CUET UG Mathematics Booster Test 1 - Expansion Methods and Evaluation of Determinants -

🚀

CUET UG Mathematics Booster Test 1 - 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 1 - Expansion Methods and Evaluation of Determinants

1 / 20

QUESTION 1 OF 20

When expanding a 3×3 determinant along the first row (R₁), each element is multiplied by its minor and which sign factor?

Answer Review

1 When expanding a 3×3 determinant along the first row (R₁), each element is multiplied by its minor and which sign factor?

A (-1)^i

B (-1)^(i+j)

C (-1)^j

D Always positive

💡 Short Explanation

Cofactor sign depends on row and column positions Expansion uses minors with alternating signs Formula for sign is (-1)^(i+j)

✅ Correct Answer Explanation

While expanding a determinant, each element is multiplied by its cofactor. The cofactor includes the sign factor (-1)^(i+j). Option B correctly gives the general cofactor sign rule. Option A depends only on row number, Option C only on column number, and Option D ignores sign alternation completely.

❓ Why Other Options Are Incorrect

Option A → (-1)^i considers only row index and ignores column position, so it cannot generate the correct checkerboard sign pattern.
Option C → (-1)^j depends only on the column number and fails to account for row variation in cofactor signs.
Option D → Signs are not always positive; determinant expansion follows alternating positive and negative signs.

🎯 Strategy

Used

Elimination

Application:

Use the standard cofactor formula and eliminate options missing either row or column dependence.

Final Logic:

Cofactor signs always follow (-1)^(i+j), so Option B is correct.

🧠 Memory Trick

"Cofactor = Minor × Checkerboard Sign"

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

2 Consider the determinant
\(∣\begin{pmatrix}2 & 1 & 3\\ 4 & 5 & 6\\ 7 & 8 & 10\end{pmatrix}∣\)
Match the elements with the values of their corresponding minors.

List I	List II
1. M₁₁ (Minor of element 2)	a. 2
2. M₁₂ (Minor of element 1)	b. -2
3. M₂₃ (Minor of element 6)	c. 9
4. M₃₂ (Minor of element 8)	d. 0

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

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

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

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

💡 Short Explanation

A minor is obtained by deleting the corresponding row and column. The remaining elements form a 2 × 2 determinant. Evaluate each determinant separately to obtain the minor.

✅ Correct Answer Explanation

The minor of an element is the determinant obtained after deleting the row and column containing that element. For M₁₁ (Minor of element 2): Delete Row 1 and Column 1. Remaining determinant: | 5 6 | | 8 10 | M₁₁ = (5 × 10) − (6 × 8) = 50 − 48 = 2 Therefore, M₁₁ → a For M₁₂ (Minor of element 1): Delete Row 1 and Column 2. Remaining determinant: | 4 6 | | 7 10 | M₁₂ = (4 × 10) − (6 × 7) = 40 − 42 = -2 Therefore, M₁₂ → b For M₂₃ (Minor of element 6): Delete Row 2 and Column 3. Remaining determinant: | 2 1 | | 7 8 | M₂₃ = (2 × 8) − (1 × 7) = 16 − 7 = 9 Therefore, M₂₃ → c For M₃₂ (Minor of element 8): Delete Row 3 and Column 2. Remaining determinant: | 2 3 | | 4 6 | M₃₂ = (2 × 6) − (3 × 4) = 12 − 12 = 0 Therefore, M₃₂ → d Hence, the correct matching is: List I — List II 1. M₁₁ — a. 2 2. M₁₂ — b. -2 3. M₂₃ — c. 9 4. M₃₂ — d. 0 Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrect because M₁₂ equals -2, not 9, and M₂₃ equals 9, not -2.
Option C → Incorrect because M₁₁ is 2, not 9, and M₃₂ is 0, not -2.
Option D → Incorrect because M₁₁ is 2 (not -2), while M₂₃ is 9 (not 2).

🎯 Strategy

Used: Substitution

Application:

Delete the corresponding row and column for each element and evaluate the resulting 2 × 2 determinant using the formula:
ad − bc

Final Logic:

Each minor is obtained independently by evaluating the corresponding 2 × 2 determinant, giving the matching:
M₁₁ = 2,
M₁₂ = -2,
M₂₃ = 9,
M₃₂ = 0.
Hence, Option A is correct.

🧠 Memory Trick

"Delete Row + Delete Column → Solve 2 × 2"

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

3 Which of the following expressions correctly represent terms in the expansion of a determinant along the second row (R₂)?
I. \(-a_{21}(a_{12}a_{33}-a_{32}a_{13})\)
II. \(+a_{22}(a_{11}a_{33}-a_{31}a_{13})\)
III. \(-a_{23}(a_{11}a_{32}-a_{31}a_{12})\)

A I and II only

B II and III only

C I and III only

D I, II and III

💡 Short Explanation

Second-row expansion uses signs (- + -) All three expressions follow cofactor expansion Each term has correct minor structure

✅ Correct Answer Explanation

Expansion along the second row follows: \(-a_{21}M_{21}+a_{22}M_{22}-a_{23}M_{23}\) All three expressions correctly represent the required minors and corresponding signs. Hence Options I, II, and III are all valid expansion terms. Therefore Option D is correct, while other options omit valid terms.

❓ Why Other Options Are Incorrect

Option A → Excludes III, although the third term correctly follows the cofactor sign pattern.
Option B → Excludes I, even though the first term correctly carries a negative sign.
Option C → Omits II, despite the middle cofactor always being positive in second-row expansion.

🎯 Strategy

Used

Option Grouping

Application:

Check each term individually using second-row sign pattern (- + -).

Final Logic:

Since all three satisfy determinant expansion rules, Option D is correct.

🧠 Memory Trick

"Second row → Minus, Plus, Minus"

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

4 Which statement is incorrect regarding cofactors of elements in the second row?

A \(A_{21}=+M_{21}\)

B \(A_{22}=+M_{22}\)

C \(A_{23}=-M_{23}\)

D Cofactor sign depends on \({\left(-1\right)}^{\left(i,\ j\right)}\)

💡 Short Explanation

Cofactor equals sign × minor \(A_{21}=(-1)^{2+1}M_{21}\) Hence sign becomes negative

✅ Correct Answer Explanation

Cofactor formula is: \(A_{ij}=(-1)^{i+j}M_{ij}\) For \(A_{21}\): \(\left(-1)^{2+1}=-1\right.\) So, \(A_{21}=-M_{21}\) Hence Option A is incorrect. Option B and C follow proper sign rules, while Option D correctly states the general cofactor principle.

❓ Why Other Options Are Incorrect

Option B → Correct because \(\left(-1)^{2+2}=+1\right.\), so \(A_{22}=+M_{22}\).
Option C → Correct because \(\left(-1)^{2+3}=-1\right.\), giving \(A_{23}=-M_{23}\).
Option D → Correctly states the standard cofactor sign formula used in determinant expansion.

🎯 Strategy

Used

Elimination

Application:

Apply the checkerboard sign rule to each cofactor expression.

Final Logic:

Only Option A violates the cofactor sign pattern.

🧠 Memory Trick

"Second row signs: − + −"

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

5 Find the cofactor of the element 7 in the determinant:
\(∣\begin{pmatrix}2 & 3 & 5\\ 6 & 0 & 4\\ 1 & 5 & 7\end{pmatrix}∣\)

A 12

B -12

C 22

D 0

💡 Short Explanation

Element 7 is at position (3,3) Minor determinant equals -18 Cofactor sign here is positive

✅ Correct Answer Explanation

Deleting row 3 and column 3 gives: \(∣\begin{pmatrix}2 & 3\\ 6 & 0\end{pmatrix}∣=(2)(0)-(6)(3)=-18\) Since \(\left(-1)^{3+3}=+1\right.\), cofactor equals \(-18\). Thus the provided answer B) -12 is incorrect. Actual Correct Answer: \(-18\)(Not available in options)

❓ Why Other Options Are Incorrect

Option A → Minor calculation does not give positive 12.
Option B → Cofactor value is not \(-12\); determinant evaluation gives \(-18\).
Option C → No correct determinant computation leads to 22.
Option D → Minor determinant is nonzero.

🎯 Strategy

Used

Substitution

Application:

Delete corresponding row and column, then evaluate the resulting 2×2 determinant.

Final Logic:

Minor equals \(-18\) and cofactor sign remains positive.

🧠 Memory Trick

"Delete row-column, then 2×2 solve"

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

6 The points \(\left(a,b+c\right)\), \(\left(b,c+a\right)\), and \(\left(c,a+b\right)\)form a triangle. Using determinant method, the area of this triangle is:

A 1

B a+b+c

C 0 (Points are collinear)

D 2

💡 Short Explanation

Triangle area uses determinant formula Determinant simplifies to zero Zero area means collinear points

✅ Correct Answer Explanation

Using area formula: \(\frac{1}{2}∣\begin{pmatrix}a & b+c & 1\\ b & c+a & 1\\ c & a+b & 1\end{pmatrix}∣\) On simplification, determinant becomes zero. Therefore area is zero, meaning all three points lie on the same straight line. Hence Option C is correct. Other options imply nonzero area and are incorrect.

❓ Why Other Options Are Incorrect

Option A → Area cannot be constant 1 for arbitrary values of \(a,b,c\).
Option B → Determinant simplification does not produce \(a+b+c\).
Option D → Area evaluates to zero, not 2.

🎯 Strategy

Used

Elimination

Application:

Use determinant area formula and check whether rows become linearly dependent.

Final Logic:

Zero determinant implies collinearity and zero area.

🧠 Memory Trick

"Zero determinant → Collinear points"

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

7 Evaluate the determinant by expanding along the first column (C₁):
\(∣\begin{pmatrix}1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 6 & 7\end{pmatrix}∣\)

A 4

B 5

C 6

D -2

💡 Short Explanation

First column contains two zeros Expansion becomes very simple Result obtained from one minor only

✅ Correct Answer Explanation

Expanding along first column: \(1\times ∣\begin{pmatrix}4 & 5\\ 6 & 7\end{pmatrix}∣=1(28-30)=-2\) Only one nonzero term contributes. Therefore Option D is correct. Other options do not satisfy determinant calculation.

❓ Why Other Options Are Incorrect

Option A → Incorrect determinant evaluation; actual value is negative.
Option B → Does not arise from the 2×2 determinant computation.
Option C → Sign and multiplication are incorrectly handled.

🎯 Strategy

Used

Elimination

Application:

Choose the row or column with maximum zeros to simplify evaluation.

Final Logic:

Single surviving term gives determinant \(-2\).

🧠 Memory Trick

"More zeros = faster determinant"

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

8 Evaluate the determinant by expanding along the first column:
\(∣\begin{pmatrix}0 & 2 & 3\\ 1 & 4 & 5\\ 0 & 6 & 7\end{pmatrix}∣\)

A -4

B 4

C 2

D 0

💡 Short Explanation

Only middle element contributes in first column Apply cofactor sign carefully Determinant equals 4, not -4

✅ Correct Answer Explanation

Expansion along first column gives: \((-1)^{2+1}(1)∣\begin{pmatrix}2 & 3\\ 6 & 7\end{pmatrix}∣=-1(14-18)=4\) Hence determinant equals \(4\). Therefore the provided answer A) -4 is incorrect. Actual Correct Answer: B) 4

❓ Why Other Options Are Incorrect

Option A → Sign handling is incorrect; final determinant becomes positive 4.
Option C → Minor determinant does not evaluate to 2.
Option D → Matrix determinant is nonzero because rows are independent.

🎯 Strategy

Used

Substitution

Application:

Expand using the only nonzero entry in the first column.

Final Logic:

Negative cofactor sign converts \(-4\) minor into \(+4\).

🧠 Memory Trick

"Middle first-column term carries minus sign"

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

9 Find the sign factor of the element in position \(\left(1,\ 2\right)\)during expansion.

A +1

B -1

C 0

D 2

💡 Short Explanation

Cofactor sign uses (-1)^(i+j) Here i+j = 1+2 = 3 (-1)^3 = -1

✅ Correct Answer Explanation

For position \(\left(1,\ 2\right)\), \(\left(-1)^{1+2}=(-1)^{3}=-1\right.\) Hence the sign factor is negative. Therefore Option B is correct. Option A ignores odd parity, while Options C and D are unrelated to determinant sign rules.

❓ Why Other Options Are Incorrect

Option A → Positive sign occurs only when \(i+j\) is even.
Option C → Sign factors are never zero in cofactor expansion.
Option D → Cofactor sign can only be \(+1\) or \(-1\), not 2.

🎯 Strategy

Used

Substitution

Application:

Substitute row and column values directly into \({\left(-1\right)}^{i+j}\).

Final Logic:

Odd exponent gives negative sign.

🧠 Memory Trick

"Odd sum → negative sign"

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

10 Find the value of \(k\) if the area of the triangle with vertices \(\left(2,-6\right)\), \(\left(5,\ 4\right)\), \(\left(k,\ 4\right)\)is 35 sq units.

A 12, -2

B -2, -12

C 12, 2

D -12, 2

💡 Short Explanation

Use determinant area formula Base points share same y-coordinate Solve modulus equation for \(k\)

✅ Correct Answer Explanation

Area formula: \(\frac{1}{2}∣2(4-4)+5(4+6)+k(-6-4)∣=35\frac{1}{2}∣50-10k∣=35∣50-10k∣=7050-10k=\pm 70\) Thus: \(k=-2 or 12\) Therefore the correct option should be A) 12, -2. The provided answer C) 12, 2 is incorrect.

❓ Why Other Options Are Incorrect

Option B → \(-12\) does not satisfy the area equation.
Option C → \(2\) gives area 15, not 35.
Option D → Neither value satisfies the determinant condition simultaneously.

🎯 Strategy

Used

Substitution

Application:

Apply triangle area determinant formula and solve absolute value equation.

Final Logic:

Only \(k=12\) and \(k=-2\) satisfy the required area.

🧠 Memory Trick

"Area with same y-values → horizontal base shortcut"

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

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

A 2

B 0

C -1

D 1

💡 Short Explanation

Integral with same limits equals zero Entire third column becomes zero Determinant with zero column is zero

✅ Correct Answer Explanation

Each integral has identical upper and lower limits: \(\int_{a}^{a}\,f(x) dx=0\) Thus the third column becomes: \(\left[\begin{aligned}0\\ 0\\ 0\end{aligned}\right]\) A determinant containing a complete zero column always evaluates to zero. Hence Option B is correct. All other numerical options ignore the zero-column property of determinants.

❓ Why Other Options Are Incorrect

Option A → Determinant cannot become 2 because one full column contains only zeros.
Option C → Negative value is impossible when determinant directly collapses to zero.
Option D → Determinant with an entire zero column is always zero, not one.

🎯 Strategy

Used

Elimination

Application:

Identify the zero-valued integrals first before evaluating the determinant.

Final Logic:

A determinant with a complete zero column must equal zero.

🧠 Memory Trick

"Same limits → integral zero"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Determinants with Zero Row/Column

12 Assertion (A): Expanding along the third column always yields a zero determinant.
Reason (R): Every column in a zero matrix evaluates to zero.

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Expansion along any column need not give zero Zero determinant depends on entries Reason statement about zero matrix is correct

✅ Correct Answer Explanation

Assertion is false because expanding along the third column does not always produce zero. Determinant value depends on matrix entries, not on the chosen expansion column. The Reason is true since every column of a zero matrix contains only zeros, making its determinant zero. Therefore Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Reason statement is mathematically correct, so both cannot be false.
Option B → Assertion is not true because determinants are not always zero when expanded along the third column.
Option C → Although the Reason is true, it does not justify the incorrect Assertion.

🎯 Strategy

Used

Elimination

Application:

Test the assertion using simple nonzero matrices and verify the reason independently.

Final Logic:

Assertion is false, but the Reason remains mathematically true.

🧠 Memory Trick

"Column choice changes method, not determinant value"

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

13 Arrange the following matrices in increasing order of evaluation effort (least to most), based on placement of zeros:
1. A 3×3 matrix with no zeros
2. A 3×3 matrix with an entire column of zeros
3. A 3×3 matrix with two zeros in a row

A 1, 3, 2

B 2, 3, 1

C 3, 2, 1

D 2, 1, 3

💡 Short Explanation

More zeros reduce computation effort Entire zero column is easiest No-zero matrix requires maximum work

✅ Correct Answer Explanation

A determinant with an entire zero column immediately becomes zero, requiring the least effort. A row containing two zeros simplifies expansion significantly. A matrix with no zeros requires full determinant expansion and maximum calculations. Thus increasing effort order is: \(2\rightarrow 3\rightarrow 1\) Hence Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Places the hardest matrix first instead of last.
Option C → A matrix with two zeros still requires more effort than a complete zero column.
Option D → A matrix without zeros cannot require less effort than one with two zeros.

🎯 Strategy

Used

Option Grouping

Application:

Compare determinant simplification based on number and placement of zeros.

Final Logic:

More zeros imply fewer calculations and lower evaluation effort.

🧠 Memory Trick

"More zeros = less work"

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

📄 Passage

To every square matrix \(A=[a_{ij}]\)of order \(n\), we can associate a number (real or complex) called the determinant of \(A\). This can be viewed as a function \(f:M\rightarrow K\), where \(M\) is the set of square matrices and \(K\) is the set of numbers (real or complex), defined by \(f(A)=k\).

According to the passage, which of the following matrices has a determinant defined?

A A rectangular matrix

B A row matrix

C A square matrix

D A column matrix

💡 Short Explanation

Determinants exist only for square matrices Equal rows and columns are necessary Rectangular matrices lack determinant definition

✅ Correct Answer Explanation

A determinant is defined only for square matrices because determinant evaluation requires the same number of rows and columns. The passage explicitly states that determinants are associated with square matrices. Therefore Option C is correct. Row, column, and rectangular matrices generally do not satisfy the square condition.

❓ Why Other Options Are Incorrect

Option A → Rectangular matrices have unequal rows and columns, so determinants are undefined.
Option B → A row matrix is usually \(1\times n\), not necessarily square.
Option D → A column matrix is generally \(n\times 1\), hence not square.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the exact wording of the passage to identify the matrix type mentioned.

Final Logic:

The passage explicitly defines determinants only for square matrices.

🧠 Memory Trick

"Determinant needs a square shape"

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

📄 Passage

In the function \(f:M\rightarrow K\), what does the set \(K\) represent?

A Only positive integers

B Only determinants of non-singular matrices

C Only real numbers

D Set of numbers (real or complex)

💡 Short Explanation

Determinants may be real or complex K represents codomain of determinant function Passage directly defines K

✅ Correct Answer Explanation

The passage defines: \(f:M\rightarrow K\) where \(K\) is the set of numbers, either real or complex. Since determinants can belong to both categories, Option D is correct. Options A, B, and C restrict the determinant range unnecessarily and contradict the given definition.

❓ Why Other Options Are Incorrect

Option A → Determinants are not restricted to positive integers.
Option B → Singular matrices also possess determinants, though possibly zero.
Option C → Determinants may also be complex numbers, not only real numbers.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Read the codomain definition directly from the passage.

Final Logic:

The passage explicitly states that K contains real or complex numbers.

🧠 Memory Trick

"K = Known number set"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Determinant as a Function

16 For a 2×2 matrix
\(A=\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right),\)
the determinant is given by:

A \(ad+bc\)

B \(ab-cd\)

C \(ad-bc\)

D \(ac-bd\)

💡 Short Explanation

Multiply principal diagonal terms Subtract secondary diagonal product Standard 2×2 determinant formula

✅ Correct Answer Explanation

For matrix \(\left(\begin{pmatrix}a & b\\ c & d\end{pmatrix}\right),\) the determinant is: \(ad-bc\) The product of the main diagonal is subtracted by the product of the opposite diagonal. Hence Option C is correct. Other options incorrectly arrange multiplication or signs.

❓ Why Other Options Are Incorrect

Option A → Determinant uses subtraction, not addition.
Option B → Terms are incorrectly paired across positions.
Option D → Products do not follow determinant diagonal structure.

🎯 Strategy

Used

Substitution

Application:

Recall the standard determinant formula for a 2×2 matrix.

Final Logic:

Determinant equals product of main diagonal minus other diagonal.

🧠 Memory Trick

"Forward multiply, backward subtract"

📚 Class XII – NCERT Mathematics Part 1 – Chapter: Determinants – Topic: Determinant of 2×2 Matrix

17 Find the positive value of \(x\) for which
\(∣\begin{pmatrix}3 & x\\ 1 & 4\end{pmatrix}∣=∣\begin{pmatrix}3 & 2\\ 1 & 4\end{pmatrix}∣\)

A 1

B 2

C 3

D 4

💡 Short Explanation

Equate determinant values Left determinant equals \(12-x\) Solve linear equation

✅ Correct Answer Explanation

Evaluate both determinants: \(∣\begin{pmatrix}3 & x\\ 1 & 4\end{pmatrix}∣=12-x∣\begin{pmatrix}3 & 2\\ 1 & 4\end{pmatrix}∣=12-2=10\) Equating: \(12-x=10x=2\) Thus Option B is correct. Other values do not satisfy determinant equality.

❓ Why Other Options Are Incorrect

Option A → Gives determinant 11, not 10.
Option C → Gives determinant 9, unequal to the right determinant.
Option D → Gives determinant 8, not equal to 10.

🎯 Strategy

Used

Substitution

Application:

Compute determinants separately and solve the resulting equation.

Final Logic:

Only \(x=2\) makes both determinants equal.

🧠 Memory Trick

"Equal determinants → equal expressions"

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

18 Which statement about \(∣A∣=∣B∣\)is incorrect, where \(A\) and \(B\) are 2×2 matrices?

A It implies their determinant values are equal

B It does not imply matrix \(A\) is identical to matrix \(B\)

C It means all corresponding elements must be equal

D It allows forming algebraic equations

💡 Short Explanation

Equal determinants do not mean equal matrices Different matrices may share determinant value Determinant is only a scalar quantity

✅ Correct Answer Explanation

If \(∣A∣=∣B∣\), only determinant values are equal. The actual matrix entries may differ completely. Hence Option C is incorrect. Option A correctly states determinant equality, Option B correctly distinguishes matrices from determinants, and Option D is valid because determinant equations can form algebraic relations.

❓ Why Other Options Are Incorrect

Option A → Correct because equal determinant notation directly means equal scalar values.
Option B → Correct since different matrices can share the same determinant.
Option D → Determinant equality is frequently used to form and solve equations.

🎯 Strategy

Used

Elimination

Application:

Differentiate between equality of determinants and equality of matrices.

Final Logic:

Equal determinants do not require equal corresponding elements.

🧠 Memory Trick

"Equal determinants ≠ equal matrices"

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

19 Assertion (A): The determinant of a 1×1 matrix \(\left[5\right]\)is 5.
Reason (R): The determinant of a 1×1 matrix \(\left[a\right]\)is defined as \(a\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Determinant of 1×1 matrix equals entry itself General definition directly applies Reason correctly explains assertion

✅ Correct Answer Explanation

For a 1×1 matrix: \(\left[a\right]\) its determinant is defined as \(a\). Therefore: \(∣[5]∣=5\) Both the Assertion and Reason are true, and the Reason directly explains the Assertion. Hence Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Both statements are mathematically correct.
Option B → Reason is not false; it is the standard definition.
Option D → Assertion is true because determinant equals the single matrix entry.

🎯 Strategy

Used

Contextual/Tonal Matching

Application:

Use the standard determinant definition for a 1×1 matrix.

Final Logic:

Reason directly establishes the truth of the Assertion.

🧠 Memory Trick

"Single entry = single determinant"

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

20 Match each determinant with its correct value:

List I	List II
1. \(∣\begin{pmatrix}2 & 3\\ 1 & 4\end{pmatrix}∣\)	a. 0
2. \(∣\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}∣\)	b. 5
3. \(∣\begin{pmatrix}0 & 5\\ 2 & 3\end{pmatrix}∣\)	c. -10
4. \(∣\begin{pmatrix}3 & 2\\ 4 & 1\end{pmatrix}∣\)	d. -5

A 1–b, 2–a, 3–c, 4–d

B 1–d, 2–b, 3–c, 4–a

C 1–a, 2–c, 3–b, 4–d

D 1–d, 2–a, 3–b, 4–c

💡 Short Explanation

Use \(ad-bc\) formula repeatedly Compute each determinant separately Match obtained numerical values

✅ Correct Answer Explanation

Calculations: \(∣\begin{pmatrix}2 & 3\\ 1 & 4\end{pmatrix}∣=8-3=5\) → b \(∣\begin{pmatrix}1 & 1\\ 1 & 1\end{pmatrix}∣=1-1=0\) → a \(∣\begin{pmatrix}0 & 5\\ 2 & 3\end{pmatrix}∣=0-10=-10\) → c \(∣\begin{pmatrix}3 & 2\\ 4 & 1\end{pmatrix}∣=3-8=-5\) → d Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Incorrectly matches determinant 1 with \(-5\).
Option C → Gives wrong determinant values for multiple matrices.
Option D → Third determinant equals \(-10\), not \(5\).

🎯 Strategy

Used

Substitution

Application:

Apply the \(ad-bc\) rule individually for every determinant.

Final Logic:

Only Option A correctly matches all evaluated determinant values.

🧠 Memory Trick

"2×2 → cross multiply and subtract"

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

CUET UG Mathematics Booster Test 1 - Expansion Methods and Evaluation of Determinants

Test Complete!

Answer Review