CUET UG Applied Mathematics Booster Test 3 - Determinants and Their Properties -

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CUET UG Applied Mathematics Booster Test 3 - Determinants and Their Properties

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Determinants and Their Properties

1 / 20

QUESTION 1 OF 20

A functional mapping f: M_n -> R defines the determinant of a matrix A. If matrix A is structured such that m != n (where m represents rows and n represents columns), what is the valid output of the function f(A)?

Answer Review

1 A functional mapping f: M_n -> R defines the determinant of a matrix A. If matrix A is structured such that m != n (where m represents rows and n represents columns), what is the valid output of the function f(A)?

A 0

B 1

C The mapping f(A) is strictly undefined.

D The sum of the principal diagonal elements.

💡 Short Explanation

�� Determinants are defined only for square matrices. �� If m ≠ n, the matrix is rectangular. �� Therefore, determinant mapping does not exist.

✅ Correct Answer Explanation

A determinant is a scalar quantity associated only with square matrices of order n × n. The function f : Mₙ → R maps square matrices to real numbers. → If m ≠ n, then the matrix is not square, so determinant evaluation is mathematically undefined. Hence, option C is correct. → Option A is incorrect because determinants are not automatically zero for rectangular matrices. → Option B is incorrect because determinant value 1 applies only to specific matrices like identity matrices. → Option D is incorrect because the sum of diagonal elements represents the trace, not the determinant.

❓ Why Other Options Are Incorrect

�� Option A → A rectangular matrix does not have determinant value 0; its determinant is undefined.
�� Option B → Determinant equals 1 only in special cases such as identity matrices.
�� Option D → Sum of diagonal entries is the trace of a matrix, not determinant evaluation.

🎯 Strategy

Used: Elimination

Application: Eliminate options giving numerical values because determinants cannot even be computed for non-square matrices.

Final Logic: Since determinant exists only for square matrices, the mapping becomes undefined when m ≠ n.

🧠 Memory Trick

"Determinant Demands Square."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinant Definition and Square Matrix Condition

2 Match the specific matrix orders and structures with their rigorous determinant evaluation characteristics.

List I	List II
1. Matrix Row 1: (x, y), Row 2: (w, z)	a. Evaluation requires computing 3 distinct terms of 2x2 minors.
2. Matrix Row 1: (a)	b. Determinant operation is strictly not possible.
3. Matrix Row 1: (a, b, c), Row 2: (d, e, f), Row 3: (g, h, i)	c. Evaluates instantly to a single term without multiplication.
4. Matrix Row 1: (x, y, z), Row 2: (u, v, w)	d. Evaluates mathematically to exactly xz - yw.

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

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

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

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

💡 Short Explanation

�� 2×2 determinants use ad − bc. �� 1×1 determinant equals the single element. �� Rectangular matrices have no determinant.

✅ Correct Answer Explanation

Matrix (1) is a 2×2 matrix, so determinant = xz − yw → matches (d). → Matrix (2) is 1×1, so determinant equals the single entry a → matches (c). → Matrix (3) is 3×3, requiring cofactor expansion with 2×2 minors → matches (a). → Matrix (4) is 2×3, not square, so determinant is undefined → matches (b). Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrectly assigns 3×3 determinant as single-term evaluation.
�� Option B → Incorrectly matches 2×2 determinant with single-term evaluation.
�� Option D → Incorrectly claims determinant exists for a 2×3 matrix.

🎯 Strategy

Used: Option Grouping

Application: Match each matrix order with its standard determinant rule systematically.

Final Logic: Correct pairing of determinant properties uniquely gives option C.

🧠 Memory Trick

"2×2 → ad − bc."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Determinants of Orders 1, 2 and 3

3 Evaluate the conceptual traits of minor M_(ij)for a determinant of order m(where m≥2). Which of the following statements hold mathematically true?
1. Deleting the i^(th)row and the j^(th)column reduces the determinant to order (m-1)×(m-1).
2. M_(ij)is obtained from the remaining elements while preserving their relative positions.
3. M_(ij)inherently includes the algebraic sign factor (-1)^(i+j).
4. For a 3 × 3 determinant, every minor M_(ij)is evaluated by finding the determinant of the remaining 2 × 2 matrix.

A 1, 2, 3

B 1, 2, 4

C 3, 4

D 1, 2, 3, 4

💡 Short Explanation

�� A minor is obtained by deleting one row and one column. �� The remaining elements retain their relative positions. �� The sign factor belongs to the cofactor, not the minor.

✅ Correct Answer Explanation

The minor M_(ij)of an element is the determinant obtained after deleting the i^(th)row and the j^(th)column. → After deleting one row and one column from an m×m determinant, the remaining determinant is of order (m-1)×(m-1). Therefore, Statement 1 is correct. → While forming the minor, the remaining elements retain their original relative positions in the reduced determinant. Hence, Statement 2 is correct. → The algebraic sign factor (-1)^(i+j)is not a part of the minor. It is used only while calculating the cofactor: Aᵢⱼ = (-1)^(i+j)Mᵢⱼ Therefore, Statement 3 is incorrect. → In a 3 × 3 determinant, deleting one row and one column leaves a 2 × 2 determinant, which is evaluated using: ad − bc Hence, Statement 4 is correct. → Therefore, Statements 1, 2, and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect because Statement 3 is false; the sign factor belongs to the cofactor.
�� Option C → Incorrect because Statement 3 is false.
�� Option D → Incorrect because it includes Statement 3, which is not true.

🎯 Strategy

Used

Elimination

Application:

�� Eliminate every option containing Statement 3 because the algebraic sign factor is associated with cofactors, not minors.

Final Logic:

�� A minor is only the reduced determinant obtained after deleting one row and one column; the sign factor is introduced only while forming the cofactor.

🧠 Memory Trick

"Minor = Delete; Cofactor = Delete + Sign."

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

4 In the rigorous algorithmic process of minor calculation, which statement is mathematically INCORRECT?

A Evaluating the minor of an element in a 3x3 matrix leaves a precise 2x2 determinant to be solved.

B The algebraic sign of the parent element a_ij itself actively determines the final sign of the minor M_ij.

C If minor M_32 is computed, the elements occupying the 3rd row and 2nd column are entirely ignored.

D The minor of a 2x2 matrix element is simply the diagonally opposite remaining element.

💡 Short Explanation

�� Minors do not depend on algebraic sign factors. �� Sign factor belongs to cofactors. �� 3×3 minors reduce to 2×2 determinants.

✅ Correct Answer Explanation

A minor is simply the determinant obtained after deleting the relevant row and column. → The sign factor (−1)^(i+j) is associated with cofactors, not minors. Hence, option B is mathematically incorrect. → Option A is correct because deleting one row and column from a 3×3 matrix leaves a 2×2 determinant. → Option C is correct because M₃₂ ignores the 3rd row and 2nd column. → Option D is correct because in a 2×2 matrix, deleting one row and one column leaves a single remaining element.

❓ Why Other Options Are Incorrect

�� Option A → Correctly describes reduction from 3×3 to 2×2 determinant.
�� Option C → Properly applies the deletion rule for minors.
�� Option D → Correctly explains minor evaluation in 2×2 matrices.

🎯 Strategy

Used: Odd One Out

Application: Identify the statement confusing minor with cofactor properties.

Final Logic: Only option B incorrectly associates sign factor with minors.

🧠 Memory Trick

"Minor no sign, Cofactor sign."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Minor and Cofactor

5 In a complex mixture of determinant properties, you are given the minors M_12 = -46 and M_22 = -19 for a 3x3 square matrix. What is the computed algebraic sum of their cofactors A_12 + A_22?

A 65

B -65

C 27

D -27

💡 Short Explanation

�� Cofactor formula is A_ij = (−1)^(i+j) M_ij. �� A_12 changes sign because 1+2 is odd. �� A_22 retains sign because 2+2 is even.

✅ Correct Answer Explanation

Cofactor formula: A_(ij)=(-1)^(i+j)M_(ij) → For M₁₂ = −46: A₁₂ = (−1)^(1+2)(−46) = (−1)(−46) = 46 → For M₂₂ = −19: A₂₂ = (−1)^(2+2)(−19) = (+1)(−19) = −19 → Therefore, A₁₂ + A₂₂ = 46 + (−19) = 27 Hence, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Obtained from incorrect addition without sign consideration.
�� Option B → Incorrect handling of cofactor signs.
�� Option D → Uses wrong sign assignment for A₁₂.

🎯 Strategy

Used: Substitution

Application: Substitute minor values directly into cofactor formula with parity-based signs.

Final Logic: Correct sign application gives 46 − 19 = 27.

🧠 Memory Trick

"Odd → negative sign, Even → positive sign."

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

6 Under the rigid index constraint region where i != j, what firmly defines the algebraic relationship between the minor M_ij and cofactor A_ij if the specific sum (i+j) evaluates to an odd integer?

A A_ij = M_ij

B A_ij = - M_ij

C A_ij = M_ij / 2

D A_ij = 0

💡 Short Explanation

�� Cofactor formula depends on parity of i+j. �� Odd sum gives negative sign. �� Therefore cofactor becomes negative of minor.

✅ Correct Answer Explanation

Cofactor is defined as: A_(ij)=(-1)^(i+j)M_(ij) → If (i+j) is odd, then: (−1)^(odd) = −1 → Therefore: A_ij = −M_ij Hence, option B is correct. → Option A is true only when i+j is even. → Option C has no determinant property support. → Option D is incorrect because cofactors are not necessarily zero.

❓ Why Other Options Are Incorrect

�� Option A → Valid only for even parity positions.
�� Option C → Cofactor is never defined as half the minor.
�� Option D → Cofactor values depend on minors, not automatic zero.

🎯 Strategy

Used: Substitution

Application: Apply parity condition directly into cofactor sign formula.

Final Logic: Odd exponent gives negative sign, so cofactor equals negative minor.

🧠 Memory Trick

"Odd index sum → opposite sign."

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

7 Evaluate the EMI (Equated Matrix Integer) derived from equating two order 2 determinants:
| Row 1: (2x, 4), Row 2: (6, x) | = | Row 1: (2, 4), Row 2: (6, 3) |.
What are the mathematically possible values for x?

A +/- 3

B +/- square root of 3

C +/- 1

D 0

💡 Short Explanation

�� Evaluate both determinants separately. �� Form an equation in x. �� Solve quadratic equation.

✅ Correct Answer Explanation

Left determinant: ∣[2x, 4; 6, x]∣=2x^2-24 → Right determinant: ∣[2, 4; 6, 3]∣=6-24=-18 → Equating: 2x² − 24 = −18 → 2x² = 6 → x² = 3 → x = ±√3 Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → ±3 gives determinant mismatch.
�� Option C → ±1 does not satisfy the equation.
�� Option D → x = 0 gives determinant −24, not −18.

🎯 Strategy

Used: Substitution

Application: Compute determinants using ad − bc and solve resulting equation.

Final Logic: Solving 2x² − 24 = −18 gives x = ±√3.

🧠 Memory Trick

"2×2 determinant → ad − bc."

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

8 When accurately computing an order 3 determinant via moving expansion strictly along the second row, the correct structural mathematical formula is:

A a21A21 + a22A22 + a23A23

B a21M21 - a22M22 + a23M23

C -a21M21 + a22M22 - a23*M23

D Both A and C are mathematically equivalent.

💡 Short Explanation

�� Determinant expansion uses cofactors. �� Cofactor signs in second row are − + −. �� Cofactor form and signed minor form are equivalent.

✅ Correct Answer Explanation

Expansion along second row is: a_(21)A_(21)+a_(22)A_(22)+a_(23)A_(23) → Since cofactors already include signs: A₂₁ = −M₂₁, A₂₂ = M₂₂, A₂₃ = −M₂₃ → Therefore: -a_(21)M_(21)+a_(22)M_(22)-a_(23)M_(23) → Thus, options A and C are mathematically equivalent. Hence, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct but incomplete because option D includes its equivalence.
�� Option B → Incorrect sign pattern for second-row expansion.
�� Option C → Correct expression but not the most complete option.

🎯 Strategy

Used: Option Grouping

Application: Compare cofactor expansion with signed minor expansion.

Final Logic: Both expressions represent the same determinant expansion.

🧠 Memory Trick

"Second row signs: − + −."

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

9 If all rows and columns of determinant |A| are completely transposed to form |A'|, what is the exact mathematical probability that the equation |A| - |A'| = 0 holds true?

A 100%

B 50%

C 0%

D Depends conditionally on the matrix order

💡 Short Explanation

�� Determinant remains unchanged under transpose. �� |A| = |A'| always holds. �� Hence difference becomes zero.

✅ Correct Answer Explanation

One fundamental determinant property states: ∣A∣=∣A^'∣ → Therefore: |A| − |A′| = 0 always. → Hence the probability is 100%, making option A correct. → Option B and C are incorrect because there is no uncertainty in this property. → Option D is incorrect because the transpose property holds for all square matrices irrespective of order.

❓ Why Other Options Are Incorrect

�� Option B → Determinant equality under transpose is not conditional.
�� Option C → Difference is never nonzero for transpose property.
�� Option D → Matrix order does not affect transpose determinant equality.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recognize the universal transpose property of determinants.

Final Logic: Since |A| always equals |A′|, the equation always evaluates to zero.

🧠 Memory Trick

"Transpose keeps determinant same."

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

10 If three vector rows sequentially shift via two consecutive distinct row interchanges (R1 swapped with R2, then subsequently R2 swapped with R3), what is the final algebraic sign of the determinant compared to its original state?

A It reverses its sign once.

B It reverts completely to its original sign.

C It permanently becomes zero.

D It becomes squared.

💡 Short Explanation

�� One row interchange changes sign. �� Two interchanges multiply determinant by (−1)². �� Final sign becomes original sign.

✅ Correct Answer Explanation

Interchanging two rows multiplies determinant by −1. → First interchange: determinant becomes −|A|. → Second interchange: determinant becomes (−1)(−|A|) = |A| → Thus after two swaps, the determinant regains its original sign. Hence, option B is correct. → Option A is incorrect because one sign reversal is canceled by the second interchange. → Option C is incorrect because row interchange alone does not force determinant to zero. → Option D is incorrect because determinants are not squared under swapping.

❓ Why Other Options Are Incorrect

�� Option A → True only for a single interchange.
�� Option C → Zero determinant occurs for dependent rows, not swapping.
�� Option D → No squaring property exists for row interchange.

🎯 Strategy

Used: Substitution

Application: Apply the sign-change rule twice sequentially.

Final Logic: Two negative sign multiplications restore the original sign.

🧠 Memory Trick

"Two swaps → sign snaps back."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Row Interchange Property

11 If the Area of a geometric triangle formed by coordinates is calculated as exactly 0 using the standard determinant formula (1/2)|A|, what structural definition immediately applies to matrix A?

A Identity Matrix

B Non-Singular Matrix

C Singular Matrix

D Adjoint Matrix

💡 Short Explanation

Zero area implies determinant equals zero. A matrix with determinant 0 is singular. Singular matrices are non-invertible.

✅ Correct Answer Explanation

The area of a triangle using determinants is given by: Area=1/2∣A∣ If the area equals zero, then: ∣A∣=0 A matrix whose determinant is zero is called a singular matrix. Option C is correct because determinant zero directly defines singularity. Option A is incorrect because identity matrices have determinant 1. Option B is incorrect because non-singular matrices require non-zero determinant. Option D is incorrect because adjoint matrices are unrelated to determinant zero classification.

❓ Why Other Options Are Incorrect

Option A → Identity matrices always have determinant equal to 1.
Option B → Non-singular matrices satisfy ∣A∣≠0.
Option D → Adjoint refers to a matrix operation, not determinant classification.

🎯 Strategy

Used: Elimination

Application: The condition "Area = 0" immediately implies determinant zero, eliminating all non-singular possibilities.

Final Logic: ∣A∣=0⇒matrix is singular.

🧠 Memory Trick

"Zero area → zero determinant → singular."

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

12 For a system to possess an integral unique solution via matrix inversion, it must act as a Non-Singular matrix. If A = | Row 1: (x, 2), Row 2: (1, 2) | is strictly Non-Singular, which condition must x confidently satisfy?

A x = 1

B x != 1

C x = 0

D x != 0

💡 Short Explanation

Non-singular means determinant is non-zero. Compute determinant of the matrix. Solve condition ∣A∣≠0.

✅ Correct Answer Explanation

For the matrix A=∣[x, 2; 1, 2]∣ the determinant is: ∣A∣=2x-2 For A to be non-singular: 2x-2≠0x≠1 Option B is correct. Option A makes the determinant zero. Option C is allowed because determinant becomes -2. Option D is unnecessary since x=0 still gives non-zero determinant.

❓ Why Other Options Are Incorrect

Option A → x=1 gives determinant zero, making the matrix singular.
Option C → x=0 still produces a valid non-zero determinant.
Option D → Non-zero x is not required; only determinant non-zero matters.

🎯 Strategy

Used: Substitution

Application: Evaluate determinant and apply the non-singular condition directly.

Final Logic: 2x-2≠0⇒x≠1.

🧠 Memory Trick

"Non-singular ⇒ determinant never zero."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Condition for Invertibility

13 If two columns of a determinant are perfectly identical, applying the column operation C1 -> C1 - C2 mathematically reduces the entire determinant to zero based fundamentally on which underlying law?

A Identical Row/Column Rule

B Zero Row/Column Rule

C Proportionality Rule

D Switching Property

💡 Short Explanation

Identical columns become a zero column after subtraction. Determinant with a zero column equals zero. Operation uses zero row/column property.

✅ Correct Answer Explanation

If two columns are identical and we apply: C_1→C_1-C_2 then every entry in C_1 becomes zero. A determinant containing an entire zero row or column evaluates to zero. Thus, the result fundamentally relies on the Zero Row/Column Rule. Option B is correct. Option A identifies the original condition but not the operational rule used after subtraction. Option C concerns proportional rows/columns. Option D changes sign only during interchange.

❓ Why Other Options Are Incorrect

Option A → Identical columns imply zero determinant, but the operation specifically creates a zero column.
Option C → Proportionality is not directly applied here.
Option D → Switching rows/columns only changes sign.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: The operation explicitly creates a zero column, pointing directly to the zero-column property.

Final Logic: Zero column formed ⇒ determinant becomes zero.

🧠 Memory Trick

"Zero row or column kills determinant."

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

14 If all elements comprising the principal diagonal and the entire upper triangular region of a 3x3 matrix are zeroes, what does the determinant evaluate to?

A The product of diagonal elements (which evaluates to 0)

B 1

C -1

D Cannot be determined without knowing the lower elements

💡 Short Explanation

Matrix becomes triangular. Determinant equals product of diagonal entries. Zero diagonal entries force determinant zero.

✅ Correct Answer Explanation

For triangular matrices, the determinant equals the product of the principal diagonal elements. Since all diagonal elements are zero: ∣A∣=0×0×0=0 Option A is correct. Option B and Option C are impossible because determinant is clearly zero. Option D is incorrect because lower triangular entries do not affect this conclusion.

❓ Why Other Options Are Incorrect

Option B → Determinant cannot become 1 with zero diagonal entries.
Option C → No sign property produces -1 here.
Option D → Triangular determinant depends only on diagonal elements.

🎯 Strategy

Used: Elimination

Application: Recognize triangular determinant property immediately.

Final Logic: Zero diagonal product ⇒ determinant zero.

🧠 Memory Trick

"Triangular matrix ⇒ multiply diagonals."

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

15 If R1 = 5 * R3 inside an order 3 determinant, the determinant computes to zero by the Proportionality Rule. This mathematical occurrence is conceptually identical to:

A Factoring out the scalar 5 from R1, leaving R1 perfectly identical to R3.

B Adding the scalar 5 to R3.

C Multiplying the entire determinant by 5.

D Transposing the entire matrix structure.

💡 Short Explanation

Proportional rows are linearly dependent. Factoring out scalar makes rows identical. Identical rows force determinant zero.

✅ Correct Answer Explanation

If R_1=5R_3 then factor 5 from R_1: ∣A∣=5×(determinant with R_1=R_3) Now two rows become identical, making the determinant zero. Option A is correct. Option B changes row values incorrectly. Option C ignores row dependence. Option D does not affect proportionality.

❓ Why Other Options Are Incorrect

Option B → Adding a scalar to a row is not determinant proportionality.
Option C → Determinant is not merely multiplied by 5 here.
Option D → Transposition preserves determinant value but does not explain zero result.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret proportional rows through scalar factoring.

Final Logic: Proportional rows ⇒ identical rows after factoring ⇒ determinant zero.

🧠 Memory Trick

"Proportional rows become same rows."

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

16 Applying the transformation R2 -> R2 + kR1 to a determinant D produces a new determinant D'. Which mathematical statement is strictly true regarding their relationship?

A D' = k*D

B D' = D + k

C D' = D

D D' = D / k

💡 Short Explanation

Adding a multiple of one row to another preserves determinant. Elementary row replacement does not change value. Determinant remains invariant.

✅ Correct Answer Explanation

One of the key determinant properties states: Replacing a row by itself plus a multiple of another row leaves the determinant unchanged. Thus: R_2→R_2+kR_1 implies: D^'=D Option C is correct. Option A incorrectly assumes multiplication scaling. Option B is not a determinant property. Option D has no mathematical basis.

❓ Why Other Options Are Incorrect

Option A → Scaling occurs only when a whole row is multiplied.
Option B → Determinants are not altered additively.
Option D → No division property applies.

🎯 Strategy

Used: Substitution

Application: Apply the standard row operation rule directly.

Final Logic: Row replacement operation leaves determinant unchanged.

🧠 Memory Trick

"Row replace, determinant stays."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Elementary Transformations

17 If a square matrix A is of order n=4 and its determinant |A| = 3, what is the exact evaluated value of the determinant |2A| using scalar multiple properties?

A 6

B 24

C 48

D 81

💡 Short Explanation

Determinant scales by k^n. Here k=2 and n=4. Multiply 2^4 with determinant 3.

✅ Correct Answer Explanation

For an n×n matrix: ∣kA∣=k^n∣A∣ Given: n=4,k=2,∣A∣=3 So: ∣2A∣=2^4×3=16×3=48 Option C is correct. Option A incorrectly scales linearly. Option B partially scales. Option D incorrectly cubes the determinant.

❓ Why Other Options Are Incorrect

Option A → Uses only single multiplication by 2.
Option B → Uses incomplete power scaling.
Option D → No cubic relation exists.

🎯 Strategy

Used: Substitution

Application: Apply the determinant scaling formula k^n∣A∣.

Final Logic: 2^4×3=48.

🧠 Memory Trick

"Order n ⇒ scale by kⁿ."

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

18 If A and B are square matrices of the exact same order where |A| = -4 and |AB| = 36, calculate the precise determinant of |B'| (the transpose of matrix B).

A -9

B 9

C -144

D 1/9

💡 Short Explanation

Use ∣AB∣=∣A∣∣B∣. Transpose preserves determinant. Solve for ∣B∣.

✅ Correct Answer Explanation

Given: ∣AB∣=∣A∣∣B∣36=(-4)∣B∣∣B∣=-9 Also: ∣B^'∣=∣B∣ Therefore: ∣B^'∣=-9 Option A is correct. Option B ignores the negative sign. Option C multiplies incorrectly. Option D incorrectly divides.

❓ Why Other Options Are Incorrect

Option B → Determinant sign is lost incorrectly.
Option C → Product rule misapplied.
Option D → Reciprocal rule does not apply.

🎯 Strategy

Used: Substitution

Application: Use determinant product and transpose properties sequentially.

Final Logic: ∣B∣=36/(-4)=-9, and transpose preserves determinant.

🧠 Memory Trick

"Transpose keeps determinant same."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Product and Transpose Properties

📄 Passage

The advanced matrix splitting technique, heavily grounded in the Summation Property, allows complex matrices to be systematically broken down into simpler, additive variants. If some or all elements of a column of a determinant are mathematically expressed as a sum of two terms, the entire determinant can be explicitly expressed as a standard sum of two independent determinants. This critical mathematical operation strictly preserves all columns that do not possess the summed terms exactly as they were in the original structure, ensuring the total numerical equality remains securely invariant.

Question:
If a 3x3 determinant possesses binomials (x+a, y+b, z+c) strictly located in Column 2, and trinomials (p+q+r, s+t+u, v+w+x) strictly in Column 3, how many simple (non-summed) independent determinants can this matrix eventually be split into?

A 2

B 3

C 5

D 6

💡 Short Explanation

Binomial column splits into 2 determinants. Trinomial column splits into 3 determinants. Total splits multiply: 2×3=6.

✅ Correct Answer Explanation

The determinant summation property allows splitting along summed columns. Column 2 has binomials ⇒ 2 determinant components. Column 3 has trinomials ⇒ 3 determinant components. Total independent determinants: 2×3=6 Option D is correct. Option A ignores one column split. Option B considers only the trinomial split. Option C incorrectly adds instead of multiplying.

❓ Why Other Options Are Incorrect

Option A → Only one split considered.
Option B → Ignores binomial contribution.
Option C → Splits combine multiplicatively, not additively.

🎯 Strategy

Used: Option Grouping

Application: Treat each column split independently and multiply total combinations.

Final Logic: 2×3=6 determinants.

🧠 Memory Trick

"Independent splits multiply."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Sum/Splitting Property

📄 Passage

Question:
When the Summation Property splits a determinant along a specific summed row, the strict preservation of mathematical equality relies completely on which of the following maintaining identical placement?

A The principal diagonal elements only.

B The non-summed elements located in the alternate rows.

C The cofactors of the summed elements.

D The scalar multipliers applied post-split.

💡 Short Explanation

Only summed row/column changes. Remaining rows/columns stay unchanged. Equality depends on structural preservation.

✅ Correct Answer Explanation

In determinant splitting: The row/column containing sums is separated. All other rows and columns remain exactly identical in every split determinant. This preservation ensures algebraic equality. Option B is correct. Option A is too restrictive. Option C concerns expansion, not splitting. Option D is unrelated.

❓ Why Other Options Are Incorrect

Option A → Equality depends on all non-summed elements, not only diagonal entries.
Option C → Cofactors are unrelated to structural duplication.
Option D → Scalar multipliers are not the key preservation factor.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Focus on the phrase "maintaining identical placement."

Final Logic: Non-summed rows/columns must remain unchanged across all split determinants.

🧠 Memory Trick

"Only summed row changes."

📚 Class XII – NCERT Mathematics – Chapter: Determinants – Topic: Sum Property of Determinants

CUET UG Applied Mathematics Booster Test 3 - Determinants and Their Properties

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