CUET UG Mathematics Booster Test 2 - Minors, Cofactors and Cofactor Expansion -

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CUET UG Mathematics Booster Test 2 - 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 2 - Minors, Cofactors and Cofactor Expansion

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QUESTION 1 OF 20

For a 3×3 matrix \(A\) representing vectors in \(R^{3}\), the minor \(M_{ij}\)of an element is related geometrically to:

Answer Review

1 For a 3×3 matrix \(A\) representing vectors in \(R^{3}\), the minor \(M_{ij}\)of an element is related geometrically to:

A Area of the parallelogram formed by the remaining two row (or column) vectors

B Volume of the 3D parallelepiped

C Sum of the vectors

D Magnitude of a single vector

💡 Short Explanation

Minor is a 2×2 determinant. 2×2 determinants represent planar area. Remaining vectors form a parallelogram.

✅ Correct Answer Explanation

In a 3×3 determinant, removing one row and one column leaves a 2×2 determinant called a minor. Geometrically, a 2×2 determinant represents the signed area of a parallelogram formed by two vectors. Hence Option A is correct. Option B corresponds to the full 3×3 determinant, not the minor.

❓ Why Other Options Are Incorrect

Option B → Volume of a parallelepiped is represented by the complete 3×3 determinant, not a minor.
Option C → Minors are determinants, not vector sums.
Option D → Magnitude of a single vector is unrelated to determinant minors.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate determinant order with its geometric interpretation.

Final Logic:

A 2×2 minor corresponds to planar area geometry.

🧠 Memory Trick

"Minor → smaller area determinant."

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

2 Assertion (A): The submatrix used to calculate \(M_{32}\)in a 4×4 matrix is a 3×3 matrix.
Reason (R): Deleting row 3 and column 2 leaves a 2×2 matrix.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Minor reduces order by one. 4×4 minor gives 3×3 submatrix. Reason incorrectly states 2×2 matrix.

✅ Correct Answer Explanation

To compute \(M_{32}\)in a 4×4 matrix, delete row 3 and column 2. The remaining matrix has order 3×3. Thus Assertion is true. Reason is false because deleting one row and one column from a 4×4 matrix cannot produce a 2×2 matrix. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Assertion is true because order reduces from 4×4 to 3×3.
Option C → Reason is incorrect, so it cannot explain the Assertion.
Option D → Assertion is correct since a 3×3 submatrix remains after deletion.

🎯 Strategy

Used: Elimination

Application:

Use the determinant order reduction rule carefully.

Final Logic:

Deleting one row and column reduces order by exactly one.

🧠 Memory Trick

"Minor order = original order − 1."

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

3 Let
\(A=\left(\begin{pmatrix}x & 1 & 2\\ 1 & x & 3\\ 2 & 3 & x\end{pmatrix}\right)\)
The determinant \(∣A∣\)is a polynomial in \(x\). Find the coefficient of \(x^{2}\).

A 1

B 2

C 3

D 4

💡 Short Explanation

Expand determinant algebraically. Collect powers of \(x\). Coefficient of \(x^{2}\)becomes 3.

✅ Correct Answer Explanation

Expanding determinant: \(∣A∣=x(x^{2}-9)-1(x-6)+2(3-2x)=x^{3}-9x-x+6+6-4x=x^{3}-14x+12\) Thus coefficient of \(x^{2}\)is actually 0, not 3. Therefore the provided answer C is incorrect and none of the options are correct.

❓ Why Other Options Are Incorrect

Option A → Coefficient of \(x^{2}\)is not 1 after expansion.
Option B → No quadratic term appears in the determinant polynomial.
Option C → Expansion incorrectly assumes a nonzero \(x^{2}\)coefficient.
Option D → Determinant simplification does not generate coefficient 4.

🎯 Strategy

Used: Substitution

Application:

Expand determinant systematically and compare polynomial coefficients.

Final Logic:

The determinant polynomial contains no \(x^{2}\)term.

🧠 Memory Trick

"Expand fully before reading coefficients."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Polynomial Determinants

4 For a 3×3 matrix, what is the order of any minor?

A 1×1

B 2×2

C 3×3

D Depends on elements

💡 Short Explanation

Minor removes one row and one column. Order reduces by one. 3×3 becomes 2×2.

✅ Correct Answer Explanation

A minor is obtained by deleting one row and one column from a matrix. Therefore the order reduces by one. In a 3×3 matrix, every minor is a determinant of order 2×2. Hence Option B is correct. Matrix elements do not affect the order of minors.

❓ Why Other Options Are Incorrect

Option A → A 1×1 determinant comes from minors of a 2×2 matrix.
Option C → Minor cannot retain the original order.
Option D → Minor order depends only on matrix size, not entries.

🎯 Strategy

Used: Elimination

Application:

Apply the standard minor-definition rule directly.

Final Logic:

Deleting one row and column reduces order from 3×3 to 2×2.

🧠 Memory Trick

"Minor means one order lower."

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

5 In the cross product of two vectors using determinant form, the components correspond to:

A Cofactors of the first row

B Minors of the second row

C Diagonal products

D Vector sums

💡 Short Explanation

Cross product uses determinant expansion. Unit vectors occupy first row. Components arise from cofactors.

✅ Correct Answer Explanation

The vector cross product is represented using determinant expansion: \(∣\begin{pmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_{1} & a_{2} & a_{3}\\ b_{1} & b_{2} & b_{3}\end{pmatrix}∣\) The resulting components come from cofactors of the first row containing unit vectors. Hence Option A is correct. Other options do not represent the determinant structure of vector products.

❓ Why Other Options Are Incorrect

Option B → Cross-product components are not obtained from second-row minors alone.
Option C → Determinant expansion is more than simple diagonal multiplication.
Option D → Cross product is not formed through vector addition.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect determinant expansion with vector cross-product representation.

Final Logic:

Cross-product components arise from first-row cofactors.

🧠 Memory Trick

"\(\hat{i},\hat{j},\hat{k}\) → cofactor expansion."

📚 Class XII – Mathematics Part 1 – Chapter: Vector Algebra – Topic: Cross Product Using Determinants

6 Arrange the following positions in increasing order of sign factor \({\left(-1\right)}^{\left(i,\ j\right)}\):
1. \(\left(1,\ 2\right)\)
2. \(\left(2,\ 2\right)\)
3. \(\left(2,\ 3\right)\)
4. \(\left(1,\ 1\right)\)

A 1, 3, 2, 4

B 1, 3, 4, 2

C 3, 1, 2, 4

D 2, 4, 1, 3

💡 Short Explanation

Odd \(i+j\) gives negative sign. Even \(i+j\) gives positive sign. Arrange negatives before positives.

✅ Correct Answer Explanation

Evaluate signs: \((1,2)\Rightarrow (-1)^{3}=-1(2,3)\Rightarrow (-1)^{5}=-1(1,1)\Rightarrow (-1)^{2}=+1(2,2)\Rightarrow (-1)^{4}=+1\) Thus increasing order is negative values first, then positive values: 1,3,4,2. Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Places a positive-sign position before another positive incorrectly.
Option C → Incorrect ordering among negative-sign positions.
Option D → Starts with positive-sign entries instead of negative ones.

🎯 Strategy

Used: Substitution

Application:

Compute \(i+j\) for each position and determine parity.

Final Logic:

Odd sums give −1 and even sums give +1.

🧠 Memory Trick

"Odd sum → minus, even sum → plus."

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

7 Let \(S_{n}\)be the sum of all sign factors \({\left(-1\right)}^{\left(i,\ j\right)}\)in an \(n\times n\) matrix.
For \(n=1,2,3,4\), the values of \(S_{n}\)are:

A 1,0,1,0

B 1,0,-1,0

C 1,1,1,1

D 0,0,0,0

💡 Short Explanation

Checkerboard signs alternate. Even-order matrices balance signs equally. Odd-order matrices leave one extra positive.

✅ Correct Answer Explanation

The cofactor sign pattern alternates like a checkerboard: \(\begin{pmatrix}+ & - & +\\ - & + & -\\ + & - & +\end{pmatrix}\) For even \(n\), positives and negatives cancel, giving zero. For odd \(n\), one extra positive remains. Hence: \(S_{1}=1, S_{2}=0, S_{3}=1, S_{4}=0\) Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → \(S_{3}\)is positive 1, not −1.
Option C → Even-order matrices do not retain positive excess.
Option D → Odd-order matrices always leave one extra positive term.

🎯 Strategy

Used: Odd One Out

Application:

Observe checkerboard parity behavior between odd and even orders.

Final Logic:

Odd orders leave one extra positive sign.

🧠 Memory Trick

"Odd leaves +1, even cancels."

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

8 The positions where \(\left(-1)^{\left(i,\ j\right)}=+1\right.\)in a matrix correspond to:

A Only diagonal elements

B Only corners

C Only middle row

D Positions where \(i+j\) is even

💡 Short Explanation

Sign depends on parity of \(i+j\). Even sum gives positive sign. Odd sum gives negative sign.

✅ Correct Answer Explanation

The cofactor sign rule is: \({\left(-1\right)}^{i+j}\) If \(i+j\) is even, exponent becomes even and sign equals +1. If odd, sign becomes −1. Therefore Option D is correct. Positive signs are not restricted only to diagonal or corner positions.

❓ Why Other Options Are Incorrect

Option A → Positive signs also appear outside the diagonal.
Option B → Corners are not the only positive positions.
Option C → Positive signs are distributed throughout the checkerboard pattern.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Use parity interpretation of exponent \(i+j\).

Final Logic:

Even exponent gives positive sign factor.

🧠 Memory Trick

"Even sum → positive cofactor."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Sign Convention

9 Match expressions with meaning:

List I	List II
1. \(∣A∣=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\)	a. Column 1 expansion
2. \(a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0\)	b. Zero property
3. \(∣A∣=a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}\)	c. Row 1 expansion

A 1–c, 2–b, 3–a

B 1–a, 2–c, 3–b

C 1–b, 2–a, 3–c

D 1–c, 2–a, 3–b

💡 Short Explanation

First expression expands along row 1. Second represents orthogonality-type zero sum. Third expands along column 1.

✅ Correct Answer Explanation

The first expression uses elements and cofactors from row 1, so it is row-1 expansion. The second expression multiplies elements of one row with cofactors of another, giving zero property. The third expression uses column-1 elements and cofactors. Hence matching becomes: 1–c, 2–b, 3–a.

❓ Why Other Options Are Incorrect

Option B → Incorrectly identifies row expansion as column expansion.
Option C → Misplaces the zero property and row expansion meanings.
Option D → Third expression clearly represents column expansion, not zero property.

🎯 Strategy

Used: Option Grouping

Application:

Identify whether indices vary by row or by column in each expression.

Final Logic:

Fixed first index → row expansion; fixed second index → column expansion.

🧠 Memory Trick

"Same row indices → row expansion."

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

10 Expansion along the \(j^{th}\)column is:
\(∣A∣=\sum_{i=1}^{n}\,a_{ij}A_{ij}\)
Which statements are correct?
1. Reduces an \(n^{th}\)order determinant to \({\left(n-1\right)}^{th}\)order
2. Represents a linear combination of cofactors
3. Depends on the column chosen

A 2 and 3

B 1 and 2

C 1 and 3

D All of the above

💡 Short Explanation

Cofactor expansion reduces determinant order. Expansion forms linear combination. Determinant value is independent of chosen column.

✅ Correct Answer Explanation

Cofactor expansion reduces determinant order by one because each cofactor contains an \(\left(n-1)\times (n-1\right)\)minor. It is also a linear combination of elements and cofactors. However, determinant value does not depend on the chosen row or column. Hence statements 1 and 2 are correct, while statement 3 is false.

❓ Why Other Options Are Incorrect

Option A → Statement 3 is incorrect because determinant value remains unchanged across columns.
Option C → Ignores the linear-combination nature of cofactor expansion.
Option D → Expansion method may vary, but determinant value is independent of column choice.

🎯 Strategy

Used: Elimination

Application:

Separate determinant value from computational convenience.

Final Logic:

Expansion row/column changes method, not determinant value.

🧠 Memory Trick

"Any row, same determinant."

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

11 Identify the incorrect statement regarding expansion
\(∣A∣=a_{i1}A_{i1}+a_{i2}A_{i2}+a_{i3}A_{i3}\)

A Cofactors include positional signs \({\left(-1\right)}^{\left(i,\ j\right)}\)

B Determinant value is independent of the row chosen

C The sum equals the trace of the cofactor matrix

D It involves evaluation of second-order determinants

💡 Short Explanation

Cofactor expansion gives determinant value. Trace is unrelated to determinant expansion. Minors of order 2 appear in 3×3 expansion.

✅ Correct Answer Explanation

The given expression represents cofactor expansion along the \(i^{th}\)row. Cofactors contain sign factors: \(A_{ij}=(-1)^{i+j}M_{ij}\) Expansion value remains same for every row or column. The process uses second-order minors. However, the sum does not represent the trace of the cofactor matrix. Hence Option C is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct because cofactors always include positional sign factors.
Option B → Determinant remains invariant under choice of expansion row or column.
Option D → 3×3 determinant expansion reduces to evaluation of 2×2 minors.

🎯 Strategy

Used: Elimination

Application:

Separate determinant concepts from unrelated matrix concepts like trace.

Final Logic:

Trace of cofactor matrix is unrelated to determinant expansion formula.

🧠 Memory Trick

"Expansion gives determinant, not trace."

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

12 If \(∣A∣=10\) for a 3×3 matrix, and each element of \(A\) is multiplied by 2, what is the new determinant value?

A 20

B 40

C 60

D 80

💡 Short Explanation

Scaling affects determinant multiplicatively. For 3×3 matrix, factor becomes \(2^{3}\). New determinant equals \(8\times 10\).

✅ Correct Answer Explanation

For an \(n\times n\) matrix: \(∣kA∣=k^{n}∣A∣\) For a 3×3 matrix: \(∣2A∣=2^{3}∣A∣=8(10)=80\) Hence Option D is correct. Other options arise from multiplying by incorrect scaling powers.

❓ Why Other Options Are Incorrect

Option A → Uses only single multiplication by 2 instead of cubic scaling.
Option B → Uses incorrect factor \(2^{2}\).
Option C → Determinant scaling for 3×3 matrices is not linear.

🎯 Strategy

Used: Substitution

Application:

Apply determinant scaling property directly.

Final Logic:

\(2^{3}\times 10=80\).

🧠 Memory Trick

"3×3 matrix → cube the factor."

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

📄 Passage

For efficient calculation, expand along the row/column with maximum zeros.

Sign factor: \({\left(-1\right)}^{\left(i,\ j\right)}\).
If elements of one row are multiplied with cofactors of another row, the sum is zero.
Why does
\(a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0\)
hold?

A It represents a determinant with two identical rows

B Sign factors cancel

C Minors are zero

D Cofactors are zero

💡 Short Explanation

Expansion uses cofactors of another row. Resulting determinant has repeated rows. Determinant with identical rows equals zero.

✅ Correct Answer Explanation

When elements of one row are multiplied by cofactors of another row, the resulting determinant effectively contains two identical rows. A determinant with identical rows is always zero. Hence: \(a_{11}A_{21}+a_{12}A_{22}+a_{13}A_{23}=0\) Therefore Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Sign cancellation alone does not explain the general zero property.
Option C → Individual minors need not be zero.
Option D → Cofactors themselves are generally nonzero.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Relate the cofactor identity to repeated-row determinant property.

Final Logic:

Identical-row determinants always vanish.

🧠 Memory Trick

"Same rows → determinant zero."

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

📄 Passage

For efficient calculation, expand along the row/column with maximum zeros.

Sign factor: \({\left(-1\right)}^{\left(i,\ j\right)}\).
If elements of one row are multiplied with cofactors of another row, the sum is zero.
If a 3×3 matrix has an entire row of zeros, its determinant is best evaluated by:

A Using column operations

B Expanding along that row (gives zero directly)

C Computing all minors

D Expanding along another row

💡 Short Explanation

Zero row simplifies cofactor expansion. Every term becomes zero immediately. Determinant directly evaluates to zero.

✅ Correct Answer Explanation

Expanding along a row containing all zeros is the fastest method because every term in the cofactor expansion becomes zero: \(∣A∣=0\) Thus determinant is obtained instantly without evaluating minors. Hence Option B is correct. Other approaches unnecessarily increase calculations.

❓ Why Other Options Are Incorrect

Option A → Column operations are unnecessary when a zero row already exists.
Option C → Computing minors wastes effort because determinant is immediately zero.
Option D → Another row may involve nonzero terms and longer calculations.

🎯 Strategy

Used: Elimination

Application:

Choose the expansion direction minimizing computation.

Final Logic:

A complete zero row gives determinant zero instantly.

🧠 Memory Trick

"Expand where zeros live."

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

15 For a triangle with vertices on the unit circle at angles \({\theta}_{1},{\theta}_{2},{\theta}_{3}\), the determinant-based area involves expressions of the form:

A Difference of cosines

B Sine of angle differences

C Product of cosines

D Area of sector

💡 Short Explanation

Unit-circle coordinates involve sine and cosine. Area determinants simplify using trig identities. Sine differences naturally appear.

✅ Correct Answer Explanation

Coordinates on the unit circle are: \(\left(cos⁡\theta ,sin⁡\theta \right)\) When substituted into determinant area formulas, trigonometric simplification produces terms involving: \(sin⁡({\theta}_{i}-{\theta}_{j})\) Hence Option B is correct. Products or simple cosine differences alone do not fully represent determinant-based area expressions.

❓ Why Other Options Are Incorrect

Option A → Cosine differences alone are insufficient for determinant area simplification.
Option C → Product forms do not directly arise from area determinants.
Option D → Sector area is unrelated to determinant-based triangle area.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Connect unit-circle coordinates with determinant trigonometric identities.

Final Logic:

Determinant simplification naturally produces sine differences.

🧠 Memory Trick

"Circle area determinants love sine."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Trigonometric Coordinate Area

16 For matrix
\(A=\left(\begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i\end{pmatrix}\right),\)
the cofactor \(A_{11}\)is:

A \(ei-fh\)

B \(fh-ei\)

C \(di-fg\)

D \(ei-fh\)

💡 Short Explanation

Delete row 1 and column 1. Remaining 2×2 determinant gives minor. Sign factor is positive at (1,1).

✅ Correct Answer Explanation

Deleting row 1 and column 1 leaves: \(∣\begin{pmatrix}e & f\\ h & i\end{pmatrix}∣=ei-fh\) Since: \(\left(-1)^{1+1}=+1\right.\) the cofactor equals: \(A_{11}=ei-fh\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Reverses determinant subtraction order incorrectly.
Option C → Uses elements from wrong rows and columns.
Option D → Duplicate of Option A; mathematically same expression.

🎯 Strategy

Used: Substitution

Application:

Form the minor first, then apply cofactor sign.

Final Logic:

Top-left cofactor has positive sign.

🧠 Memory Trick

"(1,1) starts with plus."

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

17 In
\(∣\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}∣,\)
the minor of element 1 is:

A 4

B 2

C 3

D 1

💡 Short Explanation

Remove row and column of element 1. Remaining determinant is 1×1. Minor equals 4.

✅ Correct Answer Explanation

Element 1 is at position (1,1). Deleting row 1 and column 1 leaves: \(\left[4\right]\) Thus the corresponding minor equals 4. Hence Option A is correct. Other values correspond to remaining matrix elements, not the required minor.

❓ Why Other Options Are Incorrect

Option B → Element 2 is removed during minor formation.
Option C → Element 3 does not remain after deletion.
Option D → The original element itself is not the minor.

🎯 Strategy

Used: Elimination

Application:

Delete the row and column containing the target element.

Final Logic:

Only element 4 survives after deletion.

🧠 Memory Trick

"Delete row-column, keep leftover."

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

18 For
\(A=\left(\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right),\)
the cofactor matrix is:

A \(\left(\begin{pmatrix}4 & 2\\ 3 & 1\end{pmatrix}\right)\)

B \(\left(\begin{pmatrix}4 & -3\\ -2 & 1\end{pmatrix}\right)\)

C \(\left(\begin{pmatrix}4 & -2\\ -3 & 1\end{pmatrix}\right)\)

D \(\left(\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right)\)

💡 Short Explanation

Compute minors first. Apply checkerboard sign pattern. Arrange cofactors in same positions.

✅ Correct Answer Explanation

Minors are: \(M_{11}=4,M_{12}=3,M_{21}=2,M_{22}=1\) Applying signs: \(\left(\begin{pmatrix}+ & -\\ - & +\end{pmatrix}\right)\) gives cofactor matrix: \(\left(\begin{pmatrix}4 & -3\\ -2 & 1\end{pmatrix}\right)\) Hence Option B is correct. The provided answer C is incorrect.

❓ Why Other Options Are Incorrect

Option A → Ignores cofactor sign changes completely.
Option C → Incorrectly swaps signs of off-diagonal cofactors.
Option D → Original matrix is not equal to its cofactor matrix.

🎯 Strategy

Used: Substitution

Application:

Calculate minors and apply checkerboard sign rule systematically.

Final Logic:

Off-diagonal cofactors must be negative.

🧠 Memory Trick

"Checkerboard signs: + − / − +."

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

19 If \(∣A∣=5\) for a 3×3 matrix, find \(∣adj(A)∣\).

A 5

B 25

C 125

D 625

💡 Short Explanation

Use adjoint determinant property. For \(n\times n\) matrix: \(∣adj(A)∣=∣A∣^{n-1}\) Here \(n=3\).

✅ Correct Answer Explanation

For an \(n\times n\) matrix: \(∣adj(A)∣=∣A∣^{n-1}\) For a 3×3 matrix: \(∣adj(A)∣=5^{2}=25\) Hence Option B is correct. Other options use incorrect exponents or direct copying of determinant value.

❓ Why Other Options Are Incorrect

Option A → Uses determinant directly instead of adjoint property.
Option C → Incorrectly cubes the determinant.
Option D → Uses exponent four, which is invalid here.

🎯 Strategy

Used: Substitution

Application:

Apply the standard adjoint-determinant relation carefully.

Final Logic:

For order 3, exponent becomes 2.

🧠 Memory Trick

"Adjoint power = \(n-1\)."

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

20 Let \(A\) be a 3×3 diagonal matrix with all diagonal elements equal to 3.
Find the cofactor \(A_{11}\).

A 3

B 6

C 9

D 27

💡 Short Explanation

Delete first row and column. Remaining determinant is diagonal 2×2. Cofactor equals product \(3\times 3\).

✅ Correct Answer Explanation

The matrix is: \(\left(\begin{pmatrix}3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 3\end{pmatrix}\right)\) Deleting row 1 and column 1 leaves: \(∣\begin{pmatrix}3 & 0\\ 0 & 3\end{pmatrix}∣=9\) Since sign at position (1,1) is positive, cofactor: \(A_{11}=9\) Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Uses only one diagonal entry instead of the determinant of remaining matrix.
Option B → No determinant computation gives value 6.
Option D → 27 is the determinant of the whole matrix, not the cofactor.

🎯 Strategy

Used: Substitution

Application:

Form the required minor by deleting row and column.

Final Logic:

Remaining 2×2 diagonal determinant equals 9.

🧠 Memory Trick

"Diagonal minor → multiply leftovers."

📚 Class XII – Mathematics Part 1 – Chapter: Determinants – Topic: Cofactors of Diagonal Matrices

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