 2x² + 8x - 4 = 0

By comparing the equation 2x² + 8x - 4 = 0 with ax² + bx + c = 0, we get a = 2, b = 8 and c = - 4.

Concept:

Properties of determinants:​

For a n×n matrix A, det(kA) = kn det(A).

Calculation:

Given:

|A| = 5

k = 5

From the properties of the determinants, we know that |KA| = Kn |A|, where n is the order of the determinant.

Here, n = 2, therefore, the answer is K2 |A|.

|5A| = 52|A|

|5A| = 52 × 5 = 125