CUET UG Mathematics Booster Test 2 - Expansion Methods and Evaluation of Determinants -

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CUET UG Mathematics Booster Test 2 - Expansion Methods and Evaluation of Determinants

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 2 - Expansion Methods and Evaluation of Determinants

1 / 20

QUESTION 1 OF 20

Evaluate the determinant:
\(∣\begin{pmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 3 & 6 & 9\end{pmatrix}∣\)

Answer Review

1 Evaluate the determinant:
\(∣\begin{pmatrix}1 & 2 & 3\\ 2 & 4 & 6\\ 3 & 6 & 9\end{pmatrix}∣\)

A 1

B 3

C 0

D 6

💡 Short Explanation

Rows are proportional to each other. Determinant becomes zero for dependent rows. Matrix is singular.

✅ Correct Answer Explanation

The second row is twice the first row and the third row is three times the first row. Since all rows are linearly dependent, determinant becomes zero. Hence Option C is correct. Options A, B, and D are impossible because a determinant with proportional rows must always evaluate to zero.

❓ Why Other Options Are Incorrect

Option A → A nonzero determinant cannot occur when rows are proportional.
Option B → Linear dependence among rows guarantees determinant zero, not three.
Option D → Six is impossible because the matrix loses independence due to proportional rows.

🎯 Strategy

Used: Elimination

Application:

Check whether any rows are multiples of each other before expanding the determinant.

Final Logic:

Proportional rows directly imply determinant zero.

🧠 Memory Trick

"Same pattern rows → determinant zero."

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

2 Match each minor with its numerical value for the matrix
\(\left(\begin{pmatrix}1 & 2 & 3\\ 0 & 4 & 5\\ 2 & 1 & 3\end{pmatrix}\right)\)

List I	List II
1. \(M_{11}\)	a. 7
2. \(M_{12}\)	b. -10
3. \(M_{22}\)	c. -8
4. \(M_{31}\)	d. -2

A 1–a, 2–c, 3–d, 4–b

B 1–a, 2–c, 3–b, 4–d

C 1–b, 2–a, 3–c, 4–d

D 1–c, 2–b, 3–a, 4–d

💡 Short Explanation

Minors are determinants after deleting row and column. Compute each 2×2 determinant carefully. Matching gives Option B.

✅ Correct Answer Explanation

\(M_{11}=∣\begin{pmatrix}4 & 5\\ 1 & 3\end{pmatrix}∣=12-5=7M_{12}=∣\begin{pmatrix}0 & 5\\ 2 & 3\end{pmatrix}∣=0-10=-10M_{22}=∣\begin{pmatrix}1 & 3\\ 2 & 3\end{pmatrix}∣=3-6=-3M_{31}=∣\begin{pmatrix}2 & 3\\ 4 & 5\end{pmatrix}∣=10-12=-2\) Hence the correct matching should actually be: 1–a, 2–b, 4–d. Since \(M_{22}=-3\) is absent in options, the question contains an error. The provided answer B is incorrect.

❓ Why Other Options Are Incorrect

Option A → Incorrectly assigns \(M_{12}=-8\), while actual value is −10.
Option C → Swaps values for \(M_{11}\)and \(M_{12}\), giving incorrect pairings.
Option D → Gives wrong determinant values for several minors.

🎯 Strategy

Used: Substitution

Application:

Calculate every minor individually using the 2×2 determinant formula.

Final Logic:

Correct minor values do not fully match any option because the question has a data error.

🧠 Memory Trick

"Delete row-column → small determinant."

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

3 For the matrix
\(\left(\begin{pmatrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{pmatrix}\right)\)
which of the following expressions correctly represent the determinant when expanded along the second row?
I. \(-4(2⋅9-8⋅3)\)
II. \(+5(1⋅9-7⋅3)\)
III. \(-6(1⋅8-7⋅2)\)

A I and II only

B I and III only

C II and III only

D I, II and III

💡 Short Explanation

Row 2 sign pattern is −,+,−. Each term uses corresponding minor. All three expressions are correct.

✅ Correct Answer Explanation

Expansion along second row follows cofactors: \(-,+,-\) Thus determinant becomes: \(-4M_{21}+5M_{22}-6M_{23}\) Each given expression correctly represents its corresponding cofactor expansion term. Therefore all three statements are valid and Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Omits the valid third term involving element 6.
Option B → Excludes the valid middle cofactor term involving element 5.
Option C → Ignores the correctly signed first cofactor term.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use row-wise cofactor sign pattern to verify each expansion term.

Final Logic:

Second-row expansion uses −,+,− signs for all three cofactors.

🧠 Memory Trick

"Second row → minus plus minus."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Expansion Along a Row

4 For the matrix
\(\left(\begin{pmatrix}2 & 1 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{pmatrix}\right)\)
which statement is INCORRECT during expansion?

A Signs in row 2 are -,+,-

B Minor includes sign automatically

C Cofactor uses \({\left(-1\right)}^{\left(i,\ j\right)}\)

D Expansion along any row gives same value

💡 Short Explanation

Minor is unsigned determinant. Cofactor includes sign factor. Expansion value remains same across rows/columns.

✅ Correct Answer Explanation

A minor is simply the determinant obtained after deleting a row and column. It does not include any sign. Sign is introduced only in the cofactor: \(C_{ij}=(-1)^{i+j}M_{ij}\) Hence Option B is incorrect. Options A, C, and D are correct determinant properties.

❓ Why Other Options Are Incorrect

Option A → Correct because second-row cofactor signs are −,+,−.
Option C → Correct standard formula for cofactors.
Option D → Determinant value remains same regardless of expansion row or column chosen.

🎯 Strategy

Used: Elimination

Application:

Differentiate clearly between minor and cofactor definitions.

Final Logic:

Minor has no sign; cofactor introduces sign factor.

🧠 Memory Trick

"Minor plain, cofactor signed."

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

5 Find the value of k if the area of triangle with vertices \(\left(k,0),(4,0),(0,2\right)\)is 4 sq units.

A 0 only

B 8 only

C 0, 8

D 4, 8

💡 Short Explanation

Use determinant area formula. Area equation gives \(∣4-k∣=4\). Solutions are k=0 and k=8.

✅ Correct Answer Explanation

Using triangle area formula: \(Area=\frac{1}{2}∣∣\begin{pmatrix}k & 0 & 1\\ 4 & 0 & 1\\ 0 & 2 & 1\end{pmatrix}∣∣\) Area simplifies to: \(∣4-k∣=4\) Thus: \(k=0ork=8\) Hence Option C is correct. Other options include only one valid solution or an incorrect value.

❓ Why Other Options Are Incorrect

Option A → Misses the second valid solution k=8.
Option B → Ignores k=0, which also satisfies the area condition.
Option D → k=4 gives zero area because points become collinear on the x-axis.

🎯 Strategy

Used: Substitution

Application:

Apply determinant area formula and solve the resulting absolute-value equation.

Final Logic:

\(∣4-k∣=4\) gives k=0 or 8.

🧠 Memory Trick

"Area fixed → absolute equation."

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

6 To find the equation of the line passing through \(\left(1,\ 2\right)\)and \(\left(3,\ 6\right)\)using determinants, the area of the triangle formed with a general point \(\left(x,\ y\right)\)must be:

A 1

B 2

C -1

D 0

💡 Short Explanation

Three collinear points form zero area. General point on line must satisfy collinearity. Determinant equals zero.

✅ Correct Answer Explanation

A point \(\left(x,\ y\right)\)lies on the line joining two points when the three points are collinear. Area of a triangle formed by collinear points is always zero. Therefore determinant equation is set equal to zero. Hence Option D is correct. Other numerical values would imply non-collinearity.

❓ Why Other Options Are Incorrect

Option A → Nonzero area indicates the point does not lie on the line.
Option B → Collinear points cannot form area two.
Option C → Area is never treated as negative in geometry.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recognize that line equations from determinants arise from collinearity condition.

Final Logic:

Collinear points always give zero area.

🧠 Memory Trick

"Same line → zero area."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Equation of Line Using Determinants

7 Evaluate the determinant by expanding along the first column:
\(∣\begin{pmatrix}1 & 2 & 3\\ 0 & 1 & 4\\ 0 & 2 & 8\end{pmatrix}∣\)

A 0

B 1

C 2

D -1

💡 Short Explanation

First column simplifies expansion. Only first cofactor contributes. Remaining 2×2 determinant is zero.

✅ Correct Answer Explanation

Expanding along the first column: \(1∣\begin{pmatrix}1 & 4\\ 2 & 8\end{pmatrix}∣=1(8-8)=0\) Thus determinant equals zero. Hence Option A is correct. Other options result from incorrect arithmetic or incomplete cofactor expansion.

❓ Why Other Options Are Incorrect

Option B → Ignores subtraction inside the minor determinant.
Option C → No valid determinant computation gives value two.
Option D → Negative determinant does not arise after correct evaluation.

🎯 Strategy

Used: Elimination

Application:

Choose the column with maximum zeros to simplify determinant calculation.

Final Logic:

Minor determinant becomes 8−8 = 0.

🧠 Memory Trick

"More zeros → easier determinant."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Expansion Along a Column

8 A matrix represents a transition probability system. If the determinant of the matrix is non-zero, then the matrix is invertible. What is the relation between the determinant and invertibility?

A If ∣A∣=0, then \(A^{-1}\)exists

B If ∣A∣≠0, then \(A^{-1}\)exists

C ∣A∣ does not affect invertibility

D Inverse exists only for probability matrices

💡 Short Explanation

Nonzero determinant implies nonsingular matrix. Only nonsingular matrices are invertible. Determinant directly controls inverse existence.

✅ Correct Answer Explanation

A square matrix is invertible only if its determinant is nonzero: \(∣A∣\neq 0\Rightarrow A^{-1} exists\) Hence Option B is correct. Option A reverses the rule. Option C ignores determinant importance. Option D is unrelated because invertibility depends on determinant, not matrix application type.

❓ Why Other Options Are Incorrect

Option A → Zero determinant means singular matrix with no inverse.
Option C → Determinant is the key condition for invertibility.
Option D → Inverse existence applies to all nonsingular square matrices, not only probability matrices.

🎯 Strategy

Used: Elimination

Application:

Use the standard determinant-inverse condition to reject contradictory statements.

Final Logic:

Nonzero determinant guarantees invertibility.

🧠 Memory Trick

"Nonzero det → inverse set."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Invertible Matrices

9 The volume of a parallelepiped is given by a scalar triple product (determinant of order 3). If two columns (say \(C_{2}\)and \(C_{3}\)) are interchanged, the determinant becomes:

A Unchanged

B Zero

C Multiplied by -1

D Doubled

💡 Short Explanation

Interchanging columns changes determinant sign. Magnitude remains same. Determinant becomes negative of original.

✅ Correct Answer Explanation

One property of determinants states that interchanging any two rows or columns changes the sign of the determinant: \(∣A^{'}∣=-∣A∣\) Therefore swapping \(C_{2}\)and \(C_{3}\)multiplies the determinant by −1. Hence Option C is correct. The determinant neither becomes zero nor doubles automatically.

❓ Why Other Options Are Incorrect

Option A → Interchanging columns always changes determinant sign.
Option B → Determinant becomes zero only under dependent rows or columns.
Option D → Swapping columns does not affect determinant magnitude.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall determinant transformation properties involving row/column interchange.

Final Logic:

Column interchange changes determinant sign only.

🧠 Memory Trick

"Swap once → change sign."

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

10 If three points are collinear, then the area of the triangle formed using determinant method is:

A Positive

B Negative

C Imaginary

D Zero

💡 Short Explanation

Collinear points lie on same line. No enclosed region is formed. Determinant area becomes zero.

✅ Correct Answer Explanation

Area of a triangle from determinant formula becomes zero when all three points are collinear because they do not enclose any region. Hence Option D is correct. Positive area requires non-collinear points. Negative or imaginary areas are not geometrically meaningful in this context.

❓ Why Other Options Are Incorrect

Option A → Positive area occurs only when points are non-collinear.
Option B → Geometric area is not taken as negative.
Option C → Triangle area from coordinates is always real-valued.

🎯 Strategy

Used: Elimination

Application:

Recognize the geometric meaning of collinearity before evaluating determinants.

Final Logic:

Collinear points produce zero enclosed area.

🧠 Memory Trick

"One line → no triangle."

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

11 Evaluate the determinant:
\(∣\begin{pmatrix}1 & 2 & \int_{0}^{1}\,x dx\\ 3 & 4 & \int_{0}^{1}\,x^{2} dx\\ 5 & 6 & \int_{0}^{1}\,0 dx\end{pmatrix}∣\)

A 1/2

B 1/3

C 1

D 0

💡 Short Explanation

Evaluate definite integrals first. Convert determinant into numerical matrix. Apply 3×3 determinant expansion.

✅ Correct Answer Explanation

First evaluate integrals: \(\int_{0}^{1}\,x dx=\frac{1}{2},\int_{0}^{1}\,x^{2} dx=\frac{1}{3},\int_{0}^{1}\,0 dx=0\) Thus determinant becomes: \(∣\begin{pmatrix}1 & 2 & 1/2\\ 3 & 4 & 1/3\\ 5 & 6 & 0\end{pmatrix}∣\) Expanding gives: \(1\left(4⋅0-6⋅\frac{1}{3}\right)-2\left(3⋅0-5⋅\frac{1}{3}\right)+\frac{1}{2}(18-20)=-2+\frac{10}{3}-1=\frac{1}{3}\) Hence the correct answer is actually B) 1/3. The provided answer A is incorrect.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic after determinant expansion leads to 1/2 instead of 1/3.
Option C → Determinant evaluation does not simplify to one.
Option D → Determinant is nonzero because rows are not linearly dependent.

🎯 Strategy

Used: Substitution

Application:

Evaluate the integrals first and substitute them into the determinant before expansion.

Final Logic:

After integration and expansion, determinant value becomes 1/3.

🧠 Memory Trick

"Integrate first, determinant later."

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

12 Assertion (A): Determinant of a matrix can be expanded along any row or column.
Reason (R): Determinant can only be calculated along rows.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Determinants allow row and column expansion. Expansion is not restricted to rows only. Assertion true, Reason false.

✅ Correct Answer Explanation

A determinant may be expanded along any row or any column and the final value remains unchanged. Therefore Assertion is true. However, Reason is false because determinants are not restricted only to row expansions. Column expansions are equally valid. Hence Option B correctly represents the logical relationship.

❓ Why Other Options Are Incorrect

Option A → Assertion is correct because row and column expansion both work.
Option C → Reason is false, so both statements cannot be true together.
Option D → Assertion is not false since determinants can indeed be expanded along any row or column.

🎯 Strategy

Used: Elimination

Application:

Compare the universal determinant expansion rule with the restrictive statement in the Reason.

Final Logic:

Determinants expand along rows and columns alike.

🧠 Memory Trick

"Any row, any column."

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

13 Arrange the matrices based on number of non-zero minor calculations required (highest effort → lowest effort):
1. Matrix with no zeros
2. Matrix with one zero
3. Matrix with two zeros in \(R_{1}\)
4. Matrix with three zeros in \(C_{1}\)

A 4, 3, 2, 1

B 1, 2, 4, 3

C 1, 2, 3, 4

D 4, 1, 2, 3

💡 Short Explanation

More zeros reduce determinant calculations. No-zero matrix needs maximum work. Entire zero column gives least effort.

✅ Correct Answer Explanation

Determinant expansion becomes easier when more zeros are present because fewer minors need evaluation. A matrix with no zeros requires maximum calculations. One zero reduces work slightly. Two zeros in a row simplify expansion significantly. A column with all zeros gives determinant directly as zero with minimum effort. Hence descending effort order is 1,2,3,4.

❓ Why Other Options Are Incorrect

Option A → Reverses the effort logic completely by placing easiest case first.
Option B → Incorrectly places a fully zero column before the two-zero row case.
Option D → A column with three zeros requires the least effort, not the greatest.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use the principle that zeros simplify determinant expansion calculations.

Final Logic:

More zeros mean fewer cofactors and less calculation effort.

🧠 Memory Trick

"More zeros → less work."

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

📄 Passage

A square matrix is singular if ∣A∣=0 and non-singular if ∣A∣≠0.

If A and B are non-singular matrices of the same order, then AB and BA are also non-singular.
Also, ∣AB∣=∣A∣⋅∣B∣.
If A and B are non-singular matrices of order 3, then ∣AB∣ is:

A 0

B ∣A∣+∣B∣

C ∣A∣-∣B∣

D ∣A∣⋅∣B∣

💡 Short Explanation

Determinant of product equals product of determinants. Non-singular matrices have nonzero determinants. Multiplication rule applies directly.

✅ Correct Answer Explanation

For square matrices of same order: \(∣AB∣=∣A∣⋅∣B∣\) This is a standard determinant property. Since A and B are nonsingular, their determinants are nonzero and determinant multiplication law applies directly. Hence Option D is correct. Addition or subtraction rules do not apply to determinant products.

❓ Why Other Options Are Incorrect

Option A → Product of two nonsingular matrices cannot have determinant zero.
Option B → Determinants of matrix products are multiplied, not added.
Option C → No determinant property states subtraction for products.

🎯 Strategy

Used: Substitution

Application:

Directly apply the determinant multiplication property from the passage.

Final Logic:

\(∣AB∣=∣A∣∣B∣\)is the standard determinant rule.

🧠 Memory Trick

"Product matrix → product determinant."

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

📄 Passage

A square matrix is singular if ∣A∣=0 and non-singular if ∣A∣≠0.

If A and B are non-singular matrices of the same order, then AB and BA are also non-singular.
Also, ∣AB∣=∣A∣⋅∣B∣.
If matrix A is singular, then which of the following is true?

A ∣A∣=0

B \(A^{-1}\)exists

C ∣A∣=1

D A must be identity

💡 Short Explanation

Singular matrices have zero determinant. Singular matrices are noninvertible. Identity matrices are nonsingular.

✅ Correct Answer Explanation

A square matrix is singular precisely when its determinant equals zero. Therefore Option A is correct. Singular matrices do not possess inverses, so Option B is incorrect. Option C is unrelated. Identity matrices always have determinant one, making Option D incorrect.

❓ Why Other Options Are Incorrect

Option B → Inverse exists only for nonsingular matrices with nonzero determinant.
Option C → Determinant one specifically represents special nonsingular cases.
Option D → Identity matrices are nonsingular, not singular.

🎯 Strategy

Used: Elimination

Application:

Use the determinant definition of singular matrices directly from the passage.

Final Logic:

Singular matrix means determinant zero.

🧠 Memory Trick

"Singular → determinant zero."

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

16 If A is a square matrix of order 3, then \(∣A^{T}∣\)is equal to:

A \(-∣A∣\)

B \(∣A∣\)

C \(∣A∣^{2}\)

D 0

💡 Short Explanation

Determinant remains unchanged under transpose. Row-column interchange preserves value. Applicable for all square matrices.

✅ Correct Answer Explanation

The determinant of a matrix and its transpose are always equal: \(∣A^{T}∣=∣A∣\) Transpose only interchanges rows and columns without changing determinant value. Therefore Option B is correct. The determinant neither changes sign nor squares automatically after transposition.

❓ Why Other Options Are Incorrect

Option A → Sign changes only under row or column interchange, not transpose.
Option C → Determinant is not squared during transposition.
Option D → Transpose does not force determinant to become zero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Recall the determinant-transpose property directly.

Final Logic:

Transpose preserves determinant value.

🧠 Memory Trick

"Transpose, same determinant."

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

17 If
\(A=\left(\begin{pmatrix}2 & 3\\ 4 & 5\end{pmatrix}\right),\)
evaluate \(4∣A∣\).

A -12

B 12

C -24

D 24

💡 Short Explanation

Compute determinant using ad−bc. Determinant equals −2. Multiply by 4 to get −8.

✅ Correct Answer Explanation

For matrix: \(A=\left(\begin{pmatrix}2 & 3\\ 4 & 5\end{pmatrix}\right)\) determinant is: \(∣A∣=2(5)-3(4)=10-12=-2\) Thus: \(4∣A∣=4(-2)=-8\) Hence none of the given options are correct. The provided answer C) −24 is incorrect.

❓ Why Other Options Are Incorrect

Option A → −12 results from incorrect determinant multiplication.
Option B → Sign error occurs because determinant itself is negative.
Option C → Determinant value was wrongly computed before multiplying.
Option D → Positive 24 contradicts the negative determinant obtained.

🎯 Strategy

Used: Substitution

Application:

Use the 2×2 determinant formula and multiply the final result by four.

Final Logic:

\(∣A∣=-2\Rightarrow 4∣A∣=-8\).

🧠 Memory Trick

"ad−bc first, multiply later."

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

18 Consider a system of linear equations with coefficient determinant Δ, and determinants Δx, Δy (Cramer's Rule). Which statement is incorrect?

A If Δ≠0, the system has a unique solution

B If Δ≠0, then \(x={\Delta}_{x}/\Delta , y={\Delta}_{y}/\Delta\)

C If Δ=0 and Δx≠0, the system has no solution

D If Δ=0, the system is always consistent

💡 Short Explanation

Δ=0 does not guarantee consistency. System may be inconsistent or dependent. Cramer's Rule needs determinant conditions.

✅ Correct Answer Explanation

If Δ≠0, a unique solution exists and Cramer's Rule applies directly, making Options A and B correct. If Δ=0 and Δx≠0, the system is inconsistent, so Option C is correct. However, Δ=0 alone does not ensure consistency; the system may have infinitely many or no solutions. Hence Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct because nonzero determinant guarantees uniqueness.
Option B → Correct formula from Cramer's Rule for unique solutions.
Option C → Correct because Δ=0 with nonzero numerator determinant implies inconsistency.

🎯 Strategy

Used: Elimination

Application:

Use Cramer's Rule cases to identify the universally false statement.

Final Logic:

Zero determinant alone cannot determine consistency.

🧠 Memory Trick

"Δ=0 → check more."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Cramer's Rule

19 Assertion (A): Area of a triangle formed by collinear points is 1.
Reason (R): Determinant of an identity matrix is 0.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Collinear points form zero area. Identity determinant equals one. Both statements are incorrect.

✅ Correct Answer Explanation

Area of a triangle formed by collinear points is always zero, not one. Also, determinant of the identity matrix equals one: \(∣I∣=1\) Therefore both Assertion and Reason are false. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Assertion is false because collinear points produce zero area.
Option C → Neither statement is true, so explanation relationship cannot exist.
Option D → Reason is false since identity determinant equals one, not zero.

🎯 Strategy

Used: Extreme Word Filter

Application:

Check standard determinant and geometry facts against the given absolute claims.

Final Logic:

Collinear area is zero and identity determinant is one.

🧠 Memory Trick

"Identity → determinant one."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Area and Identity Matrix

20 Match determinant conditions with system outcomes (Cramer's Rule):

List I	List II
1. Δ≠0	a. Inconsistent (no solution)
2. Δ=0, Δx≠0	b. Inconsistent (no solution)
3. Δ=0, Δx=0, Δy=0	c. Consistent with infinitely many solutions
4. Δ=0, Δx=0, Δy≠0	d. Consistent with unique solution

A 1–d, 2–b, 3–c, 4–a

B 1–b, 2–a, 3–c, 4–d

C 1–a, 2–c, 3–b, 4–d

D 1–c, 2–b, 3–a, 4–d

💡 Short Explanation

Nonzero Δ gives unique solution. Zero Δ with nonzero numerator gives inconsistency. All zero determinants imply infinitely many solutions.

✅ Correct Answer Explanation

By Cramer's Rule: • Δ≠0 → unique solution • Δ=0 with any numerator determinant nonzero → inconsistent system • Δ=0 and all numerator determinants zero → infinitely many solutions Thus matching becomes: 1–d, 2–b, 3–c, 4–a Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly assigns unique solution to a zero-determinant case.
Option C → Reverses consistent and inconsistent cases incorrectly.
Option D → Δ≠0 cannot produce infinitely many solutions.

🎯 Strategy

Used: Option Grouping

Application:

Group determinant conditions into unique, inconsistent, and dependent-system categories.

Final Logic:

Only nonzero Δ guarantees uniqueness.

🧠 Memory Trick

"Δ nonzero → one solution."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Cramer's Rule

CUET UG Mathematics Booster Test 2 - Expansion Methods and Evaluation of Determinants

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