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

Last updated on Mar 11, 2025

பெறு Evaluation of Determinants பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Evaluation of Determinants MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Evaluation of Determinants MCQ Objective Questions

Top Evaluation of Determinants MCQ Objective Questions

Evaluation of Determinants Question 1:

If the determinant \(\begin{vmatrix}a&b&aα+b\\\ b&c&bα+c\\\ aα+b&bα+c&0\end{vmatrix}\) is equal to zero, then consider the following statements

I. a, b, c are in A.P.

II. α is a root of the equation ax2 + 2bx + c = 0.

Which of the above statements is/are correct?

  1. Only I
  2. Only II
  3. I and II
  4. Neither I nor II

Answer (Detailed Solution Below)

Option 2 : Only II

Evaluation of Determinants Question 1 Detailed Solution

Concept:

Three non-zero terms a, b, c are in GP if and only if b2 = ac

For a quadratic equation ax2 + bx + c = 0,

If α is a root then aα2 + bα + c = 0

Calculation:

Given: Δ = \(\begin{vmatrix}a&b&aα+b\\\ b&c&bα+c\\\ aα+b&bα+c&0\end{vmatrix}\) = 0

C3 → C- C1α  - C2

⇒ Δ = \(\begin{vmatrix}a&b&0\\\ b&c&0\\\ aα+b&bα+c&-(aα^2+bα+bα+c)\end{vmatrix}=0\)

⇒ (aα2 + 2bα + c) (ac - b2) = 0

⇒ ac - b2 = 0 or aα2 + 2bα + c = 0

a, b, c are in GP and (x - α) is the factor ax2 + 2bx + c

∴ Only statement II is correct.

Evaluation of Determinants Question 2:

For what value of x is the matrix \(A = \left[ {\begin{array}{*{20}{c}} {3 - 2x}&{x + 1}\\ 2&4 \end{array}} \right]\)is singular ?

  1. 1
  2. 5
  3. 3
  4. 7

Answer (Detailed Solution Below)

Option 1 : 1

Evaluation of Determinants Question 2 Detailed Solution

CONCEPT:

  • If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)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: \(A = \left[ {\begin{array}{*{20}{c}} {3 - 2x}&{x + 1}\\ 2&4 \end{array}} \right]\) is a singular matrix.

As we know that, if \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\) 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.

Evaluation of Determinants Question 3:

Find the determinant of the matrix \(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|\) ?

  1. 274
  2. 387
  3. 873
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Evaluation of Determinants Question 3 Detailed Solution

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: 

\(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|\)

Apply C2 → 9C2 + C1

\(= \left| {\begin{array}{*{20}{c}} 2&65&65\\ 3&75&75\\ 5&86&86 \end{array}} \right|\)

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.

\(\left| {\begin{array}{*{20}{c}} 2&7&65\\ 3&8&75\\ 5&9&86 \end{array}} \right|=0\)

Evaluation of Determinants Question 4:

The element in the ith row and the jth column of a determinant of third order is equal to 2(i + j). What is the value of the determinant?

  1. 0
  2. 2
  3. 4
  4. 6

Answer (Detailed Solution Below)

Option 1 : 0

Evaluation of Determinants Question 4 Detailed Solution

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,

\(\begin{bmatrix} 2(1+1) & 2(1+2) &2(1+3) \\ 2(2+1) &2(2+2) & 2(2+3)\\ 2(3+1)&2(3+2) & 2(3+3) \end{bmatrix}\)

\(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 2+1 & 2+2 &2+3 \\ 3+1 & 3+2 & 3+3 \end{bmatrix}\)

R3 → R3 - Rand R2 → R2 - R1

\(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 1 & 1 &1 \\ 2 & 2 & 2 \end{bmatrix}\)

R3 → R-2R2

\(2\times \begin{bmatrix} 1+1 &1+2 & 1+3\\ 1 & 1 &1 \\ 0 & 0 & 0 \end{bmatrix}\)

= 0

So, the value of the determinant is 0

Evaluation of Determinants Question 5:

If A and B are square matrices of order 3 such that |A| = -1, |B| = 3, then |3AB| =

  1. -9
  2. -81
  3. -27
  4. 81

Answer (Detailed Solution Below)

Option 2 : -81

Evaluation of Determinants Question 5 Detailed Solution

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.

Evaluation of Determinants Question 6:

Find the value of x, if \(\left| {\begin{array}{*{20}{c}} x&4\\ 9&x \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 8&2\\ 16&4 \end{array}} \right|\) ?

  1. 6
  2. 3
  3. -3
  4. 8

Answer (Detailed Solution Below)

Option 1 : 6

Evaluation of Determinants Question 6 Detailed Solution

CONCEPT:

  • If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)is a square matrix of order 2, then determinant of A is given by |A| = (a­11 × a22) – (a12 – a21)


CALCULATION:

Given: \(\left| {\begin{array}{*{20}{c}} x&4\\ 9&x \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 8&2\\ 16&4 \end{array}} \right|\)

As we know that, \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}\\ {{a_{21}}}&{{a_{22}}} \end{array}} \right]\)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.

Evaluation of Determinants Question 7:

Let \(A = \left| {\begin{array}{*{20}{c}} p&q\\ r&s \end{array}} \right|\)

where p, q, r and s are any four different prime numbers less than 20. What is the maximum value of the determinant?

  1. 215
  2. 311
  3. 317
  4. 323

Answer (Detailed Solution Below)

Option 3 : 317

Evaluation of Determinants Question 7 Detailed Solution

Concept:

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

\(\rm A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}\)

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

Calculation:

\(A = \left| {\begin{array}{*{20}{c}} p&q\\ r&s \end{array}} \right|\)

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

\(\rm A = \left| {\begin{array}{*{20}{c}} 17&2\\3&19 \end{array}} \right|\)

|A| = [17 \(\times\) 19 - 2 \(\times\) 3]

= 323 - 6

= 317

Evaluation of Determinants Question 8:

Comprehension:

Direction: Consider the following for the next 02 (two) items:

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

What is det (3AB-1) equal to?

  1. 12
  2. 18
  3. 36
  4. 48

Answer (Detailed Solution Below)

Option 3 : 36

Evaluation of Determinants Question 8 Detailed Solution

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)

\(\Rightarrow {\rm{\;det}}\left( {3A{B^{ - 1}}} \right) = 27 \times 4 \times \frac{1}{3} = 36\)

Evaluation of Determinants Question 9:

If x2 + y2 + z2 = 1, then what is the value 

\(\begin{vmatrix}1&z&-y\\\ -z&1&x\\\ y&-x&1\end{vmatrix}=?\)

  1. 0
  2. 1
  3. 2
  4. 2 - 2xyz

Answer (Detailed Solution Below)

Option 3 : 2

Evaluation of Determinants Question 9 Detailed Solution

Calculation:

Given:

x2 + y2 + z2 = 1

then, \(​\begin{vmatrix}1&\rm z&\rm -y\\\ \rm -z&1& \rm x\\\ \rm y& \rm -x&1\end{vmatrix}\)

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"

Evaluation of Determinants Question 10:

If \(x = \frac{a}{b-c}\)\(y = \frac{b}{c - a}\)\(z = \frac{c}{a - b}\) then what is the value of the following?

\(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\)

  1. 0
  2. 1
  3. abc
  4. ab + bc + ca

Answer (Detailed Solution Below)

Option 1 : 0

Evaluation of Determinants Question 10 Detailed Solution

Formula used:

\(\rm A = \begin{vmatrix} a_{1} & a_{2} & a_{1}\\ b_{1} & b_{2} & b_{3}\\ c_{1} & c_{2} & c_{3}\ \end{vmatrix}\)

det(A) = \(\rm a_{1}\begin{vmatrix} b_{2} & b_{3}\\ c_{2} & c_{3} \end{vmatrix} - a_{2} \begin{vmatrix} b_{1} & b_{3}\\ c_{1} & c_{3} \end{vmatrix} + a_{3} \begin{vmatrix} b_{1} & b_{2}\\ c_{1} & c_{2} \end{vmatrix}\)

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

Calculation:

\(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\)

⇒  \(\begin{vmatrix} 1 & -\frac{a}{b - c} & \frac{a}{b - c}\\ 1 & 1 & -\frac{b}{c - a}\\ 1 & \frac{c}{a -b} & 1 \end{vmatrix}\) = Δ 

⇒ Δ = 1(1 + \(\rm \frac{bc}{(a - b)(c - a)}\)) + \(\rm \frac{a}{b - c}\)(\(\rm 1 + \frac{b}{c - a} \)) + \(\rm \frac{a}{b - c}\) \(\rm (\frac{c}{a - b} - 1)\)

⇒ Δ = 1 + \(\rm \frac{bc}{(a - b)(c - a)} + \frac{ab}{(b - c)(c - a)} + \frac{ac}{(b - c)(a - b)}\)

⇒ Δ = 1 + \( \frac{a}{(b - c)} \times \frac{(b - c)(a - b - c)}{(a - b)(c - a)} + \frac{bc}{(a - b)(c - a)}\)

⇒ Δ = \( \frac{ac - a^2 - bc + ab+ a^2 - ab - ac + bc}{(a - b)(c - a)}\)

∴ Δ =  0

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

⇒ \(x = \frac{a}{b-c}\) = 0, \(y = \frac{b}{c - a}\) = 1/2, \(z = \frac{c}{a - b}\) = -2

Let, \(\begin{vmatrix} 1 & -x & x\\ 1 & 1 & -y\\ 1 & z & 1 \end{vmatrix}\) = Δ 

⇒  Δ = \(\begin{vmatrix} 1 & 0 & 0\\ 1 & 1 & -1/2\\ 1 & -2 & 1 \end{vmatrix}\)

⇒  Δ = 1[1 - (-1/2)(-2)] 

∴   Δ = 0

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