Evaluation of Determinants MCQ Quiz in தமிழ் - Objective Question with Answer for Evaluation of Determinants - இலவச PDF ஐப் பதிவிறக்கவும்

Calculation:

Given: 

⇒ 2(5k - 12) - 3(20 - 6) + 1(16 - 2k) = 6

⇒ 10k - 24 - (3 × 14) + 16 - 2k = 6

⇒ 8k - 24 - 42 + 16 = 6

⇒ 8k = 56

∴ k = 7

CONCEPT:

• If is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)
• If A is a singular matrix of order n then |A| = 0

CALCULATION:

Given:  is a singular matrix.

As we know that, if  is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

⇒ |A| = 4(3 - 2x) - 2(x + 1)

⇒ |A| = 12 - 8x - 2x - 2

⇒ |A| = -10x + 10

∵ A is a singular matrix so |A| = 0

⇒ |A| = -10x + 10 = 0

⇒ x = 1

Hence, the correct option is 1.

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, then 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 → 9C2 + C1

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.

Concept:

If all the elements of a row or column are zeroes, then the value of the determinant is zero.

To evaluate the determinant row or column operation is done.

Calculation:

The determinant can be written as,

= 

R₃ → R₃ - R₁ and R2 → R2 - R1

= 

R₃ → R₃ -2R₂

= 

= 0

So, the value of the determinant is 0

CONCEPT:

• When the determinant operation is applied to the product of two matrices then the determination will be distributed.

|XY| = |X||Y|

Also, 

|kX| = kⁿ|A| 

Where n is the order of the determinant and k is any constant.

CALCULATION

Given:

|A| = -1, |B| = 3

 ⇒ |3AB| = 3³ |AB| = 27  |A||B| 

⇒ |3AB| =  27 × (-1) × 3 = - 81     [Since |KA| = Kⁿ |A|] 

So, the correct answer is option 2.

CONCEPT:

• If is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

CALCULATION:

Given: 

As we know that, is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)

⇒ x² - 36 = 32 - 32 = 0

⇒ x² = 36

⇒ x = 6

Hence, correct option is 1.

Concept:

Suppose A is a square matrix of 2 rows and 2 columns

The determinant is; |A| = [ad - bc]

Calculation:

p, q, r and s are any four different prime numbers less than 20.

Here we have to find the maximum value of the determinant so we have to take maximum value for p and s and minimum value for q and r

All prime numbers less than 20 = 2, 3, 5, 7, 11, 13, 17, 19

So, p = 17, q = 2, r = 3 and s = 19

|A| = [17  19 - 2  3]

= 323 - 6

= 317

Calculation:

Given:

x² + y² + z² = 1

then, 

Expanding along R₁, we get-

1(1 + x²) + z(xy + z) − y(xz − y)

1(1 + x2) - z(-z - xy) − y(xz − y)

= 1 + x² + xyz + z² − xyz + y²

= 1 + x² + y² + z²

= 1 + 1   (∵ x2 + y2 + z2 = 1 (given))

= 2

The correct answer is option "3"

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

Calculation:

Given: Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (3AB-1) = 33 × det (A) × det (B- 1)

As we know that, If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

Formula used:

det(A) = 

= a1(b2 c3 - b3 c2) - a2(b1c3 - b3c1) + a3 (b1c2 - b2c1)

Calculation:

⇒   = Δ 

⇒ Δ = 1(1 + ) + () +  

⇒ Δ = 1 + 

⇒ Δ = 1 + 

⇒ Δ = 

∴ Δ =  0

Shortcut TrickPut x = 0, b = 1 and c = 2

⇒  = 0,  = 1/2,  = -2

Let,  = Δ 

⇒  Δ = 

⇒  Δ = 1[1 - (-1/2)(-2)] 

∴   Δ = 0