CUET UG Mathematics Booster Test 1 - Fundamentals and Basic Properties of Determinants -

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CUET UG Mathematics Booster Test 1 - 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 1 - Fundamentals and Basic Properties of Determinants

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QUESTION 1 OF 20

A system of linear equations can be written in matrix form as AX=B. What is matrix A?

Answer Review

1 A system of linear equations can be written in matrix form as AX=B. What is matrix A?

A Matrix of variables

B Matrix of coefficients

C Identity matrix

D Zero matrix

💡 Short Explanation

In AX = B, A contains coefficients of variables X represents variables B represents constants

✅ Correct Answer Explanation

In matrix representation AX = B, matrix A stores coefficients of variables from the linear equations. Matrix X contains unknown variables, while B contains constants. Hence, option B is correct. Option A refers to X, not A. Options C and D are special matrices unrelated to general coefficient representation.

❓ Why Other Options Are Incorrect

Option A → Matrix of variables corresponds to X, not A.
Option C → Identity matrix is a special square matrix with ones on the principal diagonal, not the general coefficient matrix.
Option D → Zero matrix contains all zero entries and cannot generally represent coefficients of a system of equations.

🎯 Strategy

Used

Elimination

Application:

Identify the role of each matrix in AX = B and eliminate options not matching coefficient representation.

Final Logic:

A always stores coefficients of variables in matrix equations.

🧠 Memory Trick

"A = Amounts before variables"

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Matrix Representation of Linear Equations

2 Consider a system of equations. Determine whether the system has a unique solution using the determinant condition.

A No, value is 0

B Yes, value is 1

C Yes, value is -1

D No, value is 3

💡 Short Explanation

Unique solution exists if determinant ≠ 0 Negative determinant is also non-zero Hence solution is unique

✅ Correct Answer Explanation

A system of linear equations has a unique solution when the determinant of the coefficient matrix is non-zero. Since −1 is non-zero, the system possesses a unique solution. Hence option C is correct. Option A is incorrect because determinant 0 gives no unique solution. Option B is also valid mathematically, but the intended numerical answer given is −1.

❓ Why Other Options Are Incorrect

Option A → Determinant 0 implies either infinitely many solutions or no solution, not a unique solution.
Option B → Determinant 1 also gives a unique solution, but it does not match the intended evaluated determinant.
Option D → Determinant 3 is non-zero, so saying "No" is conceptually incorrect.

🎯 Strategy

Used

Elimination

Application:

Use the determinant condition det(A) ≠ 0 for uniqueness and remove contradictory statements.

Final Logic:

Since −1 ≠ 0, the system has a unique solution.

🧠 Memory Trick

"Non-zero determinant → unique answer"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Consistency and Unique Solution

3 (Match the Following | Numerical)

List I	List II
1. \(\left(\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right)\)(Determinant)	a. \(-5\)
2. Minor of element 4	b. \(1\)
3. Cofactor of element 4	c. \(-2\)
4. \(\left(\begin{pmatrix}2 & 5\\ 1 & 3\end{pmatrix}\right)\)(Determinant)	d. \(1\)

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

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

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

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

💡 Short Explanation

Determinant of first matrix = −2 Minor of 4 equals 1 Second determinant equals −5

✅ Correct Answer Explanation

For \(\left(\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right)\)(Determinant= (1×4 − 2×3) = −2, matching c. Minor of element 4 is obtained after deleting second row and second column, leaving 1, matching b. Cofactor = (−1)^(2+2) × 1 = 1, matching d. Determinant of second matrix equals (2×3 − 5×1) = 1, not −5. Hence the provided answer key is incorrect. Correct matching should be: 1-c, 2-b, 3-d, 4-b

❓ Why Other Options Are Incorrect

Option A → First determinant is not 1.
Option B → Minor of 4 is not −2.
Option D → Determinant of second matrix is incorrectly matched to −5 instead of 1.

🎯 Strategy

Used

Substitution

Application:

Directly compute determinants, minors, and cofactors using formulas.

Final Logic:

Evaluate each expression numerically and match systematically.

🧠 Memory Trick

"ad − bc for 2×2 determinants"

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

4 Identify theIncorrect Statement

A ∣A∣represents modulus of matrix

B ∣A∣represents determinant

C Determinant can be negative

D Defined only for square matrices

💡 Short Explanation

|A| denotes determinant Determinants may be negative Only square matrices have determinants

✅ Correct Answer Explanation

The notation |A| represents the determinant of matrix A, not modulus. Determinants are defined only for square matrices and can take positive, negative, or zero values. Hence option A is the incorrect statement. Options B, C, and D are standard determinant properties from NCERT.

❓ Why Other Options Are Incorrect

Option B → Correct because |A| is the standard determinant notation.
Option C → Correct since determinant values can be negative depending on entries.
Option D → Correct because only square matrices possess determinants.

🎯 Strategy

Used

Extreme Word Filter

Application:

Check for mathematically invalid terminology in determinant notation.

Final Logic:

|A| refers to determinant, not modulus.

🧠 Memory Trick

"Vertical bars mean determinant in matrices"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Definition and Notation

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

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 rows equal columns. Among the given matrices, 2×2 and 4×4 are square matrices, while 3×4 and 1×3 are rectangular matrices. Therefore only two matrices have determinants. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Ignores one valid square matrix.
Option C → Includes a non-square matrix incorrectly.
Option D → Assumes all matrices possess determinants, which is false.

🎯 Strategy

Used

Odd One Out

Application:

Separate square matrices from rectangular matrices.

Final Logic:

Only matrices with equal rows and columns have determinants.

🧠 Memory Trick

"Square only → determinant possible"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Square Matrix Rule

6 If a 2×2matrix has determinant 0, how many inverses does it have?

A 1

B 2

C 0

D Infinite

💡 Short Explanation

Inverse exists only if determinant ≠ 0 Determinant here equals 0 Hence inverse does not exist

✅ Correct Answer Explanation

A square matrix is invertible only when its determinant is non-zero. If determinant equals zero, the matrix is singular and has no inverse. Therefore the number of inverses is zero. Hence option C is correct. Other options contradict the invertibility condition.

❓ Why Other Options Are Incorrect

Option A → A singular matrix cannot have exactly one inverse.
Option B → A matrix can never have two distinct inverses.
Option D → Infinite inverses are impossible for matrices.

🎯 Strategy

Used

Elimination

Application:

Apply the inverse condition det(A) ≠ 0.

Final Logic:

Zero determinant means no inverse exists.

🧠 Memory Trick

"Zero determinant → zero inverse"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Singular and Non-Singular Matrices

7 For the matrix [ⓜ-5]:
(i) Determinant = -5
(ii) Determinant is negative
(iii) Determinant is always positive
(iv) Determinant equals the element

A (i) only

B (i) and (iii)

C (ii) and (iii)

D (i), (ii), and (iv)

💡 Short Explanation

Determinant of 1×1 matrix equals its element Here determinant = −5 Negative determinant is possible

✅ Correct Answer Explanation

For a 1×1 matrix [−5], the determinant equals the only element present, which is −5. Therefore statements (i), (ii), and (iv) are true. Statement (iii) is false because determinants can be negative. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits valid statements (ii) and (iv).
Option B → Statement (iii) is false because determinants are not always positive.
Option C → Statement (iii) is incorrect, making the option invalid.

🎯 Strategy

Used

Elimination

Application:

Check each statement using determinant property of 1×1 matrices.

Final Logic:

The determinant equals the element itself.

🧠 Memory Trick

"1×1 determinant = single entry"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinant of Order One

8 Evaluate the determinant: [7]

A 7

B 0

C 1

D -7

💡 Short Explanation

Determinant of 1×1 matrix equals its element Single element is 7 Hence determinant = 7

✅ Correct Answer Explanation

For any 1×1 matrix [a], the determinant equals a itself. Here the only element is 7, so the determinant equals 7. Hence option A is correct. Options B, C, and D do not follow the determinant definition for order one matrices.

❓ Why Other Options Are Incorrect

Option B → Determinant is not automatically zero.
Option C → Determinant need not always equal one.
Option D → The sign changes only if the entry itself is negative.

🎯 Strategy

Used

Substitution

Application:

Apply the direct rule for 1×1 determinants.

Final Logic:

Single-element matrix determinant equals the same number.

🧠 Memory Trick

"One box → same value"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinant of Order One

9 Evaluate:
\(∣\begin{pmatrix}5 & 2\\ 1 & 3\end{pmatrix}∣\)

A 13

B 15

C 10

D 11

💡 Short Explanation

Use ad − bc formula (5×3) − (2×1) Result = 13

✅ Correct Answer Explanation

For a 2×2 determinant \(∣\begin{pmatrix}a & c\\ d & b\end{pmatrix}∣\) determinant = ad − bc. Here determinant = (5×3) − (2×1) = 15 − 2 = 13. Therefore option A is correct. Other options arise from incorrect multiplication or addition operations.

❓ Why Other Options Are Incorrect

Option B → Obtained by multiplying diagonal terms only and ignoring subtraction.
Option C → Incorrect arithmetic evaluation.
Option D → Results from wrong subtraction or calculation.

🎯 Strategy

Used

Substitution

Application:

Directly substitute values into ad − bc formula.

Final Logic:

15 − 2 = 13.

🧠 Memory Trick

"Main diagonal minus side diagonal"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinant of Order Two

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

A 5

B 3

C 4

D 6

💡 Short Explanation

Apply ad − bc (2×2) − (1×−1) Result = 5

✅ Correct Answer Explanation

Using determinant formula ad − bc for 2×2 matrices: determinant = (2×2) − (1×−1) = 4 − (−1) = 5. Therefore option A is correct. Other options result from ignoring the negative sign or performing incorrect arithmetic.

❓ Why Other Options Are Incorrect

Option B → Obtained if subtraction sign is mishandled.
Option C → Ignores contribution of −1 term.
Option D → Incorrect addition and multiplication.

🎯 Strategy

Used

Substitution

Application:

Insert matrix entries into determinant formula carefully.

Final Logic:

Subtracting a negative increases the value to 5.

🧠 Memory Trick

"Minus negative becomes plus"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinant of Order Two

11 Assertion (A): A 3×3matrix has 6 possible expansions.
Reason (R): We can expand along 3 rows and 3 columns.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Expansion possible along every row and column 3 rows + 3 columns = 6 expansions Reason correctly explains assertion

✅ Correct Answer Explanation

A 3×3 determinant can be expanded along any one of its 3 rows or 3 columns. Therefore, total possible expansions are 6. Hence Assertion (A) is true. Reason (R) correctly explains this fact because determinant expansion is allowed row-wise and column-wise. Therefore option C is correct. The provided answer key was incorrect.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct, so this option is invalid.
Option B → Reason is not false; it correctly explains the assertion.
Option D → Assertion is true because 6 expansions are indeed possible.

🎯 Strategy

Used

Substitution

Application:

Count allowable expansions directly using rows and columns.

Final Logic:

3 rows + 3 columns = 6 valid determinant expansions.

🧠 Memory Trick

"3 rows + 3 columns = 6 expansions"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Expansion of Determinants

12 How many zeros are present in the matrix
\(\left(\begin{pmatrix}0 & 2 & 0\\ 1 & 0 & 3\\ 0 & 4 & 5\end{pmatrix}\right)\)
making it suitable for easy expansion?

A 2

B 3

C 4

D 5

💡 Short Explanation

Count all zero entries in matrix Zeros appear at three positions More zeros simplify expansion

✅ Correct Answer Explanation

The matrix contains zeros at positions (1,1), (1,3), and (3,1). Thus total zeros = 3. Matrices with more zeros are easier for determinant expansion because terms involving zeros vanish. Therefore option B is correct.

❓ Why Other Options Are Incorrect

Option A → Misses one zero entry.
Option C → Includes a non-zero element incorrectly.
Option D → Overcounts zero entries in the matrix.

🎯 Strategy

Used

Substitution

Application:

Directly inspect and count zero entries.

Final Logic:

Exactly three entries are zero.

🧠 Memory Trick

"More zeros → easier determinant"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Expansion Using Convenient Row/Column

13 Arrange the steps to evaluate the determinant along the first row of
\(∣\begin{pmatrix}2 & 1 & 0\\ 3 & 4 & 5\\ 1 & 2 & 3\end{pmatrix}∣\)
1. Multiply 1 with its sign and minor
2. Multiply 2 with its sign and minor
3. Add all obtained values
4. Multiply 0 with its sign and minor

A 1, 2, 4, 3

B 2, 1, 4, 3

C 3, 2, 1, 4

D 4, 1, 2, 3

💡 Short Explanation

Expand from left to right in first row Process entries 2, then 1, then 0 Finally add all terms

✅ Correct Answer Explanation

During expansion along the first row, elements are processed sequentially: first 2, then 1, then 0. Each element is multiplied by its sign and corresponding minor. After calculating all terms, they are added together. Hence the correct order is 2, 1, 4, 3 corresponding to option B.

❓ Why Other Options Are Incorrect

Option A → Begins with element 1 instead of the first-row first element 2.
Option C → Addition cannot occur before computing terms.
Option D → Starts incorrectly with the zero term.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Follow the natural left-to-right determinant expansion process.

Final Logic:

Compute row terms first, then add them.

🧠 Memory Trick

"Expand → multiply → add"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Row Expansion

14 In the matrix
\(\left(\begin{pmatrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{pmatrix}\right)\)
what is the sign factor for the element in position a_(2,3)?

A +1

B -1

C (-1)^(2+3)

D 0

💡 Short Explanation

Cofactor sign factor formula is (-1)^(i+j) Here i = 2 and j = 3 Sign factor = (-1)^(5)

✅ Correct Answer Explanation

The sign factor used in cofactors is given by (−1)^(i+j). For element a_(2,3), i = 2 and j = 3. Hence the sign factor becomes (−1)^(2+3). Therefore option C is correct. Numerically this equals −1, but the question specifically asks for the sign-factor expression.

❓ Why Other Options Are Incorrect

Option A → Incorrect because exponent sum is odd, giving negative sign.
Option B → Numerically correct but not the exact sign-factor form requested.
Option D → Sign factors are never zero.

🎯 Strategy

Used

Substitution

Application:

Insert row and column indices into sign-factor formula.

Final Logic:

Use (−1)^(i+j) directly.

🧠 Memory Trick

"Row + column decides sign"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Cofactors and Sign Rule

15 Find the sign multiplier for the element in position a_(3,2).

A 1

B -1

C 5

D 0

💡 Short Explanation

Sign factor = (−1)^(i+j) Here (3+2)=5 is odd Odd exponent gives −1

✅ Correct Answer Explanation

For element a_(3,2), the sign multiplier equals (−1)^(3+2) = (−1)^5 = −1. Hence option B is correct. Sign depends on parity of row and column indices. Odd sums produce negative signs while even sums produce positive signs.

❓ Why Other Options Are Incorrect

Option A → Even exponent gives +1, but exponent here is odd.
Option C → 5 is only the exponent sum, not the sign multiplier.
Option D → Sign multipliers cannot be zero.

🎯 Strategy

Used

Substitution

Application:

Substitute row and column numbers into cofactor sign formula.

Final Logic:

Odd sum of indices gives negative sign.

🧠 Memory Trick

"Odd sum → minus sign"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Sign Pattern of Cofactors

16 For the element 4 in the matrix
\(\left(\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right)\)
the cofactor depends on:
(i) Minor of 4
(ii) Sign factor (-1)^(2+2)
(iii) Signed minor
(iv) Absolute value of minor

A (ii) and (iii)

B (i) and (iv)

C (ii) only

D (i), (ii), and (iii)

💡 Short Explanation

Cofactor = sign × minor Signed minor forms cofactor Absolute value is irrelevant

✅ Correct Answer Explanation

The cofactor of an element is defined as Cij = (−1)^(i+j) × Mij, where Mij is the minor. Therefore cofactor depends on the minor, the sign factor, and the resulting signed minor. Absolute value of the minor has no role. Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the minor itself, which is essential.
Option B → Absolute value is not used in cofactor definition.
Option C → Sign factor alone cannot determine cofactor value.

🎯 Strategy

Used

Elimination

Application:

Apply direct cofactor definition and remove irrelevant statements.

Final Logic:

Cofactor requires both sign and minor.

🧠 Memory Trick

"Cofactor = sign × minor"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Cofactors

17 If the determinant of a matrix is 5, what is the value obtained by multiplying elements of one row with cofactors of another row?

A 5

B 0

C 1

D -5

💡 Short Explanation

Row elements with own cofactors give determinant Different-row cofactors give zero Property of determinants

✅ Correct Answer Explanation

A determinant property states that the sum of products of elements of one row with cofactors of another row equals zero. Since cofactors belong to a different row here, the value is 0 irrespective of determinant value 5. Hence option B is correct.

❓ Why Other Options Are Incorrect

Option A → Determinant value appears only with cofactors of the same row.
Option C → No determinant property gives value 1 here.
Option D → Negative determinant value is irrelevant in this property.

🎯 Strategy

Used

Elimination

Application:

Recall determinant-cofactor multiplication properties.

Final Logic:

Different row and cofactor combinations always produce zero.

🧠 Memory Trick

"Different rows → zero result"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Properties of Cofactors

18 Evaluate the determinant:
\(∣\begin{pmatrix}1 & 2 & 2\\ 3 & 4 & 4\\ 5 & 6 & 6\end{pmatrix}∣\)

A 6

B 14

C 0

D 3

💡 Short Explanation

Two columns are identical Determinant with equal columns is zero Hence determinant = 0

✅ Correct Answer Explanation

The second and third columns of the matrix are identical: [2,4,6]. A determinant becomes zero if any two rows or columns are equal. Therefore determinant = 0. Hence option C is correct. Other numerical values ignore this basic determinant property.

❓ Why Other Options Are Incorrect

Option A → Incorrect because equal columns force determinant zero.
Option B → Ignores determinant property of identical columns.
Option D → Wrong numerical evaluation.

🎯 Strategy

Used

Odd One Out

Application:

Identify repeated columns immediately.

Final Logic:

Equal columns imply determinant equals zero.

🧠 Memory Trick

"Same columns → determinant zero"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Properties of Determinants

📄 Passage

To every square matrix of order n, we associate a number (real or complex) called its determinant. This can be viewed as a function mapping square matrices to numbers.

If A is a matrix and k is a scalar, then:
Answer: D) \(det⁡(kA)=k^{n}det⁡(A)\)

A \(det⁡(kA)=k det⁡(A)\)

B \(det⁡(kA)=k^{2}det⁡(A)\)

C \(det⁡(kA)=det⁡(A)\)

D \(det⁡(kA)=k^{n}det⁡(A)\)

💡 Short Explanation

Scalar multiplication affects every row Determinant scales n times Formula becomes k^n det(A)

✅ Correct Answer Explanation

If A is an n×n matrix and every element is multiplied by scalar k, then each row contributes a factor k to the determinant. Since there are n rows, total factor becomes k^n. Therefore det(kA)=k^n det(A). Hence option D is correct.

❓ Why Other Options Are Incorrect

Option A → Missing power n in determinant scaling.
Option B → Valid only for specific order-2 matrices, not general order n.
Option C → Determinants do change under scalar multiplication.

🎯 Strategy

Used

Elimination

Application:

Recall the standard determinant scaling property.

Final Logic:

Each row contributes one factor of k.

🧠 Memory Trick

"n rows → k^n factor"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Scaling Property

📄 Passage

To every square matrix of order n, we associate a number (real or complex) called its determinant. This can be viewed as a function mapping square matrices to numbers.

If matrix \(A\) is of order 2 and \(det⁡(A)=-4\), find \(det⁡(2A)\).

A -32

B -16

C -8

D -64

💡 Short Explanation

Matrix order = 2 det(kA)=k^n det(A) det(2A)=2^2(−4)=−16

✅ Correct Answer Explanation

For a matrix of order n, determinant scales as det(kA)=k^n det(A). Here n=2 and k=2. Therefore det(2A)=2^2×(−4)=4×(−4)=−16. Hence option B is correct. The provided answer key was incorrect because it used an incorrect scaling factor.

❓ Why Other Options Are Incorrect

Option A → Uses wrong scaling factor 8 instead of 4.
Option C → Multiplies determinant by 2 instead of 2².
Option D → Incorrectly applies a higher power of 2.

🎯 Strategy

Used

Substitution

Application:

Substitute n=2 and k=2 into determinant scaling formula.

Final Logic:

4 × (−4) = −16.

🧠 Memory Trick

"Order 2 → square the scalar"

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Scalar Multiplication Property

CUET UG Mathematics Booster Test 1 - Fundamentals and Basic Properties of Determinants

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