CUET UG Mathematics Booster Test 2 - Fundamentals and Basic Properties of Determinants -

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CUET UG Mathematics Booster Test 2 - Fundamentals and Basic Properties of Determinants

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Mathematics Booster Test 2 - Fundamentals and Basic Properties of Determinants

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QUESTION 1 OF 20

The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is ₹60. The equation in matrix form uses coefficients of quantities. The first row of the matrix is:

Answer Review

1 The cost of 4 kg onion, 3 kg wheat, and 2 kg rice is ₹60. The equation in matrix form uses coefficients of quantities. The first row of the matrix is:

A [1" " 1" " 1]

B [60" " 0" " 0]

C [4" " 3" " 2]

D [2" " 3" " 4]

💡 Short Explanation

Matrix coefficients come from variable quantities. Coefficients are 4, 3, and 2. Constant 60 belongs to RHS matrix.

✅ Correct Answer Explanation

In matrix form, coefficients of variables form the matrix A. Here onion, wheat, and rice quantities are 4, 3, and 2 respectively, so the row becomes [4 3 2]. Option A uses identity-like entries. Option B contains the constant term. Option D changes the correct coefficient order given in the equation.

❓ Why Other Options Are Incorrect

Option A → [1 1 1] does not represent the actual coefficients from the equation. It ignores the quantities of onion, wheat, and rice.
Option B → 60 is the constant term on the RHS of the equation, not part of the coefficient matrix row.
Option D → Although it contains the same numbers, the order is incorrect. Matrix entries must follow the variable order exactly.

🎯 Strategy

Used: Substitution

Application:

Identify coefficients directly from the equation and place them in the same variable order.

Final Logic:

The coefficients of onion, wheat, and rice are 4, 3, and 2 respectively.

🧠 Memory Trick

"Matrix row = coefficients in order."

📚 Class XII – Mathematics Part 1 – Chapter: Matrices – Topic: Matrix Representation of Equations

2 For a system of linear equations AX=B, if Ais singular (det⁡(A)=0)and the system is inconsistent, then:

A Consistent with unique solution

B Consistent with infinitely many solutions

C Solvable using inverse matrix

D Inconsistent with no solution

💡 Short Explanation

Singular matrix has determinant zero. Inconsistent systems have no solution. Inverse method cannot be used.

✅ Correct Answer Explanation

If det(A)=0, the matrix is singular and has no inverse. A system is inconsistent when equations contradict each other, producing no common solution. Therefore the system has no solution. Option A is for nonsingular matrices. Option B applies to dependent consistent systems. Option C is impossible because inverse exists only when determinant is nonzero.

❓ Why Other Options Are Incorrect

Option A → Unique solutions occur only when det(A) ≠ 0. Singular matrices cannot guarantee uniqueness.
Option B → Infinitely many solutions occur for consistent dependent systems, not inconsistent systems.
Option C → Inverse matrix method requires A⁻¹ to exist. Singular matrices do not have inverses.

🎯 Strategy

Used: Elimination

Application:

Remove options involving inverse or consistency because the question explicitly states singular and inconsistent.

Final Logic:

"Inconsistent" directly implies "no solution."

🧠 Memory Trick

"Inconsistent = Impossible intersection."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Consistency of Linear Equations

3 Assertion (A): Any rectangular matrix can have a determinant.
Reason (R): Determinant is obtained by multiplying all elements.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Determinants exist only for square matrices. Determinant is not product of all entries. Both statements are incorrect.

✅ Correct Answer Explanation

A determinant is defined only for square matrices. Rectangular matrices do not possess determinants. Also, determinants are calculated using specific expansion rules, not by multiplying all matrix elements. Therefore both Assertion and Reason are false. The determinant involves signed combinations of products, not a direct multiplication of every element.

❓ Why Other Options Are Incorrect

Option B → Assertion is false because rectangular matrices do not have determinants.
Option C → Both statements are not true. The reason itself is mathematically incorrect.
Option D → Reason is false because determinant evaluation follows expansion formulas, not direct multiplication of all elements.

🎯 Strategy

Used: Extreme Word Filter

Application:

Words like "any" and "all" often indicate incorrect universal statements in mathematics.

Final Logic:

Only square matrices have determinants, and determinants are not simple products.

🧠 Memory Trick

"Square only for determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Definition of Determinant

4 If
\(A=\left(\begin{pmatrix}2 & 1\\ 3 & 4\end{pmatrix}\right)\)
then what does ∣A∣ represent numerically?

A 5

B 8

C 10

D 11

💡 Short Explanation

Use 2×2 determinant formula. Formula is ad−bc. 2×4−1×3 = 5.

✅ Correct Answer Explanation

For a 2×2 matrix, determinant is calculated using: \(∣\begin{pmatrix}a & b\\ c & d\end{pmatrix}∣=ad-bc\) Substituting values gives: 2×4 − 1×3 = 8 − 3 = 5. Hence Option A is correct. Option B is only ad. Option C adds terms incorrectly. Option D results from wrong arithmetic operations.

❓ Why Other Options Are Incorrect

Option B → 8 comes from multiplying diagonal entries only and ignores subtraction of cross products.
Option C → 10 is not obtained from the determinant formula and reflects incorrect addition.
Option D → 11 results from adding products instead of subtracting them according to determinant rules.

🎯 Strategy

Used: Substitution

Application:

Directly apply the standard 2×2 determinant formula.

Final Logic:

Using ad−bc gives 8−3 = 5.

🧠 Memory Trick

"Main diagonal minus cross diagonal."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Determinant of Order 2

5 How many of the following matrices have determinants?
• 2×2
• 3×5
• 4×4
• 1×3

A 1

B 2

C 3

D 4

💡 Short Explanation

Determinants exist only for square matrices. 2×2 and 4×4 are square. Total = 2 matrices.

✅ Correct Answer Explanation

A determinant exists only for square matrices where number of rows equals number of columns. Among the given matrices, 2×2 and 4×4 are square matrices. The matrices 3×5 and 1×3 are rectangular and do not have determinants. Therefore exactly two matrices possess determinants.

❓ Why Other Options Are Incorrect

Option A → Incorrect because there are two square matrices, not one.
Option C → Includes one rectangular matrix incorrectly as having a determinant.
Option D → Determinants are not defined for all matrices listed because some are rectangular.

🎯 Strategy

Used: Odd One Out

Application:

Separate square matrices from rectangular matrices to identify valid determinant cases.

Final Logic:

Only 2×2 and 4×4 satisfy rows = columns.

🧠 Memory Trick

"Square means determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Square Matrix Condition

6 Match the following

List 1	List 2
1. ⎡1 2⎤ ⎣2 4⎦	a. Has inverse
2. ⎡2 1⎤ ⎣3 2⎦	b. Has no inverse
3. det(A) = 0	c. Singular matrix
4. det(A) ≠ 0	d. Nonsingular matrix

A 1-a, 2-b, 3-d, 4-c

B 1-b, 2-a, 3-c, 4-d

C 1-c, 2-d, 3-a, 4-b

D 1-b, 2-a, 3-d, 4-c

💡 Short Explanation

First matrix determinant is zero. Second matrix determinant is nonzero. det(A)=0 → singular; det(A)≠0 → nonsingular.

✅ Correct Answer Explanation

For the first matrix, determinant = 1×4−2×2 = 0, so it has no inverse. For the second matrix, determinant = 2×2−1×3 = 1 ≠ 0, so inverse exists. Also, det(A)=0 indicates singular matrix, while det(A)≠0 indicates nonsingular matrix. Hence Option B is correct. The provided answer D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Reverses inverse properties and singular/nonsingular meanings incorrectly.
Option C → Confuses determinant conditions with inverse properties. A singular matrix cannot have an inverse.
Option D → Incorrectly matches det(A)=0 with nonsingular and det(A)≠0 with singular.

🎯 Strategy

Used: Elimination

Application:

Use determinant values to eliminate impossible inverse and singularity pairings.

Final Logic:

Zero determinant means singular and no inverse; nonzero means inverse exists.

🧠 Memory Trick

"Zero det → zero inverse."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Singular and Nonsingular Matrices

7 If the dot product of two orthogonal vectors is placed in a 1×1matrix, what is its determinant?

A 0

B 1

C -1

D Undefined

💡 Short Explanation

Orthogonal vectors have zero dot product. Determinant of 1×1 matrix equals entry itself. Hence determinant equals 0.

✅ Correct Answer Explanation

Orthogonal vectors satisfy: \(\vec{a}⋅\vec{b}=0\) A 1×1 determinant equals the single element inside it. Therefore determinant of [0] is 0. Option B and C are unrelated numerical values. Option D is incorrect because determinants of 1×1 matrices are well defined.

❓ Why Other Options Are Incorrect

Option B → Orthogonal vectors do not produce dot product 1 in general.
Option C → No negative value arises because orthogonal vectors have zero scalar product.
Option D → Determinants of order 1 are properly defined and equal the only entry present.

🎯 Strategy

Used: Substitution

Application:

Replace the dot product value using orthogonality property.

Final Logic:

Orthogonal vectors give dot product 0, so determinant is 0.

🧠 Memory Trick

"Orthogonal → dot product zero."

📚 Class XI – Mathematics – Chapter: Vector Algebra – Topic: Orthogonal Vectors and Dot Product

8 Evaluate the determinant: [ⓜ-3]

A -3

B 3

C 0

D 1

💡 Short Explanation

This is a 1×1 determinant. Determinant equals the single entry. Therefore value is −3.

✅ Correct Answer Explanation

For a determinant of order 1, the determinant is simply the number itself. Thus: \(∣-3∣=-3\) Here the vertical bars denote determinant notation, not modulus. Hence Option A is correct. Options B, C, and D ignore the determinant definition for 1×1 matrices.

❓ Why Other Options Are Incorrect

Option B → Confuses determinant notation with modulus notation and changes sign incorrectly.
Option C → There is no operation leading to zero.
Option D → Determinant of a 1×1 matrix equals the entry itself, not necessarily one.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize determinant notation instead of absolute value notation.

Final Logic:

A 1×1 determinant equals its only element.

🧠 Memory Trick

"One box = same value."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Determinant of Order 1

9 Find the value(s) of x such that
\(∣\begin{pmatrix}x & 2\\ 3 & x\end{pmatrix}∣=0\)

A x=2

B x=3

C x=±√6

D x=6

💡 Short Explanation

Use determinant formula ad−bc. Equation becomes x²−6=0. Solutions are ±√6.

✅ Correct Answer Explanation

Using the determinant formula: \(∣\begin{pmatrix}x & 2\\ 3 & x\end{pmatrix}∣=x^{2}-6\) Setting determinant equal to zero gives: \(x^{2}-6=0x=\pm \sqrt{6}\) Hence Option C is correct. Options A, B, and D do not satisfy the determinant equation because substituting them does not produce zero.

❓ Why Other Options Are Incorrect

Option A → Substituting x=2 gives determinant 4−6 = −2, not zero.
Option B → Substituting x=3 gives determinant 9−6 = 3, not zero.
Option D → Substituting x=6 gives determinant 36−6 = 30, not zero.

🎯 Strategy

Used: Substitution

Application:

Apply determinant formula and solve resulting quadratic equation.

Final Logic:

x²−6 = 0 gives x = ±√6.

🧠 Memory Trick

"ad−bc → quadratic."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Determinant Equations

10 Evaluate the determinant:
\(∣\begin{pmatrix}4 & 2\\ 1 & 3\end{pmatrix}∣\)

A 10

B 12

C 8

D 10

💡 Short Explanation

Use determinant formula ad−bc. 4×3−2×1 = 12−2. Determinant equals 10.

✅ Correct Answer Explanation

Using the determinant formula for order 2: \(∣\begin{pmatrix}a & b\\ c & d\end{pmatrix}∣=ad-bc\) we get: \(4\times 3-2\times 1=12-2=10\) Hence determinant equals 10. Both Option A and D display the same value numerically, but the first correct occurrence is Option A. Option B and C ignore subtraction.

❓ Why Other Options Are Incorrect

Option B → 12 is only the product of the main diagonal and ignores subtraction of cross products.
Option C → 8 comes from incorrect arithmetic manipulation.
Option D → Numerically correct but duplicated; conventionally first correct option is chosen.

🎯 Strategy

Used: Substitution

Application:

Insert entries into the determinant formula directly.

Final Logic:

ad−bc = 12−2 = 10.

🧠 Memory Trick

"Multiply diagonals, then subtract."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Evaluation of 2×2 Determinants

11 For the matrix
\(\left(\begin{pmatrix}1 & 2 & 3\\ 0 & 4 & 5\\ 1 & 0 & 6\end{pmatrix}\right)\)
which methods can be used to find its determinant?
(i) Row expansion
(ii) Column expansion
(iii) Cofactor method
(iv) Sarrus rule

A (i), (ii), and (iv)

B (i) and (iii)

C (ii) and (iii)

D (iii) and (iv)

💡 Short Explanation

3×3 determinants allow several methods. Row and column expansions are valid. Sarrus rule works only for 3×3 matrices.

✅ Correct Answer Explanation

A determinant of order 3 can be evaluated by row expansion, column expansion, cofactor expansion, or Sarrus rule. Since cofactor expansion is included within row/column expansion ideas, the correct combination here is Option A containing row expansion, column expansion, and Sarrus rule. The provided matrix is specifically a 3×3 determinant where all these methods are applicable.

❓ Why Other Options Are Incorrect

Option B → Excludes column expansion and Sarrus rule, both of which are valid for 3×3 determinants.
Option C → Omits row expansion even though determinant expansion along rows is a standard method.
Option D → Leaves out row and column expansion methods, which are fundamental determinant evaluation techniques.

🎯 Strategy

Used: Option Grouping

Application:

Identify all universally valid determinant methods for 3×3 matrices and eliminate incomplete combinations.

Final Logic:

3×3 determinants support row, column, cofactor, and Sarrus methods.

🧠 Memory Trick

"3×3 → Sarrus works."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Expansion of Determinants

12 How many terms are obtained when expanding the determinant of a 3×3 matrix?

A 3

B 6

C 9

D 12

💡 Short Explanation

3×3 determinant expansion has factorial terms. Number of terms = 3! Therefore total terms = 6.

✅ Correct Answer Explanation

A determinant of order n contains n! terms in its complete expansion. For a 3×3 determinant: \(3!=6\) Hence six terms appear in the expanded form. Option A undercounts the terms. Option C and D exceed the actual number generated by determinant permutations.

❓ Why Other Options Are Incorrect

Option A → Three terms occur only in cofactor expansion form, not in the complete determinant expansion.
Option C → Nine terms would imply all element combinations, which determinant expansion does not use.
Option D → Twelve exceeds the factorial count for a 3×3 determinant.

🎯 Strategy

Used: Substitution

Application:

Use factorial rule for determinant expansion directly.

Final Logic:

For order 3, total terms = 3! = 6.

🧠 Memory Trick

"Order n → n! terms."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Expansion of Determinants

13 Arrange the following matrices in descending order of their determinant values:
1. \(\left(\begin{pmatrix}5 & 2\\ 1 & 4\end{pmatrix}\right)\)
2. \(\left(\begin{pmatrix}7 & 3\\ 2 & 6\end{pmatrix}\right)\)
3. \(\left(\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}\right)\)
4. \(\left(\begin{pmatrix}3 & 2\\ 1 & 2\end{pmatrix}\right)\)

A 2, 4, 1, 3

B 3, 1, 4, 2

C 2, 1, 4, 3

D 1, 2, 3, 4

💡 Short Explanation

Compute determinants using ad−bc. Values are 36, 18, 4, and 0. Descending order gives 2,1,4,3.

✅ Correct Answer Explanation

Determinants are: \(5(4)-2(1)=187(6)-3(2)=361(1)-1(1)=03(2)-2(1)=4\) Descending order is: 36 > 18 > 4 > 0 Hence arrangement becomes 2,1,4,3. Therefore Option A is correct. The provided answer C is incorrect because it places determinant 18 before 36 incorrectly.

❓ Why Other Options Are Incorrect

Option B → Places the zero determinant first, which cannot be greatest.
Option C → Incorrectly swaps first two matrices despite determinant 36 being أكبر than 18.
Option D → Does not follow determinant magnitude order at all.

🎯 Strategy

Used: Substitution

Application:

Evaluate each determinant individually and compare numerically.

Final Logic:

36 > 18 > 4 > 0 gives order 2,1,4,3.

🧠 Memory Trick

"Compute first, arrange later."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Evaluation and Comparison of Determinants

14 Find the area of the triangle with vertices (1,2),(2,4),(3,6) using determinant method.

A 1

B 2

C 0

D 3

💡 Short Explanation

Points are collinear. Collinear points form zero area triangle. Determinant value becomes zero.

✅ Correct Answer Explanation

Area of a triangle using determinants is: \(Area=\frac{1}{2}∣\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1\end{pmatrix}∣\) The given points satisfy the same linear relation and lie on one straight line. Hence determinant becomes zero, making the triangle's area zero. Therefore Option C is correct. Nonzero areas in other options contradict collinearity.

❓ Why Other Options Are Incorrect

Option A → Area cannot be 1 because collinear points do not enclose any region.
Option B → Determinant evaluation for these points gives zero, not two.
Option D → A positive area requires non-collinear vertices, which is not true here.

🎯 Strategy

Used: Elimination

Application:

Recognize that all points lie on the same line, immediately implying zero area.

Final Logic:

Collinear points always form zero-area triangles.

🧠 Memory Trick

"Straight line → zero area."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Area of Triangle Using Determinants

15 Assertion (A): The minor and cofactor always have opposite signs.
Reason (R): The cofactor is given by \(C_{ij}=(-1)^{\left(i,\ j\right)}M_{ij}\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Cofactor sign depends on position. Minor and cofactor may have same sign. Formula for cofactor is correct.

✅ Correct Answer Explanation

The cofactor of an element is: \(C_{ij}=(-1)^{i+j}M_{ij}\) This formula is correct, so Reason is true. However, minor and cofactor are not always opposite in sign. When i+j is even, sign remains positive and cofactor equals the minor. Therefore Assertion is false while Reason is true. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason is mathematically correct because it gives the standard cofactor formula.
Option B → Assertion itself is false since signs are not always opposite.
Option C → Assertion is false, so both statements cannot be true together.

🎯 Strategy

Used: Extreme Word Filter

Application:

The word "always" signals possible exception cases in sign patterns.

Final Logic:

When i+j is even, minor and cofactor have the same sign.

🧠 Memory Trick

"Even sum → same sign."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Minors and Cofactors

16 Given a matrix-based data structure with defined cofactors, the evaluated operational sum is:

A -24

B 0

C 90

D -66

💡 Short Explanation

No matrix or cofactor values are provided. Determinant operations need explicit entries. Numerical result cannot be evaluated.

✅ Correct Answer Explanation

The question does not provide the actual matrix, cofactors, determinant expression, or operational details. Without numerical values or a defined formula, no mathematical evaluation can be performed. Therefore none of the options can be verified logically. The provided answer B) 0 is unsupported because insufficient information is available to compute any operational sum.

❓ Why Other Options Are Incorrect

Option A → No calculation or matrix values justify −24.
Option B → Zero cannot be concluded without determinant or cofactor information.
Option C → The value 90 is unsupported because the expression itself is missing.
Option D → −66 also lacks any computational basis from the question statement.

🎯 Strategy

Used: Elimination

Application:

Check whether sufficient mathematical data exists before attempting computation.

Final Logic:

A determinant-related value cannot be evaluated without the actual matrix or formula.

🧠 Memory Trick

"No matrix → no determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Cofactors and Expansion

17 If two rows of a matrix become identical, the determinant of the matrix is:

A 1

B -1

C 0

D Not defined

💡 Short Explanation

Equal rows make rows linearly dependent. Determinant becomes zero. Such matrices are singular.

✅ Correct Answer Explanation

One important determinant property states that if any two rows or columns are identical, then the determinant equals zero. Identical rows create linear dependence, meaning the matrix loses invertibility. Therefore Option C is correct. Options A and B are arbitrary values. Option D is wrong because determinants remain defined for square matrices.

❓ Why Other Options Are Incorrect

Option A → Determinant does not become one due to identical rows.
Option B → No determinant property gives −1 for equal rows.
Option D → The determinant exists but evaluates to zero for such matrices.

🎯 Strategy

Used: Elimination

Application:

Use the standard determinant property regarding identical rows.

Final Logic:

Equal rows imply determinant zero.

🧠 Memory Trick

"Same rows → zero determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Properties of Determinants

18 If two columns of a matrix are identical, what is the probability that its inverse exists?

A 1

B 0.5

C 0.25

D 0

💡 Short Explanation

Identical columns give determinant zero. Zero determinant means singular matrix. Singular matrices have no inverse.

✅ Correct Answer Explanation

When two columns are identical, determinant becomes zero because columns are linearly dependent. A matrix with determinant zero is singular and therefore noninvertible. Since inverse cannot exist in such cases, the probability of existence of inverse is zero. Hence Option D is correct. All other probabilities contradict determinant properties.

❓ Why Other Options Are Incorrect

Option A → Inverse never exists when determinant equals zero.
Option B → There is no partial probability concept here; inverse existence is definite.
Option C → Determinant properties give a strict conclusion, not a fractional chance.

🎯 Strategy

Used: Elimination

Application:

Apply determinant-zero condition directly to inverse existence.

Final Logic:

Identical columns ⇒ determinant zero ⇒ no inverse.

🧠 Memory Trick

"Same columns, inverse gone."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Properties of Determinants

19 If A is a 3×3 matrix and det⁡(kA)=27det⁡(A), find k.

A 27

B 9

C 3

D 81

💡 Short Explanation

For n×n matrices, det(kA)=kⁿdet(A). Here n=3. So k³=27 giving k=3.

✅ Correct Answer Explanation

For an n×n matrix: \(det⁡(kA)=k^{n}det⁡(A)\) Since A is a 3×3 matrix: \(k^{3}det⁡(A)=27det⁡(A)\) Thus: \(k^{3}=27k=3\) Hence Option C is correct. The provided answer A is incorrect because 27 is the determinant scaling factor, not k itself.

❓ Why Other Options Are Incorrect

Option A → 27 equals k³, not the value of k.
Option B → 9³ gives 729, not 27.
Option D → 81³ is far larger than the required scaling factor.

🎯 Strategy

Used: Substitution

Application:

Use determinant scaling property directly for 3×3 matrices.

Final Logic:

k³ = 27 gives k = 3.

🧠 Memory Trick

"3×3 → cube scaling."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Multiplication Property of Determinants

20 If \(det⁡(A)=3\) and the matrix is scaled appropriately, find the new determinant value.

A 27

B 108

C 4

D 12

💡 Short Explanation

Scaling factor is not specified. Determinant scaling depends on matrix order and scalar. Numerical answer cannot be uniquely found.

✅ Correct Answer Explanation

To determine the new determinant after scaling, both the scalar multiplier k and the matrix order n are necessary because: \(det⁡(kA)=k^{n}det⁡(A)\) The question provides only det(A)=3 but omits the scaling factor and matrix order. Therefore no unique numerical answer can be computed. The provided answer B) 108 is unsupported due to insufficient information.

❓ Why Other Options Are Incorrect

Option A → No given scaling information produces 27 uniquely.
Option B → 108 could arise only under specific missing conditions.
Option C → The determinant cannot be fixed as 4 without matrix details.
Option D → No valid determinant scaling data supports 12 uniquely.

🎯 Strategy

Used: Elimination

Application:

Verify whether all required determinant-scaling information is available before calculation.

Final Logic:

Without matrix order and scalar multiplier, determinant scaling cannot be evaluated.

🧠 Memory Trick

"No k, no new determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Scaling Property of Determinants

CUET UG Mathematics Booster Test 2 - Fundamentals and Basic Properties of Determinants

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