UG Mathematics Booster Test 2 - Operations on Matrices -

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UG Mathematics Booster Test 2 - Operations on Matrices

⏱️ 30 Minutes

❓ 20 Questions

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

UG Mathematics Booster Test 2 - Operations on Matrices

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QUESTION 1 OF 20

Match the matrix pairs with whether addition is defined:

List I	List II
1. \(A(2\times 3),B(2\times 3)\)	a. Defined, result \(3\times 3\)
2. \(A(3\times 2),B(2\times 3)\)	b. Defined, result \(2\times 3\)
3. \(A(1\times 4),B(4\times 1)\)	c. Not defined (dimension mismatch)
4. \(A(3\times 3),B(3\times 3)\)	d. Not defined (row–column swapped)

Answer Review

1 Match the matrix pairs with whether addition is defined:

List I	List II
1. \(A(2\times 3),B(2\times 3)\)	a. Defined, result \(3\times 3\)
2. \(A(3\times 2),B(2\times 3)\)	b. Defined, result \(2\times 3\)
3. \(A(1\times 4),B(4\times 1)\)	c. Not defined (dimension mismatch)
4. \(A(3\times 3),B(3\times 3)\)	d. Not defined (row–column swapped)

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

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

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

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

💡 Short Explanation

Same order is necessary for addition. Different dimensions prevent addition. Result keeps common order.

✅ Correct Answer Explanation

Matrices can be added only if their orders are identical. Thus: \(2\times 3+2\times 3\)→ defined \(3\times 2+2\times 3\)→ not defined \(1\times 4+4\times 1\)→ row-column mismatch \(3\times 3+3\times 3\)→ defined Hence Option C is correct. The provided answer was incorrect.

❓ Why Other Options Are Incorrect

Option A → Incorrectly matches \(2\times 3\) pair with \(3\times 3\) result.
Option B → Second pair is dimension mismatch, not merely swapped interpretation.
Option D → First and fourth mappings are incorrect.

🎯 Strategy

Used: Option Grouping

Application:

Check dimension compatibility pairwise.

Final Logic:

Addition requires exactly equal orders.

🧠 Memory Trick

"Same shape → can add"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Addition of Matrices

2 Factory II produces 25 boys' and 31 girls' shoes. If production increases by 5 in each category, find the new number of girls' shoes.

A 30

B 25

C 36

D 31

💡 Short Explanation

Girls' shoes initially = 31. Increase = 5. New total = 36.

✅ Correct Answer Explanation

The girls' shoe production increases from 31 to: \(31+5=36\) Thus the updated production count becomes 36. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect arithmetic.
Option B → Represents boys' shoes count.
Option D → Original value before increase.

🎯 Strategy

Used: Substitution

Application:

Add increase directly to original value.

Final Logic:

New production = old production + increase.

🧠 Memory Trick

"31 grows to 36"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Matrix Addition Applications

3 Let
\(A=\left[\begin{pmatrix}3 & 4\\ 5 & 7\end{pmatrix}\right]\)
For \(3A\), which statements are true?
1. Top-left becomes 9
2. Bottom-right becomes 21
3. Scalar 3 multiplies all elements

A 1 only

B 2 only

C 1 and 3

D 1, 2, and 3

💡 Short Explanation

Multiply every entry by 3. \(3\times 3=9\). \(3\times 7=21\).

✅ Correct Answer Explanation

Scalar multiplication means multiplying every entry by scalar 3: \(3A=\left[\begin{pmatrix}9 & 12\\ 15 & 21\end{pmatrix}\right]\) Hence top-left becomes 9, bottom-right becomes 21, and all elements are multiplied. Therefore all three statements are correct.

❓ Why Other Options Are Incorrect

Option A → Ignores statements 2 and 3.
Option B → Ignores statements 1 and 3.
Option C → Omits correct statement 2.

🎯 Strategy

Used: Substitution

Application:

Multiply each entry by scalar 3.

Final Logic:

Scalar multiplication affects every matrix entry.

🧠 Memory Trick

"Scalar touches all entries"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Scalar Multiplication

4 Which statement is incorrect?

A Scaling by \(k=0\) gives identity matrix

B Scaling by \(k=1\) leaves matrix unchanged

C Scaling by \(k=-1\) gives \(-A\)

D Each element becomes \(k⋅a_{ij}\)

💡 Short Explanation

Multiplying by zero gives zero matrix. Identity matrix is different. Other scalar properties are correct.

✅ Correct Answer Explanation

If scalar \(k=0\), then: \(0A=O\) where \(O\) is the zero matrix, not identity matrix. Identity matrix contains diagonal ones. Hence Option A is incorrect.

❓ Why Other Options Are Incorrect

Option B → Multiplying by 1 leaves matrix unchanged.
Option C → Multiplying by \(-1\) gives negative matrix.
Option D → Standard scalar multiplication rule.

🎯 Strategy

Used: Elimination

Application:

Check standard scalar multiplication properties.

Final Logic:

Zero scalar produces zero matrix, not identity matrix.

🧠 Memory Trick

"Zero scalar → zero matrix"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Scalar Multiplication Properties

5 Let
\(A=\left[\begin{pmatrix}1 & -2\\ 3 & 4\end{pmatrix}\right]\)
Then
\(-A=\left[\begin{pmatrix}-1 & 2\\ -3 & -4\end{pmatrix}\right]\)
Assertion (A): This is the negative matrix of \(A\).
Reason (R): \(-A=(-1)⋅A\).

A Both false

B A true, R false

C Both true, R explains A

D A false, R true

💡 Short Explanation

Negative matrix changes every sign. Multiply matrix by \(-1\). Reason correctly explains assertion.

✅ Correct Answer Explanation

The negative matrix is obtained by multiplying every element by \(-1\): \(-A=(-1)A\) Thus all signs reverse correctly. Therefore both Assertion and Reason are true, and Reason explains Assertion.

❓ Why Other Options Are Incorrect

Option A → Both statements are actually correct.
Option B → Reason is true, not false.
Option D → Assertion is valid.

🎯 Strategy

Used: Elimination

Application:

Verify both matrix transformation and definition.

Final Logic:

Negative matrix equals scalar multiplication by \(-1\).

🧠 Memory Trick

"Minus flips every sign"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Negative Matrix

6 Let
\(A=\left[\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\right]\)
Arrange the correct steps to evaluate \(A+(-A)\):
1. Add corresponding elements
2. Find \(-A=\left[\begin{pmatrix}-1 & -2\\ -3 & -4\end{pmatrix}\right]\)
3. Identify matrix \(A\)
4. Conclude result is the zero matrix

A 3, 1, 2, 4

B 3, 2, 1, 4

C 1, 2, 3, 4

D 2, 3, 1, 4

💡 Short Explanation

Start with matrix identification. Form negative matrix. Add entries and conclude zero matrix.

✅ Correct Answer Explanation

Correct procedure: 1. Identify matrix \(A\). 2. Compute \(-A\). 3. Add corresponding entries. 4. Final result becomes zero matrix: \(A+(-A)=O\) Hence Option B gives proper logical sequence.

❓ Why Other Options Are Incorrect

Option A → Addition attempted before finding \(-A\).
Option C → Logical order completely disturbed.
Option D → Matrix must be identified first.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Arrange mathematical operations sequentially.

Final Logic:

Identify → negate → add → conclude.

🧠 Memory Trick

"Find minus, then add"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Additive Inverse

7 Let
\(A=\left[\begin{pmatrix}5 & 6\\ 7 & 8\end{pmatrix}\right],B=\left[\begin{pmatrix}2 & 3\\ 4 & 1\end{pmatrix}\right]\)
Find \(A-B\).
\(A-B=\left[\begin{pmatrix}3 & 3\\ 3 & 7\end{pmatrix}\right]\)

A \(\left[\begin{pmatrix}3 & 3\\ 3 & 7\end{pmatrix}\right]\)

B \(\left[\begin{pmatrix}7 & 9\\ 11 & 9\end{pmatrix}\right]\)

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

D \(\left[\begin{pmatrix}5 & 6\\ 7 & 8\end{pmatrix}\right]\)

💡 Short Explanation

Subtract corresponding entries. \(5-2=3\), \(6-3=3\). Continue element-wise subtraction.

✅ Correct Answer Explanation

Matrix subtraction is entry-wise: \(A-B=\left[\begin{pmatrix}5-2 & 6-3\\ 7-4 & 8-1\end{pmatrix}\right]=\left[\begin{pmatrix}3 & 3\\ 3 & 7\end{pmatrix}\right]\) Thus Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents matrix addition.
Option C → Only repeats matrix \(B\).
Option D → Only repeats matrix \(A\).

🎯 Strategy

Used: Substitution

Application:

Subtract entries position-wise.

Final Logic:

Matrix subtraction works element by element.

🧠 Memory Trick

"Subtract box to box"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Subtraction of Matrices

8 Let
\(A=\left[\begin{pmatrix}20 & 25\end{pmatrix}\right],B=\left[\begin{pmatrix}15 & 20\end{pmatrix}\right]\)
Find growth matrix \(A-B\).
\(A-B=\left[\begin{pmatrix}5 & 5\end{pmatrix}\right]\)

A \(\left[\begin{pmatrix}5 & 5\end{pmatrix}\right]\)

B \(\left[\begin{pmatrix}35 & 45\end{pmatrix}\right]\)

C \(\left[\begin{pmatrix}0 & 0\end{pmatrix}\right]\)

D \(\left[\begin{pmatrix}20 & 25\end{pmatrix}\right]\)

💡 Short Explanation

Growth means subtraction. Subtract corresponding entries. Both differences equal 5.

✅ Correct Answer Explanation

Compute subtraction entry-wise: \(20-15=525-20=5\) Thus: \(A-B=\left[\begin{pmatrix}5 & 5\end{pmatrix}\right]\) Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → Represents addition instead of subtraction.
Option C → Incorrect zero difference.
Option D → Merely repeats matrix \(A\).

🎯 Strategy

Used: Substitution

Application:

Apply subtraction to each corresponding entry.

Final Logic:

Growth matrix equals entry-wise increase.

🧠 Memory Trick

"Growth = subtract old from new"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Matrix Subtraction Applications

9 Let
\(u=\left[\begin{pmatrix}1 & 2\end{pmatrix}\right],v=\left[\begin{pmatrix}3 & 4\end{pmatrix}\right]\)
Given \(u+v=v+u\), find \(u+v\).
\(u+v=\left[\begin{pmatrix}4 & 6\end{pmatrix}\right]\)

A \(\left[\begin{pmatrix}1 & 2\end{pmatrix}\right]\)

B \(\left[\begin{pmatrix}4 & 6\end{pmatrix}\right]\)

C \(\left[\begin{pmatrix}3 & 4\end{pmatrix}\right]\)

D \(\left[\begin{pmatrix}2 & 1\end{pmatrix}\right]\)

💡 Short Explanation

Add corresponding entries. Order does not matter. Result becomes \(\left[4 6\right]\).

✅ Correct Answer Explanation

Using commutative property: \(u+v=v+u\) Now add corresponding entries: \(\left[\begin{pmatrix}1 & 2\end{pmatrix}\right]+\left[\begin{pmatrix}3 & 4\end{pmatrix}\right]=\left[\begin{pmatrix}4 & 6\end{pmatrix}\right]\) Therefore Option B is correct.

❓ Why Other Options Are Incorrect

Option A → Only matrix \(u\).
Option C → Only matrix \(v\).
Option D → Entries improperly arranged.

🎯 Strategy

Used: Substitution

Application:

Perform direct vector addition.

Final Logic:

Add corresponding entries only.

🧠 Memory Trick

"1+3, 2+4"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Commutative Law of Addition

10 Let
\(X=\left[\begin{aligned}2\end{aligned}\right],Y=\left[\begin{aligned}3\end{aligned}\right],Z=\left[\begin{aligned}4\end{aligned}\right]\)
Evaluate \((X+Y)+Z\) and identify the matrix for \(\left(X,\ Y\right)\).
\(X+Y=\left[\begin{aligned}5\end{aligned}\right]\)

A \(\left[\begin{aligned}7\end{aligned}\right]\)

B \(\left[\begin{aligned}4\end{aligned}\right]\)

C \(\left[\begin{aligned}6\end{aligned}\right]\)

D \(\left[\begin{aligned}5\end{aligned}\right]\)

💡 Short Explanation

First compute \(X+Y\). \(2+3=5\). Associative grouping preserves total.

✅ Correct Answer Explanation

The question specifically asks to identify matrix \(\left(X,\ Y\right)\): \(X+Y=\left[\begin{aligned}2+3\end{aligned}\right]=\left[\begin{aligned}5\end{aligned}\right]\) Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Represents total after adding \(Z\).
Option B → Incorrect arithmetic.
Option C → Partial incorrect addition.

🎯 Strategy

Used: Contextual/Tonal Matching

Application:

Carefully identify which quantity is asked.

Final Logic:

Question asks only for \(\left(X,\ Y\right)\).

🧠 Memory Trick

"First pair gives 5"

📚 Class XI – NCERT Mathematics – Chapter: Matrices – Topic: Associative Law of Addition

11 Let a region be represented by a vertex matrix
\(V=\left[\begin{pmatrix}x_{1} & y_{1}\\ x_{2} & y_{2}\\ x_{3} & y_{3}\end{pmatrix}\right]\)
If the zero matrix \(O\) of the same order is added, i.e., \(V+O\), what happens to the region?

A It remains in the same position

B It is translated to the origin

C It is reflected across the x-axis

D Its area becomes zero

💡 Short Explanation

Zero matrix acts as additive identity Adding zero changes no coordinates Region remains unchanged geometrically

✅ Correct Answer Explanation

The zero matrix is the additive identity in matrix algebra. When \(O\) is added to matrix \(V\), every element remains unchanged. Therefore, all coordinates of the region stay the same. Option A is correct. Option B incorrectly suggests translation, Option C reflection, and Option D area collapse, none of which occur.

❓ Why Other Options Are Incorrect

Option B → Adding a zero matrix does not shift coordinates to the origin. Translation requires adding non-zero values to vertex coordinates.
Option C → Reflection changes sign of coordinates. Zero matrix addition leaves every coordinate unchanged.
Option D → Area becomes zero only if all points collapse together, which does not happen with additive identity.

🎯 Strategy

Used: Elimination

Application:

Use the additive identity property of matrices to eliminate all transformation-based options.

Final Logic:

Since \(V+O=V\), the region remains unchanged.

🧠 Memory Trick

"Add zero → No change."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Zero Matrix and Additive Identity

12 Let
\(P=\left[\begin{pmatrix}0.2 & 0.8\end{pmatrix}\right]\)
be a probability distribution. Its additive inverse is \(-P\).
Which statement is correct?

A \(\left[\begin{pmatrix}-0.2 & -0.8\end{pmatrix}\right]\), valid distribution

B \(\left[\begin{pmatrix}0.2 & -0.8\end{pmatrix}\right]\), valid distribution

C \(\left[\begin{pmatrix}-0.2 & -0.8\end{pmatrix}\right]\), not a valid distribution

D \(\left[\begin{pmatrix}0 & 0\end{pmatrix}\right]\), valid distribution

💡 Short Explanation

Additive inverse multiplies entries by \(-1\) Probabilities cannot be negative Hence inverse matrix is not a valid distribution

✅ Correct Answer Explanation

The additive inverse of a matrix is obtained by multiplying each element by \(-1\). Thus, \(-P=\left[\begin{pmatrix}-0.2 & -0.8\end{pmatrix}\right]\) . However, probabilities must lie between 0 and 1. Therefore, this matrix cannot represent a valid probability distribution. Hence Option C is correct. Other options either use incorrect inverses or violate probability rules.

❓ Why Other Options Are Incorrect

Option A → Although the additive inverse is correct, negative probabilities are invalid in probability distributions.
Option B → Only one element is negated, so this is not the true additive inverse.
Option D → Zero matrix is not the additive inverse of \(P\); it also does not sum to 1 as probabilities should.

🎯 Strategy

Used: Substitution

Application:

Apply the additive inverse rule directly by multiplying every element by \(-1\).

Final Logic:

Negative entries violate probability conditions, so only Option C is valid.

🧠 Memory Trick

"Inverse flips signs."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Negative of a Matrix

📄 Passage

Matrix multiplication is not commutative (\(AB\neq BA\)). For a square matrix \(A\), there exists an identity matrix \(I\) such that \(AI=IA=A\). Also, the distributive law holds:
\(A(B+C)=AB+AC\)

Using the distributive law, evaluate \(A(B+C)\).

A \(AB+BA\)

B \(AB+AC\)

C \(BC+CA\)

D \(A+B+C\)

💡 Short Explanation

Matrix multiplication follows distributive law Multiply \(A\) with both matrices separately Add the obtained products

✅ Correct Answer Explanation

The distributive property of matrix multiplication states: \(A(B+C)=AB+AC\) . Matrix \(A\) multiplies each matrix inside the bracket individually. Hence Option B is correct. Option A incorrectly changes order, Option C omits \(A\) multiplication properly, and Option D ignores multiplication entirely.

❓ Why Other Options Are Incorrect

Option A → Matrix multiplication is generally non-commutative, so \(AB+BA\) is incorrect.
Option C → Expression does not follow distributive structure from the left side multiplication.
Option D → Matrix addition alone cannot replace distributive multiplication.

🎯 Strategy

Used: Option Grouping

Application:

Identify the option matching the standard distributive identity from matrix algebra.

Final Logic:

Distribute \(A\) across both matrices:
\(A(B+C)=AB+AC\)

🧠 Memory Trick

"Distribute A to all."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Properties of Matrix Multiplication

📄 Passage

Matrix multiplication is not commutative (\(AB\neq BA\)). For a square matrix \(A\), there exists an identity matrix \(I\) such that \(AI=IA=A\). Also, the distributive law holds:
\(A(B+C)=AB+AC\)

If a scalar \(k=0\) is multiplied with any matrix \(A\), what is the result?

A \(A\)

B \(I\)

C \(kA\)

D Zero matrix \(O\)

💡 Short Explanation

Scalar multiplication multiplies every element Multiplying by zero makes all entries zero Result becomes zero matrix

✅ Correct Answer Explanation

When scalar \(0\) multiplies a matrix, every entry becomes zero: \(0A=O\) . Therefore, the resulting matrix is the zero matrix. Option D is correct. Option A leaves matrix unchanged, Option B refers to identity matrix, and Option C only restates the expression without evaluating it.

❓ Why Other Options Are Incorrect

Option A → Matrix remains unchanged only when multiplied by scalar \(1\), not \(0\).
Option B → Identity matrix is unrelated to scalar multiplication by zero.
Option C → \(kA\) is incomplete because \(k=0\) must be evaluated.

🎯 Strategy

Used: Substitution

Application:

Substitute \(k=0\) directly into scalar multiplication definition.

Final Logic:

Every element multiplied by zero becomes zero.

🧠 Memory Trick

"Zero times matrix → Zero matrix."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Scalar Multiplication

15 Let matrix \(A\) be of order \(m\times n\) and matrix \(B\) be of order \(p\times q\).
For \(AB\) to be defined, what must hold?

A \(m=p\)

B \(n=q\)

C \(n=p\)

D \(m=q\)

💡 Short Explanation

Matrix multiplication uses row–column rule Inner dimensions must match Columns of first = rows of second

✅ Correct Answer Explanation

If matrix \(A\) is \(m\times n\) and matrix \(B\) is \(p\times q\), multiplication \(AB\) is possible only when the number of columns of \(A\) equals the number of rows of \(B\). Thus \(n=p\). Hence Option C is correct. Other options compare unrelated dimensions.

❓ Why Other Options Are Incorrect

Option A → Equality of rows is unnecessary for multiplication definition.
Option B → Number of columns need not match in both matrices.
Option D → Outer dimensions determine result order, not multiplication condition.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Compare matrix dimensions using multiplication compatibility rules.

Final Logic:

Inner dimensions must match:
\((m\times n)(p\times q)\Rightarrow n=p\)

🧠 Memory Trick

"Inside numbers must match."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Matrix Multiplication

16 Let
\(A=\left[\begin{pmatrix}1 & 2 & 3\end{pmatrix}\right],B=\left[\begin{aligned}4\\ 5\\ 6\end{aligned}\right]\)
Find the single element of \(AB\) using the row–column rule.
\(AB=(1⋅4+2⋅5+3⋅6)\)

A 32

B 15

C 24

D 18

💡 Short Explanation

Multiply corresponding entries Add all obtained products Row–column multiplication gives scalar value

✅ Correct Answer Explanation

Using row–column multiplication: \(AB=(1\times 4)+(2\times 5)+(3\times 6)=4+10+18=32\) Hence Option A is correct. Options B, C, and D arise from incomplete addition or incorrect multiplication steps.

❓ Why Other Options Are Incorrect

Option B → Obtained from partial addition without including all products.
Option C → Incorrect multiplication total.
Option D → Only one multiplication term \(3\times 6\) considered incorrectly.

🎯 Strategy

Used: Substitution

Application:

Directly substitute values into the row–column multiplication formula.

Final Logic:

\(4+10+18=32\)

🧠 Memory Trick

"Multiply then add."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Matrix Multiplication

17 If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), the element \(c_{ik}\)of \(AB\) is:

A \(\sum_{j=1}^{m}\,a_{ij}b_{jk}\)

B \(\sum_{j=1}^{p}\,a_{ij}b_{jk}\)

C \(a_{ik}b_{ik}\)

D \(\sum_{j=1}^{n}\,a_{ij}b_{jk}\)

💡 Short Explanation

Product elements use summation notation Multiply row entries with column entries Sum over common index \(j\)

✅ Correct Answer Explanation

For matrices \(A_{m\times n}\)and \(B_{n\times p}\), the product element is: \(c_{ik}=\sum_{j=1}^{n}\,a_{ij}b_{jk}\) . The summation index runs through the common dimension \(n\). Hence Option D is correct. Other options use wrong limits or incorrect multiplication structure.

❓ Why Other Options Are Incorrect

Option A → Summation upper limit should be \(n\), not \(m\).
Option B → Summation does not extend to \(p\), since \(p\) is output column count.
Option C → Matrix multiplication is not simple element-wise multiplication.

🎯 Strategy

Used: Option Grouping

Application:

Recognize the standard matrix multiplication formula directly.

Final Logic:

Summation runs across the shared dimension \(n\).

🧠 Memory Trick

"Common dimension gets summed."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Product of Matrices

18 If \(X\) is \(2\times 5\) and \(Y\) is \(5\times 2\), how many elements are in \(XY\)?

A 6

B 15

C 10

D 4

💡 Short Explanation

Product order uses outer dimensions \(2\times 5\) and \(5\times 2\) give \(2\times 2\) Total elements \(=2\times 2=4\)

✅ Correct Answer Explanation

For matrix multiplication: \((2\times 5)(5\times 2)\rightarrow 2\times 2\) . Therefore, matrix \(XY\) has 4 elements in total. Option D is correct. Other options come from incorrect multiplication of dimensions or element counts.

❓ Why Other Options Are Incorrect

Option A → Does not correspond to resulting matrix dimensions.
Option B → Incorrectly multiplies all dimensions together.
Option C → Represents sum of dimensions rather than total elements.

🎯 Strategy

Used: Dimensional/Unit Analysis

Application:

Use outer dimensions to determine result order before counting elements.

Final Logic:

Result order \(2\times 2\) gives 4 elements.

🧠 Memory Trick

"Outer dimensions survive."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Order of Product Matrix

19 Let \(A\) be \(2\times 3\) and \(B\) be \(3\times 4\).
Then \(AB\) is \(2\times 4\). What is the order of \(BA\)?

A \(3\times 3\)

B \(4\times 2\)

C \(2\times 4\)

D Not defined

💡 Short Explanation

Reverse multiplication changes compatibility \(B\) has 4 columns, \(A\) has 2 rows Inner dimensions do not match

✅ Correct Answer Explanation

Matrix \(B\) is \(3\times 4\) and \(A\) is \(2\times 3\). For \(BA\) to exist, columns of \(B\) must equal rows of \(A\). Since \(4\neq 2\), multiplication is impossible. Therefore, \(BA\) is not defined. Hence Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Product cannot exist, so no \(3\times 3\) matrix forms.
Option B → Outer dimensions cannot be used because multiplication condition fails.
Option C → This is the order of \(AB\), not \(BA\).

🎯 Strategy

Used: Elimination

Application:

Check inner dimensions before attempting multiplication.

Final Logic:

Since \(4\neq 2\), \(BA\) is undefined.

🧠 Memory Trick

"Reverse order may fail."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Non-Commutativity of Matrix Multiplication

20 If \(AB=O\)(zero matrix), what can be concluded?

A It is not necessary that \(A\) or \(B\) is zero

B \(A\) must be zero

C \(B\) must be zero

D Both must be zero

💡 Short Explanation

Product can become zero without zero matrices Non-zero matrices may produce zero product Zero-product rule differs from real numbers

✅ Correct Answer Explanation

In matrix algebra, it is possible for two non-zero matrices to multiply and produce the zero matrix. Such matrices are called zero divisors. Therefore, \(AB=O\) does not force either \(A\) or \(B\) to be zero. Hence Option A is correct.

❓ Why Other Options Are Incorrect

Option B → \(A\) need not be zero; non-zero matrices can still produce zero product.
Option C → \(B\) also need not be zero individually.
Option D → Matrix multiplication does not satisfy ordinary zero-product property.

🎯 Strategy

Used: Extreme Word Filter

Application:

Words like "must" indicate overly strict conclusions often invalid in matrix algebra.

Final Logic:

Zero product does not imply zero matrices.

🧠 Memory Trick

"Zero product ≠ zero matrix factors."

📚 Class XII – NCERT Mathematics – Chapter: Matrices – Topic: Properties of Matrix Multiplication

UG Mathematics Booster Test 2 - Operations on Matrices

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