CUET UG Mathematics Booster Test 1 - Minors, Cofactors and Cofactor Expansion -

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CUET UG Mathematics Booster Test 1 - Minors, Cofactors and Cofactor Expansion

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Mathematics Booster Test 1 - Minors, Cofactors and Cofactor Expansion

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QUESTION 1 OF 20

Assertion (A): The minor of an element in a 2×2 determinant is a 1×1 determinant.
Reason (R): The minor of an element is obtained by deleting its row and column, reducing the order by 1.

Answer Review

1 Assertion (A): The minor of an element in a 2×2 determinant is a 1×1 determinant.
Reason (R): The minor of an element is obtained by deleting its row and column, reducing the order by 1.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Minor is formed by deleting row and column Order reduces by one 2×2 determinant gives 1×1 minor

✅ Correct Answer Explanation

For a 2×2 determinant, deleting one row and one column leaves a 1×1 determinant, which is the minor. Thus the Assertion is true. The Reason correctly explains the method of obtaining minors and why the order reduces by one. Therefore both statements are true and the Reason explains the Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct according to the definition of minors.
Option B → The Reason is not false because deleting a row and column always reduces determinant order by one.
Option D → Assertion is true since a 2×2 determinant indeed produces a 1×1 minor.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Apply the standard definition of minor and observe determinant order reduction.

Final Logic:

Deleting one row and one column from a 2×2 determinant leaves a 1×1 determinant.

🧠 Memory Trick

"Minor → one order less"

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

2 For a 3×3 matrix, how many distinct 2×2 submatrices are formed to calculate all minors?

A 6

B 9

C 8

D 12

💡 Short Explanation

Each element has one minor 3×3 matrix contains 9 elements Hence 9 distinct minors exist

✅ Correct Answer Explanation

Every element of a 3×3 matrix has a corresponding minor obtained after deleting its row and column. Since the matrix contains 9 elements, there are 9 distinct 2×2 submatrices used for minor calculations. Hence Option B is correct. Other options underestimate or overestimate the total minors.

❓ Why Other Options Are Incorrect

Option A → A 3×3 matrix has more than six elements, so six minors are insufficient.
Option C → Eight minors would leave one element without a corresponding minor.
Option D → Twelve exceeds the actual number of elements in the matrix.

🎯 Strategy

Used

Substitution

Application:

Count total elements in the matrix since each element has exactly one minor.

Final Logic:

Nine elements imply nine corresponding minors.

🧠 Memory Trick

"Every element gets one minor"

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

3 Arrange the correct steps for expanding a 3×3 determinant using minors:
1. Multiply each element by its corresponding 2×2 minor
2. Choose a row or column for expansion
3. Apply the sign factor (-1)^(i+j)
4. Add all resulting terms

A 1, 2, 3, 4

B 2, 1, 4, 3

C 2, 3, 1, 4

D 3, 2, 1, 4

💡 Short Explanation

First select expansion row or column Apply cofactor sign pattern Then multiply and add terms

✅ Correct Answer Explanation

Determinant expansion begins by choosing a row or column. Next, apply the cofactor sign factor \({\left(-1\right)}^{i+j}\). Then multiply each element by its corresponding minor, and finally add all terms. This sequence matches Option C. Other options place sign application or multiplication incorrectly.

❓ Why Other Options Are Incorrect

Option A → Multiplication cannot occur before selecting a row or applying signs.
Option B → Addition must occur after applying signs and multiplication properly.
Option D → Sign factors cannot be applied before selecting the expansion row or column.

🎯 Strategy

Used

Option Grouping

Application:

Arrange determinant expansion steps in the exact operational sequence.

Final Logic:

Correct determinant expansion follows selection → sign → multiplication → addition.

🧠 Memory Trick

"Choose → Sign → Multiply → Add"

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

4 Let \(M_{11}\)be the minor of the element \(a_{11}\)in a 4×4 determinant.
What is the order of \(M_{11}\)?

A 1

B 2

C 3

D 4

💡 Short Explanation

Minor reduces determinant order by one 4×4 becomes 3×3 Deleting row and column forms minor

✅ Correct Answer Explanation

A minor is formed by deleting one row and one column from the determinant. Removing one row and one column from a 4×4 determinant leaves a 3×3 determinant. Therefore the order of \(M_{11}\)is 3, making Option C correct.

❓ Why Other Options Are Incorrect

Option A → Removing one row and column from 4×4 cannot reduce directly to 1×1.
Option B → Order reduces by exactly one, not by two.
Option D → Minor cannot have the same order as the original determinant.

🎯 Strategy

Used

Elimination

Application:

Use the rule that minors always reduce determinant order by one.

Final Logic:

A 4×4 determinant produces a 3×3 minor.

🧠 Memory Trick

"Minor = original order − 1"

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

5 Which of the following correctly describe a cofactor \(A_{ij}\)?
1. It includes the sign factor \({\left(-1\right)}^{\left(i,\ j\right)}\)
2. It equals the minor for positions where \(\left(i,\ j\right)\)is even
3. It is always non-negative

A 1 and 2

B 2 and 3

C 1 and 3

D All of the above

💡 Short Explanation

Cofactor includes sign factor Even sum gives positive sign Cofactors may be negative

✅ Correct Answer Explanation

A cofactor is defined as: \(A_{ij}=(-1)^{i+j}M_{ij}\) Thus Statement 1 is correct. When \(i+j\) is even, the sign factor becomes positive, making the cofactor equal to the minor, so Statement 2 is also correct. Statement 3 is false because cofactors can be positive or negative.

❓ Why Other Options Are Incorrect

Option B → Statement 3 is incorrect because cofactors are not always non-negative.
Option C → Although Statement 1 is true, Statement 3 remains false.
Option D → All statements cannot be correct since cofactors may become negative.

🎯 Strategy

Used

Elimination

Application:

Check each statement directly using the cofactor formula.

Final Logic:

Only Statements 1 and 2 satisfy cofactor properties.

🧠 Memory Trick

"Even sum → same sign"

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

6 In a 4×4 matrix, what is the probability that the cofactor sign factor \({\left(-1\right)}^{\left(i,\ j\right)}\)is positive?

A 1/4

B 1/2

C 1/3

D 5/16

💡 Short Explanation

Positive sign occurs for even \(i+j\) Half positions are even-sum positions Checkerboard signs split equally

✅ Correct Answer Explanation

A 4×4 matrix contains 16 positions. Cofactor signs follow a checkerboard pattern: \(+,-,+,-\) Exactly half the positions have even \(i+j\), producing positive signs. Thus 8 out of 16 positions are positive: \(\frac{8}{16}=\frac{1}{2}\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Positive signs occur more frequently than one-fourth of the entries.
Option C → Cofactor signs divide equally, not in one-third proportion.
Option D → Checkerboard distribution gives exactly half positive positions.

🎯 Strategy

Used

Substitution

Application:

Count even \(i+j\) positions in the checkerboard sign pattern.

Final Logic:

Half the entries in a 4×4 matrix carry positive signs.

🧠 Memory Trick

"Checkerboard splits equally"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Sign Pattern of Cofactors

7 For the second row (\(i=2\)) of a 3×3 matrix, the sign pattern \({\left(-1\right)}^{\left(i,\ j\right)}\)for \(j=1,2,3\) is:

A (1,-1,1)

B (-1,-1,-1)

C (-1,1,-1)

D (1,1,-1)

💡 Short Explanation

Substitute row number \(i=2\) Signs alternate in checkerboard form Second row pattern becomes −,+,−

✅ Correct Answer Explanation

Using: \({\left(-1\right)}^{i+j}\) For \(i=2\): \(\left(-1)^{3}=-1\right.\left(-1)^{4}=1\right.\left(-1)^{5}=-1\right.\) Hence the pattern becomes: \(\left(-1,1,-1\right)\) Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Represents the first-row sign pattern, not the second row.
Option B → Cofactor signs alternate and cannot all be negative.
Option D → The second entry should be positive, but the first must be negative.

🎯 Strategy

Used

Substitution

Application:

Substitute values of \(j=1,2,3\) into the cofactor sign formula.

Final Logic:

Second-row signs alternate as negative, positive, negative.

🧠 Memory Trick

"Second row starts negative"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Cofactor Sign Pattern

8 Which transformation converts a vector of minors \(\left(M_{21},\ M_{22},\ M_{23}\right)\)into cofactors \(\left(A_{21},\ A_{22},\ A_{23}\right)\)?

A Identity matrix

B Matrix with all entries −1

C Adjoint matrix

D Diagonal matrix with entries (-1,1,-1)

💡 Short Explanation

Cofactors differ only by signs Second-row sign pattern is −,+,− Diagonal sign transformation applies

✅ Correct Answer Explanation

Cofactors are obtained by multiplying minors with the sign pattern: \({\left(-1\right)}^{i+j}\) For the second row, the signs become: \(\left(-1,1,-1\right)\) Thus the transformation is equivalent to multiplying by a diagonal matrix: \(diag(-1,1,-1)\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Identity transformation leaves minors unchanged and does not create cofactors.
Option B → All cofactors are not simultaneously negative.
Option C → Adjoint matrix is constructed from cofactors and does not represent the sign transformation itself.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Relate second-row cofactor signs directly to diagonal sign multiplication.

Final Logic:

Only the diagonal matrix reproduces the required sign changes.

🧠 Memory Trick

"Second row → − + −"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Cofactor Transformation

9 Identify the incorrect statement:

A Expanding along a row with more zeros simplifies calculation

B Expansion along any row gives the same determinant value

C \(∣A∣=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\)is valid

D Expansion results in a scalar value

💡 Short Explanation

More zeros reduce calculation effort Any row expansion gives same determinant All statements are mathematically correct

✅ Correct Answer Explanation

Option A is actually correct because rows or columns containing more zeros simplify determinant expansion. Options B, C, and D are also correct determinant properties. Therefore no incorrect statement exists. The provided answer is wrong because it incorrectly labels a true statement as false.

❓ Why Other Options Are Incorrect

Option B → Determinant value remains unchanged regardless of expansion row or column.
Option C → This is the standard first-row cofactor expansion formula.
Option D → Determinants always evaluate to scalar values, not matrices.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using standard determinant properties.

Final Logic:

Since all four statements are true, no incorrect option exists.

🧠 Memory Trick

"More zeros = easier expansion"

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

10 Match column expansions of a 3×3 determinant:

List I	List II
1. Expansion along \(C_{1}\)	a. \(a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}\)
2. Expansion along \(C_{2}\)	b. \(a_{12}A_{12}+a_{22}A_{22}+a_{32}A_{32}\)
3. Expansion along \(C_{3}\)	c. \(a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}\)

A 1–a, 2–b, 3–c

B 1–b, 2–c, 3–a

C 1–c, 2–a, 3–b

D 1–b, 2–a, 3–c

💡 Short Explanation

Column expansion uses fixed column entries Each column matches same-index cofactors Standard determinant expansion property

✅ Correct Answer Explanation

Expansion along a column uses elements and cofactors from that column only: \(C_{1}\rightarrow a_{11},a_{21},a_{31}C_{2}\rightarrow a_{12},a_{22},a_{32}C_{3}\rightarrow a_{13},a_{23},a_{33}\) Thus the correct matching is: \(1-a, 2-b, 3-c\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Column indices are mismatched with incorrect cofactor groups.
Option C → Expansion terms do not correspond to the specified columns.
Option D → First-column terms are incorrectly assigned.

🎯 Strategy

Used

Option Grouping

Application:

Match each column with cofactors having the same column index.

Final Logic:

Correct column expansion preserves identical column numbering.

🧠 Memory Trick

"Column number stays fixed"

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

11 For a determinant \(∣A∣\), a student computes:
\(a_{11}A_{11}=26,a_{12}A_{12}=13,a_{13}A_{13}=13\)
Find \(∣A∣\).

A 0

B 52

C 104

D 26

💡 Short Explanation

Determinant equals sum of row products Add corresponding cofactor products \(26+13+13=52\)

✅ Correct Answer Explanation

Using first-row expansion: \(∣A∣=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\) Substituting given values: \(∣A∣=26+13+13=52\) Thus Option B is correct. Other options ignore the cofactor expansion rule or involve incorrect addition.

❓ Why Other Options Are Incorrect

Option A → Determinant is not zero because the sum of products is nonzero.
Option C → Incorrect addition doubles the actual determinant value.
Option D → Uses only one term instead of summing all cofactor products.

🎯 Strategy

Used

Substitution

Application:

Substitute given cofactor products directly into determinant expansion formula.

Final Logic:

Adding all row-product terms gives determinant value 52.

🧠 Memory Trick

"Determinant = sum of cofactor products"

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

12 A 3×3 determinant is expanded along three different rows/columns, giving values 15, 15, 15.
Find the 3-period moving average.

A 5

B 45

C 30

D 15

💡 Short Explanation

Moving average equals arithmetic mean Determinant value remains same for expansions Average of identical values stays same

✅ Correct Answer Explanation

The three obtained values are: \(15, 15, 15\) Their moving average is: \(\frac{15+15+15}{3}=15\) Thus Option D is correct. Since determinant value remains unchanged irrespective of expansion row or column, all values are identical.

❓ Why Other Options Are Incorrect

Option A → Incorrect division or averaging has been used.
Option B → Sum is 45, but the question asks for average, not total.
Option C → Arithmetic mean of identical values cannot become 30.

🎯 Strategy

Used

Substitution

Application:

Apply the moving-average formula directly to the given values.

Final Logic:

Average of three equal values remains the same value.

🧠 Memory Trick

"Same numbers → same average"

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

13 If
\(a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0,\)
this verifies which property?

A Expansion along a row

B Sum of products of elements of one row with cofactors of another row is zero

C Determinant is always positive

D Cofactors are always equal

💡 Short Explanation

Different row–cofactor products vanish Standard determinant property Same-row expansion alone gives determinant

✅ Correct Answer Explanation

A determinant property states: \(\sum a_{ij}A_{kj}=0(i\neq k)\) That means the sum of products of elements of one row with cofactors of another row equals zero. Hence Option B is correct. Option A refers to same-row expansion, while C and D are unrelated determinant properties.

❓ Why Other Options Are Incorrect

Option A → Expansion along the same row gives determinant value, not zero necessarily.
Option C → Determinants may be positive, negative, or zero.
Option D → Cofactors need not be equal in a determinant.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Match the expression with the standard orthogonality property of rows and cofactors.

Final Logic:

Different-row cofactor products always sum to zero.

🧠 Memory Trick

"Different row + cofactors = zero"

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

📄 Passage

Area of a triangle using determinants is taken as absolute value.

If determinant = 0 → points are collinear.
If determinant < 0 → take absolute value.
If the determinant formed by three points is zero, then:

A Points form equilateral triangle

B Points are collinear

C Points lie on a circle

D Area is negative

💡 Short Explanation

Zero determinant gives zero area Zero area implies collinearity Points lie on same straight line

✅ Correct Answer Explanation

The determinant formula for triangle area becomes zero when all three points lie on one straight line. The passage explicitly states: \(determinant=0\Rightarrow collinear points\) Therefore Option B is correct. Other options do not follow from zero area.

❓ Why Other Options Are Incorrect

Option A → Equilateral triangles have nonzero area.
Option C → Collinearity does not imply points lie on a circle uniquely.
Option D → Area is never taken as negative because absolute value is used.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Directly apply the determinant-area condition stated in the passage.

Final Logic:

Zero determinant means zero area and hence collinear points.

🧠 Memory Trick

"Zero determinant → straight line"

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

📄 Passage

Area of a triangle using determinants is taken as absolute value.

If determinant = 0 → points are collinear.
If determinant < 0 → take absolute value.
If determinant value is negative while computing area, we should:

A Ignore result

B Assume error

C Take absolute value

D Square the value

💡 Short Explanation

Area cannot be negative Determinant sign depends on order Absolute value gives physical area

✅ Correct Answer Explanation

Area obtained from determinant formulas is always taken in absolute value because geometric area cannot be negative. The sign only depends on the order of point arrangement. Hence Option C is correct. Ignoring or squaring the determinant is mathematically unnecessary.

❓ Why Other Options Are Incorrect

Option A → Negative sign is not ignored; absolute value must be taken.
Option B → A negative determinant does not indicate calculation error.
Option D → Squaring changes the actual area value incorrectly.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the explicit statement in the passage about handling negative determinants.

Final Logic:

Area is always represented by the absolute value of determinant.

🧠 Memory Trick

"Area = modulus of determinant"

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

16 For the determinant
\(∣\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1\end{pmatrix}∣,\)
the cofactor of the element at position \(\left(1,\ 3\right)\)is:

A \(y_{2}-y_{3}\)

B \(y_{3}-y_{2}\)

C \(x_{2}-x_{3}\)

D \(x_{3}-x_{2}\)

💡 Short Explanation

Delete first row and third column Evaluate resulting 2×2 determinant Apply positive cofactor sign

✅ Correct Answer Explanation

Deleting row 1 and column 3 gives: \(∣\begin{pmatrix}x_{2} & y_{2}\\ x_{3} & y_{3}\end{pmatrix}∣\) Minor: \(x_{2}y_{3}-x_{3}y_{2}\) Since cofactor sign at \(\left(1,\ 3\right)\)is positive, cofactor remains unchanged. Thus the provided answer is incorrect. Actual cofactor: \(x_{2}y_{3}-x_{3}y_{2}\) None of the options match exactly.

❓ Why Other Options Are Incorrect

Option A → Gives only a difference of y-values, not the determinant expression.
Option B → Incorrect sign and incomplete determinant form.
Option C → Cofactor depends on determinant products, not simple x-differences.
Option D → Does not represent the required 2×2 determinant.

🎯 Strategy

Used

Substitution

Application:

Delete corresponding row and column, then evaluate the resulting determinant.

Final Logic:

Cofactor equals the determinant \(x_{2}y_{3}-x_{3}y_{2}\).

🧠 Memory Trick

"Delete row-column → solve remaining determinant"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Cofactor Calculation

17 Find the minor of element 5 in
\(∣\begin{pmatrix}1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9\end{pmatrix}∣\)

A -3

B 5

C 2

D 4

💡 Short Explanation

Element 5 lies at position (2,2) Delete second row and column Evaluate remaining 2×2 determinant

✅ Correct Answer Explanation

Deleting row 2 and column 2 leaves: \(∣\begin{pmatrix}1 & 3\\ 7 & 9\end{pmatrix}∣\) Minor: \(1(9)-3(7)=9-21=-12\) Thus the provided answer A) -3 is incorrect. Actual Correct Answer: \(-12\)(Not available in options)

❓ Why Other Options Are Incorrect

Option A → Minor evaluates to \(-12\), not \(-3\).
Option B → 5 is the element itself, not its minor.
Option C → Incorrect determinant calculation.
Option D → Does not result from evaluating the remaining 2×2 determinant.

🎯 Strategy

Used

Substitution

Application:

Delete the row and column of the selected element and compute the determinant.

Final Logic:

Minor of element 5 equals \(-12\).

🧠 Memory Trick

"Middle element → corner determinant"

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

18 For
\(∣\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}∣,\)
find the cofactor \(A_{11}\).

A 0

B 4

C -4

D 2

💡 Short Explanation

Delete first row and column Remaining determinant equals 4 Cofactor sign at (1,1) is positive

✅ Correct Answer Explanation

Minor \(M_{11}\)is obtained by deleting row 1 and column 1: \(\left[4\right]\) Thus: \(M_{11}=4\) Cofactor: \(A_{11}=(-1)^{1+1}M_{11}=4\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Remaining determinant after deletion is not zero.
Option C → Position (1,1) has positive sign, not negative.
Option D → Cofactor value is determined directly from the remaining element 4.

🎯 Strategy

Used

Substitution

Application:

Find the minor first and then apply cofactor sign.

Final Logic:

Positive cofactor sign keeps the value equal to 4.

🧠 Memory Trick

"Top-left cofactor stays positive"

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

19 If
\(∣\begin{pmatrix}1 & x\\ 2 & 4\end{pmatrix}∣=0,\)
find the sum of all possible values of \(x\).

A 8

B 2

C 0

D 4

💡 Short Explanation

Use \(ad-bc=0\) Solve linear equation in \(x\) Only one value satisfies condition

✅ Correct Answer Explanation

Evaluate determinant: \(1(4)-2x=04-2x=0x=2\) Since only one possible value exists, the sum of all values is also 2. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → No determinant calculation gives \(x=8\).
Option C → Determinant becomes zero only at \(x=2\), not 0.
Option D → Substituting \(x=4\) gives determinant \(-4\), not zero.

🎯 Strategy

Used

Substitution

Application:

Apply the standard determinant formula and solve the equation.

Final Logic:

Only \(x=2\) satisfies the zero-determinant condition.

🧠 Memory Trick

"Zero determinant → solve linear equation"

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

20 Find the determinant of the identity matrix of order 3:

A 0

B 3

C 2

D 1

💡 Short Explanation

Identity matrix has diagonal ones Off-diagonal entries are zero Determinant always equals one

✅ Correct Answer Explanation

For the identity matrix: \(I_{3}=\left(\begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}\right)\) The determinant equals the product of diagonal entries: \(1\times 1\times 1=1\) Hence Option D is correct. Other options do not satisfy determinant properties of identity matrices.

❓ Why Other Options Are Incorrect

Option A → Identity matrices are non-singular, so determinant cannot be zero.
Option B → Determinant is not obtained by adding diagonal entries.
Option C → Product of diagonal entries equals 1, not 2.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Recall the standard determinant property of identity matrices.

Final Logic:

Determinant of every identity matrix equals one.

🧠 Memory Trick

"Identity determinant is identity itself: 1"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Identity Matrix Determinant

CUET UG Mathematics Booster Test 1 - Minors, Cofactors and Cofactor Expansion

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