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Matrices: Class 12 Maths Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A school tuck shop records the number of sandwiches (S), rolls (R) and wraps (W) sold on Monday and Tuesday. Monday sales: M = [[12, 8, 5]], Tuesday sales: T = [[15, 6, 9]] (1×3 row matrices). Selling prices: sandwich ₹30, roll ₹20, wrap ₹25, stored as column matrix P = [[30], [20], [25]].

A school tuck shop records the number of sandwiches (S), rolls (R) and wraps (W) sold on Monday and Tuesday in a matrix. The sales matrix for Monday is M = [[12, 8, 5]] and for Tuesday is T = [[15, 6, 9]] (each a 1×3 row matrix representing [S, R, W]).

The selling prices (in ₹) of a sandwich, a roll and a wrap are ₹30, ₹20 and ₹25 respectively, stored in a column matrix P = [[30], [20], [25]].

(i) Write down the combined sales matrix C = M + T and state its order. [1]
(ii) Using matrix multiplication, find the total revenue collected on Monday alone. [1]
(iii) Find the total revenue collected over both days combined, using the matrix C found in part (i). [2]

Show answer
(i)
Combined sales matrix:

C = M + T = [12 + 15, 8 + 6, 5 + 9] = [27, 14, 14]

∴ C = [27 14 14], which is a matrix of order 1×3.

(ii)
Total revenue on Monday = M × P

= [12 8 5] × [[30], [20], [25]]

= (12 × 30) + (8 × 20) + (5 × 25)

= 360 + 160 + 125

∴ Total revenue on Monday = ₹645

(iii)
Total revenue over both days = C × P

= [27 14 14] × [[30], [20], [25]]

= (27 × 30) + (14 × 20) + (14 × 25)

= 810 + 280 + 350

∴ Total revenue over both days = ₹1440
Q2Case-based4 marks

A school's science club tracks the performance of its three teams — Alpha, Beta, and Gamma — over two consecutive rounds of an inter-school quiz. The scores in Round 1 and Round 2 are recorded in matrices P and Q respectively:

P = [[4, 6, 2], [3, 5, 7]], Q = [[1, 3, 5], [6, 2, 4]]

where rows represent the two subjects (Physics and Chemistry) and columns represent the three teams (Alpha, Beta, Gamma).

A school's science club tracks the performance of its three teams — Alpha, Beta, and Gamma — over two consecutive rounds of an inter-school quiz. The scores in Round 1 and Round 2 are recorded in matrices P and Q respectively:

P = [[4, 6, 2], [3, 5, 7]], Q = [[1, 3, 5], [6, 2, 4]]

where rows represent the two subjects (Physics and Chemistry) and columns represent the three teams (Alpha, Beta, Gamma).

(i) Find the matrix representing the total scores of each team in each subject over both rounds. [1]
(ii) The club decides to give double weightage to Round 2 scores. Find the matrix representing the combined weighted score (Round 1 + 2 × Round 2). [1]
(iii) Using your answer from part (ii), find the matrix S = P + 2Q − 3I₂, where I₂ is the 2×2 identity matrix. What difficulty does this operation encounter, and why? [2]

Show answer
(i) The total scores matrix is obtained by adding P and Q.

P + Q = [[4+1, 6+3, 2+5], [3+6, 5+2, 7+4]]

∴ P + Q = [[5, 9, 7], [9, 7, 11]]

This matrix represents the total scores of each team in each subject over both rounds.

(ii) The combined weighted score matrix is P + 2Q.

2Q = 2 × [[1, 3, 5], [6, 2, 4]] = [[2, 6, 10], [12, 4, 8]]

P + 2Q = [[4+2, 6+6, 2+10], [3+12, 5+4, 7+8]]

∴ P + 2Q = [[6, 12, 12], [15, 9, 15]]

(iii) From part (ii), P + 2Q = [[6, 12, 12], [15, 9, 15]], which is a 2×3 matrix.

The identity matrix I₂ = [[1, 0], [0, 1]] is a 2×2 matrix.

For matrix subtraction (P + 2Q) − 3I₂ to be defined, both matrices must be of the same order.

However, (P + 2Q) is of order 2×3 and 3I₂ is of order 2×2.

∵ the orders 2×3 and 2×2 are not equal, the operation (P + 2Q) − 3I₂ is NOT defined.

∴ The matrix S = P + 2Q − 3I₂ cannot be computed because matrix addition/subtraction requires both matrices to have the same order, and (P + 2Q) is of order 2×3 while 3I₂ is of order 2×2 — these orders are incompatible.
Q3Case-based4 marks

A school management committee is preparing a budget report. The monthly expenditure (in ₹ thousands) on three heads — Salaries, Infrastructure, and Miscellaneous — for two departments (Science and Arts) is recorded in matrix A, and a 15% increase is proposed for the next month represented by B = 1.15A.

A = [[120, 80, 40], [90, 60, 30]]

A school management committee is preparing a budget report. The monthly expenditure (in ₹ thousands) on three heads — Salaries, Infrastructure, and Miscellaneous — for two departments (Science and Arts) is recorded in matrix A, and a 15% increase proposed for the next month is represented by matrix B = 1.15A.

The current monthly expenditure matrix is:

A = [[120, 80, 40], [90, 60, 30]]

where rows represent Science and Arts departments respectively, and columns represent Salaries, Infrastructure, and Miscellaneous.

Based on the above information, answer the following:

(i) Write the matrix B representing next month's proposed expenditure. [1]

(ii) The committee decides the total expenditure matrix for the two months combined is T = A + B. Find matrix T. [1]

(iii) If the committee wants to express T in the form T = kA for some scalar k, find the value of k. Hence verify that the total Salaries expenditure for the Science department obtained from T equals k times the corresponding entry in A. [2]

Show answer
(i) Since B = 1.15A,

B = 1.15 × [[120, 80, 40], [90, 60, 30]]

∴ B = [[138, 92, 46], [103.5, 69, 34.5]]

(ii) T = A + B

T = [[120, 80, 40], [90, 60, 30]] + [[138, 92, 46], [103.5, 69, 34.5]]

Adding corresponding elements,

∴ T = [[258, 172, 86], [193.5, 129, 64.5]]

(iii) We have T = A + B = A + 1.15A.

Using scalar multiplication and matrix addition property:

T = (1 + 1.15)A = 2.15A

∴ k = 2.15

Verification:

The (1, 1) entry of T (total Salaries for Science) = 258.

k × (1, 1) entry of A = 2.15 × 120 = 258.

∵ 258 = 258,

∴ T = kA is verified for the Salaries entry of the Science department.

∴ k = 2.15 and the total Salaries expenditure for the Science department from T equals 2.15 times the corresponding entry in A. Hence verified.
Q4Case-based4 marks

A manufacturing company tracks its monthly production and pricing data using matrices. The production matrix P (units produced) and the price matrix C (cost per unit in ₹) for two products X and Y across two plants are:

P = [[120, 80], [150, 60]] (rows = Plant 1, Plant 2; columns = Product X, Product Y)

C = [[200], [350]] (rows = Product X, Product Y)

The Revenue matrix R = PC, and the Bonus matrix B = P + Pᵀ.

A manufacturing company tracks its monthly production and pricing data using matrices. The production matrix P (units produced) and the price matrix C (cost per unit in ₹) for two products X and Y across two plants are given as:

P = [[120, 80], [150, 60]] where rows represent Plant 1 and Plant 2, columns represent Product X and Product Y.

C = [[200], [350]] where rows represent Product X and Product Y.

The Revenue matrix R is defined by R = PC, and a Bonus matrix B is defined such that B = P + Pᵀ (where Pᵀ is the transpose of P).

Based on this context, answer the following:

(i) Find the order of the Revenue matrix R = PC. Also compute R, interpreting each element in context. [2]

(ii) A quality analyst claims: 'The matrix B = P + Pᵀ is always a symmetric matrix for any matrix P.' Verify this claim analytically (without computing B explicitly), and state whether the analyst's claim is correct. [2]

Show answer
Part (i) [2 marks]

P is of order 2×2 and C is of order 2×1.

∵ Number of columns of P = Number of rows of C = 2,

∴ R = PC is defined and its order is 2×1.

Computing R:

R = [[120, 80], [150, 60]] × [[200], [350]]

R₁₁ = 120×200 + 80×350 = 24000 + 28000 = 52000

R₂₁ = 150×200 + 60×350 = 30000 + 21000 = 51000

∴ R = [[52000], [51000]]

Interpretation: Plant 1 earns a total monthly revenue of ₹52,000 and Plant 2 earns a total monthly revenue of ₹51,000 from the two products.

---

Part (ii) [2 marks]

To verify that B = P + Pᵀ is symmetric, we must show that Bᵀ = B.

We know the following standard results:

(i) (P + Q)ᵀ = Pᵀ + Qᵀ for any two matrices of same order

(ii) (Pᵀ)ᵀ = P

Now, taking the transpose of B:

Bᵀ = (P + Pᵀ)ᵀ

⟹ Bᵀ = Pᵀ + (Pᵀ)ᵀ [using property (i)]

⟹ Bᵀ = Pᵀ + P [using property (ii)]

⟹ Bᵀ = P + Pᵀ = B

∵ Bᵀ = B,

∴ B = P + Pᵀ is always a symmetric matrix for any matrix P for which Pᵀ is defined (i.e., any square or rectangular matrix).

Hence, the quality analyst's claim is correct.
Q5Case-based4 marks

A school's attendance monitoring system records daily presence of students in two sections, P and Q, over two consecutive days. The data is stored as matrices:

Day 1: A = [[30, 12], [8, 25]], Day 2: B = [[28, 15], [10, 22]]

Rows represent Sections P and Q; columns represent Boys and Girls present.

A school's attendance monitoring system records daily presence of students in two sections, P and Q, over two consecutive days. The data is stored as matrices:

Day 1: A = [[30, 12], [8, 25]] where rows represent Sections P and Q, and columns represent Boys and Girls present.

Day 2: B = [[28, 15], [10, 22]]

The school administration defines a 'Combined Attendance Matrix' C = A + B, and an 'Adjusted Matrix' D = 2A − B.

(i) Find the Combined Attendance Matrix C = A + B. [1 mark]
(ii) Find the Adjusted Matrix D = 2A − B. [1 mark]
(iii) The principal wants to award sections based on a 'Performance Score Matrix' S = C + D. Find S, and hence determine which section (P or Q) has the higher total attendance score (sum of Boys and Girls entries in that row of S). [2 marks]

Show answer
(i) Finding C = A + B:

C = A + B = [[30, 12], [8, 25]] + [[28, 15], [10, 22]]

⟹ C = [[30+28, 12+15], [8+10, 25+22]]

∴ C = [[58, 27], [18, 47]]

(ii) Finding D = 2A − B:

2A = 2 × [[30, 12], [8, 25]] = [[60, 24], [16, 50]]

⟹ D = 2A − B = [[60, 24], [16, 50]] − [[28, 15], [10, 22]]

⟹ D = [[60−28, 24−15], [16−10, 50−22]]

∴ D = [[32, 9], [6, 28]]

(iii) Finding S = C + D and comparing section scores:

S = C + D = [[58, 27], [18, 47]] + [[32, 9], [6, 28]]

⟹ S = [[58+32, 27+9], [18+6, 47+28]]

∴ S = [[90, 36], [24, 75]]

Total score for Section P (Row 1) = 90 + 36 = 126

Total score for Section Q (Row 2) = 24 + 75 = 99

∵ 126 > 99,

∴ Section P has the higher total attendance score.
Q6Case-based4 marks

A school tuck-shop records its daily sales of three items — Samosa (S), Sandwich (W) and Cold Drink (C) — for Monday and Tuesday in matrix form:

Monday: M = [[12, 8, 15], [10, 6, 20]]
Tuesday: T = [[14, 10, 12], [8, 9, 18]]

where Row 1 = Morning break sales and Row 2 = Lunch break sales.

The price (in ₹) of each item is given by the column matrix:

P = [[10], [15], [20]]

A school tuck-shop records its daily sales of three items — Samosa (S), Sandwich (W) and Cold Drink (C) — for Monday and Tuesday in matrix form:

Monday: M = [[12, 8, 15], [10, 6, 20]]
Tuesday: T = [[14, 10, 12], [8, 9, 18]]

where Row 1 = Morning break sales and Row 2 = Lunch break sales.

The price (in ₹) of each item is given by the column matrix:

P = [[10], [15], [20]]

(i) Write the matrix S = M + T that gives the total sales (item-wise, break-wise) over the two days. [1]
(ii) Find the total revenue collected during the Morning break over the two days. [1]
(iii) Find the matrix R = S × P giving the total revenue (in ₹) for each break over the two days, and hence state which break generated more revenue. [2]

Show answer
(i) Matrix addition: S = M + T

S = [[12+14, 8+10, 15+12], [10+8, 6+9, 20+18]]

∴ S = [[26, 18, 27], [18, 15, 38]]

(ii) Total revenue during Morning break over the two days

= 26 × 10 + 18 × 15 + 27 × 20
= 260 + 270 + 540
∴ Total Morning break revenue = ₹1070

(iii) R = S × P, where S is a 2×3 matrix and P is a 3×1 matrix, giving R as a 2×1 matrix.

R = [[26, 18, 27], [18, 15, 38]] × [[10], [15], [20]]

R₁₁ = 26×10 + 18×15 + 27×20 = 260 + 270 + 540 = 1070
R₂₁ = 18×10 + 15×15 + 38×20 = 180 + 225 + 760 = 1165

∴ R = [[1070], [1165]]

∵ 1165 > 1070, the Lunch break generated more revenue.

∴ Total revenue: Morning break = ₹1070, Lunch break = ₹1165. The Lunch break generated more revenue.
Q7Case-based4 marks

A school tuck shop tracks daily sales of three items — Sandwich (S), Juice (J) and Cookie (C) — over two days. Day 1 sales: S = 40, J = 25, C = 60. Day 2 sales: S = 55, J = 30, C = 45. Selling price per unit (₹): S = 30, J = 20, C = 10. Cost price per unit (₹): S = 18, J = 12, C = 6.

A school tuck shop tracks daily sales of three items — Sandwich (S), Juice (J) and Cookie (C) — over two days using matrices.

Day 1 sales (units): S = 40, J = 25, C = 60
Day 2 sales (units): S = 55, J = 30, C = 45

The selling price (in ₹) per unit is: S = 30, J = 20, C = 10.
The cost price (in ₹) per unit is: S = 18, J = 12, C = 6.

Represent the sales data as a 2×3 matrix A, and the price data as a 3×2 matrix B where the first column gives selling prices and the second column gives cost prices of each item.

(i) Write down matrices A and B. [1]
(ii) Find the matrix P = AB and state what each element of P represents in context. [1]
(iii) Hence, find the matrix representing the profit earned (in ₹) on each day, and find the total profit over both days. [2]

Show answer
(i) The sales data over two days is represented as a 2×3 matrix:

A = [40 25 60]
[55 30 45]

The price data (selling price and cost price per unit) is represented as a 3×2 matrix:

B = [30 18]
[20 12]
[10 6]

∴ A is a 2×3 matrix and B is a 3×2 matrix.

(ii) Since A is 2×3 and B is 3×2, the product P = AB is defined and is a 2×2 matrix.

P = AB = [40 25 60] × [30 18]
[55 30 45] [20 12]
[10 6]

P₁₁ = 40×30 + 25×20 + 60×10 = 1200 + 500 + 600 = 2300
P₁₂ = 40×18 + 25×12 + 60×6 = 720 + 300 + 360 = 1380
P₂₁ = 55×30 + 30×20 + 45×10 = 1650 + 600 + 450 = 2700
P₂₂ = 55×18 + 30×12 + 45×6 = 990 + 360 + 270 = 1620

∴ P = AB = [2300 1380]
[2700 1620]

Each element Pᵢⱼ gives the total revenue (when j = 1) or total cost (when j = 2) for day i.
Specifically, P₁₁ = ₹2300 is total revenue on Day 1; P₁₂ = ₹1380 is total cost on Day 1; P₂₁ = ₹2700 is total revenue on Day 2; P₂₂ = ₹1620 is total cost on Day 2.

(iii) The profit on each day is obtained by subtracting the cost column from the revenue column of P.

Let Q = [1 -1] be a 2×1...

Alternatively, the profit matrix is obtained as the difference of the two columns of P:

Profit on Day 1 = P₁₁ − P₁₂ = 2300 − 1380 = ₹920
Profit on Day 2 = P₂₁ − P₂₂ = 2700 − 1620 = ₹1080

∴ Profit matrix = [920 ]
[1080]

∴ Total profit over both days = 920 + 1080 = ₹2000.

∴ The tuck shop earned a total profit of ₹2000 over the two days.
Q8Case-based4 marks

A school is preparing merit certificates for three streams — Science, Commerce, and Arts. The number of Gold, Silver, and Bronze certificates awarded to each stream is represented by a 3×3 matrix A, and the cost of each certificate type is stored in a column matrix C. Matrix multiplication is used to compute stream-wise and overall expenditure.

A school is preparing merit certificates for students. The data for three categories — Science, Commerce, and Arts — is stored in a matrix. The number of students receiving Gold, Silver, and Bronze certificates in each stream is given by the matrix:

A = [[3, 5, 2], [4, 1, 6], [2, 3, 4]]

where rows represent Science, Commerce, and Arts respectively, and columns represent Gold, Silver, and Bronze respectively.

The cost (in ₹) of each type of certificate is stored as:
Gold = ₹150, Silver = ₹100, Bronze = ₹75

(i) Write the cost matrix C as a 3×1 column matrix. [1]
(ii) Find the product matrix AC, which gives the total certificate expenditure for each stream. [1]
(iii) The school decides to add a 'Special Merit' certificate costing ₹200 each. The number of Special Merit certificates for Science, Commerce, and Arts are 1, 2, and 3 respectively. Find the new total expenditure matrix and hence compute the grand total expenditure across all three streams. [2]

Show answer
(i) The cost matrix C is a 3×1 column matrix:

C = [[150], [100], [75]]

∴ C = [[150], [100], [75]] ... (i)

(ii) A = [[3, 5, 2], [4, 1, 6], [2, 3, 4]]

AC = [[3×150 + 5×100 + 2×75], [4×150 + 1×100 + 6×75], [2×150 + 3×100 + 4×75]]

= [[450 + 500 + 150], [600 + 100 + 450], [300 + 300 + 300]]

= [[1100], [1150], [900]]

∴ The total certificate expenditure matrix is:
Science: ₹1100, Commerce: ₹1150, Arts: ₹900 ... (ii)

(iii) Let the Special Merit certificate count matrix be:

S = [[1], [2], [3]]

Additional expenditure for Special Merit = 200 × S = [[200], [400], [600]]

New total expenditure matrix T = AC + 200·S

T = [[1100 + 200], [1150 + 400], [900 + 600]]

= [[1300], [1550], [1500]]

Grand total expenditure = 1300 + 1550 + 1500

= ₹4350

∴ The new total expenditure matrix is [[1300], [1550], [1500]] and the grand total expenditure across all three streams is ₹4350.
Q9MCQ1 mark

If A is a square matrix such that A² = A, then (I + A)² − 3A is equal to:

Show answer
Option (A) is correct.

Explanation: Expanding (I + A)² − 3A using matrix algebra,
(I + A)² − 3A = I² + IA + AI + A² − 3A
= I + A + A + A − 3A [∵ A² = A, and IA = AI = A]
= I + 3A − 3A
= I
Q10MCQ1 mark

If A = [[k, 0], [0, k]] is a scalar matrix of order 2, then A² is equal to:

Show answer
Option (C) is correct.

Explanation: A scalar matrix of order 2 has the form A = [[k, 0], [0, k]] = kI, where I is the identity matrix of order 2.

A² = A × A = [[k, 0], [0, k]] × [[k, 0], [0, k]] = [[k·k + 0·0, k·0 + 0·k], [0·k + k·0, 0·0 + k·k]] = [[k², 0], [0, k²]].
Q11MCQ1 mark

If A is any square matrix, then which of the following is always a symmetric matrix?

Show answer
Option (B) is correct.

Explanation: A matrix M is symmetric if M' = M. Let M = A + A'. Then M' = (A + A')' = A' + (A')' = A' + A = A + A' = M. ∴ A + A' is always symmetric for any square matrix A.
Q12Short Answer1 mark

Assertion (A) : For any square matrix A, the matrix (A + A') is always a symmetric matrix.

Reason (R) : A matrix B is symmetric if and only if B' = B.

Show answer
Option (a) is correct.

Explanation: R is true: by definition, a matrix B is symmetric if and only if B' = B.

To verify A, let B = A + A'. Then B' = (A + A')' = A' + (A')' = A' + A = A + A' = B. Since B' = B, the matrix (A + A') is symmetric. Thus R is the correct explanation of A.
Q13Short Answer2 marks

If A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A² − 5A + 7I = O.

Show answer
A² = A · A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

= \begin{bmatrix} 3×3 + 1×(−1) & 3×1 + 1×2 \\ (−1)×3 + 2×(−1) & (−1)×1 + 2×2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ −5 & 3 \end{bmatrix} …(i)

5A = \begin{bmatrix} 15 & 5 \\ −5 & 10 \end{bmatrix},\quad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} …(ii)

Now, A² − 5A + 7I

= \begin{bmatrix} 8 & 5 \\ −5 & 3 \end{bmatrix} − \begin{bmatrix} 15 & 5 \\ −5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

= \begin{bmatrix} 8−15+7 & 5−5+0 \\ −5+5+0 & 3−10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

∴ A² − 5A + 7I = O. Hence proved.
Q14Short Answer2 marks

If A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A² − 5A + 7I = O, where I is the identity matrix of order 2 and O is the zero matrix of order 2.

Show answer
Given: A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

Step 1: Compute A².

A² = A · A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}

= \begin{bmatrix} 3×3 + 1×(−1) & 3×1 + 1×2 \\ (−1)×3 + 2×(−1) & (−1)×1 + 2×2 \end{bmatrix}

= \begin{bmatrix} 9 − 1 & 3 + 2 \\ −3 − 2 & −1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ −5 & 3 \end{bmatrix}

Step 2: Compute A² − 5A + 7I.

A² − 5A + 7I = \begin{bmatrix} 8 & 5 \\ −5 & 3 \end{bmatrix} − 5\begin{bmatrix} 3 & 1 \\ −1 & 2 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

= \begin{bmatrix} 8 & 5 \\ −5 & 3 \end{bmatrix} + \begin{bmatrix} −15 & −5 \\ 5 & −10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

= \begin{bmatrix} 8 − 15 + 7 & 5 − 5 + 0 \\ −5 + 5 + 0 & 3 − 10 + 7 \end{bmatrix}

= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

∴ L.H.S. = R.H.S. Hence proved.
Q15Short Answer2 marks

If , find the value of .

Show answer
Equal matrices ⟹ corresponding elements are equal.

Comparing the (1,1) elements:
3x − 1 = 8
⟹ 3x = 9
⟹ x = 3

Comparing the (2,2) elements:
2y + 3 = 9
⟹ 2y = 6
⟹ y = 3

∴ x + y = 3 + 3 = 6
Q16Short Answer2 marks

If A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}, find the matrix B such that A + B = 2I, where I is the identity matrix of order 2.

Show answer
Given: A + B = 2I

⟹ B = 2I − A

⟹ B = 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} − \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}

⟹ B = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} − \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}

∴ B = \begin{bmatrix} 0 & -3 \\ 1 & -2 \end{bmatrix}
Q17Short Answer3 marks

Express the matrix as the sum of a symmetric matrix and a skew-symmetric matrix . Write and explicitly.

Show answer
Any square matrix can be written as the sum of a symmetric and a skew-symmetric matrix using:

P = ½(A + A') and Q = ½(A − A')

Step 1: Find A'.

A = [3, −2, 1; 4, 0, −3; −1, 5, 2]

∴ A' = [3, 4, −1; −2, 0, 5; 1, −3, 2]

Step 2: Find P = ½(A + A').

A + A' = [3+3, −2+4, 1+(−1); 4+(−2), 0+0, −3+5; −1+1, 5+(−3), 2+2]

⟹ A + A' = [6, 2, 0; 2, 0, 2; 0, 2, 4]

∴ P = ½(A + A') = [3, 1, 0; 1, 0, 1; 0, 1, 2]

Verification: P' = [3, 1, 0; 1, 0, 1; 0, 1, 2] = P ✓ (P is symmetric)

Step 3: Find Q = ½(A − A').

A − A' = [3−3, −2−4, 1−(−1); 4−(−2), 0−0, −3−5; −1−1, 5−(−3), 2−2]

⟹ A − A' = [0, −6, 2; 6, 0, −8; −2, 8, 0]

∴ Q = ½(A − A') = [0, −3, 1; 3, 0, −4; −1, 4, 0]

Verification: Q' = [0, 3, −1; −3, 0, 4; 1, −4, 0] = −Q ✓ (Q is skew-symmetric)

Step 4: Verify A = P + Q.

P + Q = [3+0, 1+(−3), 0+1; 1+3, 0+0, 1+(−4); 0+(−1), 1+4, 2+0]

= [3, −2, 1; 4, 0, −3; −1, 5, 2] = A ✓

∴ A = P + Q, where

P = [3, 1, 0; 1, 0, 1; 0, 1, 2] (symmetric) and Q = [0, −3, 1; 3, 0, −4; −1, 4, 0] (skew-symmetric).
Q18Short Answer3 marks

If A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} and f(x) = x² - 6x + 17, find f(A). Hence show that f(A) = 6I, where I is the identity matrix of order 2.

Show answer
Given: A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} and f(x) = x² − 6x + 17.

Then f(A) = A² − 6A + 17I.

Step 1: Compute A².

A² = A × A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}

= \begin{bmatrix} (2)(2)+(−1)(3) & (2)(−1)+(−1)(4) \\ (3)(2)+(4)(3) & (3)(−1)+(4)(4) \end{bmatrix}

= \begin{bmatrix} 4−3 & −2−4 \\ 6+12 & −3+16 \end{bmatrix}

= \begin{bmatrix} 1 & −6 \\ 18 & 13 \end{bmatrix} ...(i)

Step 2: Compute 6A.

6A = 6 \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 12 & −6 \\ 18 & 24 \end{bmatrix} ...(ii)

Step 3: Compute 17I.

17I = 17 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix} ...(iii)

Step 4: Compute f(A) = A² − 6A + 17I using (i), (ii) and (iii).

f(A) = \begin{bmatrix} 1 & −6 \\ 18 & 13 \end{bmatrix} − \begin{bmatrix} 12 & −6 \\ 18 & 24 \end{bmatrix} + \begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}

= \begin{bmatrix} 1−12+17 & −6+6+0 \\ 18−18+0 & 13−24+17 \end{bmatrix}

= \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

Now, 6I = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}

Since f(A) = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = 6I,

∴ f(A) = 6I. Hence proved.
Q19Short Answer3 marks

If A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}, verify that A² − 6A + 11I = O, where I is the 2×2 identity matrix and O is the 2×2 zero matrix.

Show answer
Given: A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}

Step 1: Compute A².

A² = A × A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}

A² = \begin{pmatrix} (2)(2)+(−1)(3) & (2)(−1)+(−1)(4) \\ (3)(2)+(4)(3) & (3)(−1)+(4)(4) \end{pmatrix}

⟹ A² = \begin{pmatrix} 4−3 & −2−4 \\ 6+12 & −3+16 \end{pmatrix} = \begin{pmatrix} 1 & −6 \\ 18 & 13 \end{pmatrix}

Step 2: Compute 6A.

6A = 6 \begin{pmatrix} 2 & −1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 12 & −6 \\ 18 & 24 \end{pmatrix}

Step 3: Compute 11I.

11I = 11 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}

Step 4: Compute A² − 6A + 11I.

A² − 6A + 11I = \begin{pmatrix} 1 & −6 \\ 18 & 13 \end{pmatrix} − \begin{pmatrix} 12 & −6 \\ 18 & 24 \end{pmatrix} + \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}

⟹ = \begin{pmatrix} 1−12+11 & −6+6+0 \\ 18−18+0 & 13−24+11 \end{pmatrix}

⟹ = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O

∴ A² − 6A + 11I = O. Hence proved.
Q20Short Answer3 marks

If A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}, show that A + A' = 2\cos\theta \cdot I, where I is the identity matrix of order 2. Hence find the value of \theta \in [0, 2\pi) for which A + A' = \sqrt{3}\, I.

Show answer
Given: A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

Then, A' = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}

Now,
A + A' = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} + \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}

= \begin{bmatrix} 2\cos\theta & 0 \\ 0 & 2\cos\theta \end{bmatrix}

= 2\cos\theta \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

= 2\cos\theta \cdot I

∴ A + A' = 2\cos\theta \cdot I Hence proved.

Now, for A + A' = √3 I:

2\cos\theta \cdot I = √3 I

⟹ 2\cos\theta = √3

⟹ \cos\theta = √3/2

∵ \theta \in [0, 2\pi) and \cos\theta = √3/2 > 0, \theta lies in the first or fourth quadrant.

∴ \theta = \pi/6 or \theta = 11\pi/6
Q21Short Answer3 marks

If A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}, show that A² − 5A + 7I = O, where I is the identity matrix of order 2 and O is the zero matrix of order 2. Hence find A⁻¹.

Show answer
Given: A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}

Step 1: Compute A².

A² = A · A = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}

= \begin{pmatrix} (2)(2)+(-1)(1) & (2)(-1)+(-1)(3) \\ (1)(2)+(3)(1) & (1)(-1)+(3)(3) \end{pmatrix}

= \begin{pmatrix} 4-1 & -2-3 \\ 2+3 & -1+9 \end{pmatrix}

= \begin{pmatrix} 3 & -5 \\ 5 & 8 \end{pmatrix}

Step 2: Compute A² − 5A + 7I.

A² − 5A + 7I

= \begin{pmatrix} 3 & -5 \\ 5 & 8 \end{pmatrix} − 5\begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} + 7\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

= \begin{pmatrix} 3 & -5 \\ 5 & 8 \end{pmatrix} + \begin{pmatrix} -10 & 5 \\ -5 & -15 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix}

= \begin{pmatrix} 3-10+7 & -5+5+0 \\ 5-5+0 & 8-15+7 \end{pmatrix}

= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O

∴ A² − 5A + 7I = O. Hence proved.

Step 3: Find A⁻¹ using the relation.

From A² − 5A + 7I = O, pre-multiplying both sides by A⁻¹,

A⁻¹(A² − 5A + 7I) = A⁻¹ · O

⟹ A − 5I + 7A⁻¹ = O

⟹ 7A⁻¹ = 5I − A

⟹ A⁻¹ = (1/7)(5I − A)

= (1/7)\left[5\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} − \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}\right]

= (1/7)\begin{pmatrix} 5-2 & 0+1 \\ 0-1 & 5-3 \end{pmatrix}

= (1/7)\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}

∴ A⁻¹ = (1/7)\begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}
Q22Short Answer3 marks

Let A = [aᵢⱼ] be a 3×3 matrix where aᵢⱼ = 2i − j. Express A as the sum of a symmetric matrix P and a skew-symmetric matrix Q. Write only the matrices P and Q explicitly.

Show answer
First, construct A using aᵢⱼ = 2i − j.

For i = 1, 2, 3 and j = 1, 2, 3:

a₁₁ = 2(1)−1 = 1, a₁₂ = 2(1)−2 = 0, a₁₃ = 2(1)−3 = −1
a₂₁ = 2(2)−1 = 3, a₂₂ = 2(2)−2 = 2, a₂₃ = 2(2)−3 = 1
a₃₁ = 2(3)−1 = 5, a₃₂ = 2(3)−2 = 4, a₃₃ = 2(3)−3 = 3

∴ A = [1, 0, −1; 3, 2, 1; 5, 4, 3] …(i)

A' = [1, 3, 5; 0, 2, 4; −1, 1, 3] …(ii)

Using the result: every square matrix A can be expressed as A = P + Q,
where P = ½(A + A') is symmetric and Q = ½(A − A') is skew-symmetric.

P = ½(A + A') = ½([1+1, 0+3, −1+5; 3+0, 2+2, 1+4; 5+(−1), 4+1, 3+3])

= ½([2, 3, 4; 3, 4, 5; 4, 5, 6])

∴ P = [1, 3/2, 2; 3/2, 2, 5/2; 2, 5/2, 3]

Q = ½(A − A') = ½([1−1, 0−3, −1−5; 3−0, 2−2, 1−4; 5−(−1), 4−1, 3−3])

= ½([0, −3, −6; 3, 0, −3; 6, 3, 0])

∴ Q = [0, −3/2, −3; 3/2, 0, −3/2; 3, 3/2, 0]

Verification: P + Q = [1+0, 3/2+(−3/2), 2+(−3); 3/2+3/2, 2+0, 5/2+(−3/2); 2+3, 5/2+3/2, 3+0] = [1, 0, −1; 3, 2, 1; 5, 4, 3] = A ✓

∴ A = P + Q, where P = [1, 3/2, 2; 3/2, 2, 5/2; 2, 5/2, 3] is symmetric and Q = [0, −3/2, −3; 3/2, 0, −3/2; 3, 3/2, 0] is skew-symmetric.
Q23Case-based4 marks

A school library tracks books borrowed by Classes 10, 11, and 12 across Fiction, Science, and History categories over two weeks. The borrowing data is stored in 3×3 matrices W₁ and W₂. The library uses matrix operations to analyse borrowing patterns and award reward points.

A school library is tracking the number of books borrowed across three categories — Fiction (F), Science (S), and History (H) — for two consecutive weeks. The data is recorded in matrix form as follows:

Week 1: W₁ = [[12, 8, 5], [7, 15, 10], [9, 6, 14]]

Week 2: W₂ = [[10, 12, 7], [5, 9, 13], [11, 8, 6]]

where rows represent Classes 10, 11, and 12 respectively, and columns represent Fiction, Science, and History.

Based on this information, answer the following sub-questions:

(i) Find the total number of books borrowed by each class over both weeks (compute W₁ + W₂).

(ii) The librarian wants to find the difference in borrowing between Week 1 and Week 2, i.e., W₁ − W₂. Compute this matrix and identify which class borrowed MORE Science books in Week 1 compared to Week 2.

(iii) If the library assigns a reward-point matrix R = [[2], [3], [1]] (points per book for Fiction, Science, History respectively), find the total reward points earned by each class in Week 1 by computing W₁ · R. Also verify the order of the resulting matrix.

Show answer
(i) W₁ + W₂:

W₁ + W₂ = [[12+10, 8+12, 5+7], [7+5, 15+9, 10+13], [9+11, 6+8, 14+6]]

∴ W₁ + W₂ = [[22, 20, 12], [12, 24, 23], [20, 14, 20]]

This represents the total books borrowed by Classes 10, 11, and 12 across all three categories over both weeks.

[1 mark]

(ii) W₁ − W₂:

W₁ − W₂ = [[12−10, 8−12, 5−7], [7−5, 15−9, 10−13], [9−11, 6−8, 14−6]]

∴ W₁ − W₂ = [[2, −4, −2], [2, 6, −3], [−2, −2, 8]]

The Science column (second column) of W₁ − W₂ gives: 8−12 = −4 for Class 10, 15−9 = 6 for Class 11, 6−8 = −2 for Class 12.

∵ The entry for Class 11 in the Science column is 6 > 0,

∴ Class 11 borrowed MORE Science books in Week 1 compared to Week 2.

[1 mark]

(iii) Here W₁ is of order 3×3 and R is of order 3×1.

∵ Number of columns of W₁ = Number of rows of R = 3,

∴ W₁ · R is defined and the resulting matrix is of order 3×1.

W₁ · R = [[12, 8, 5], [7, 15, 10], [9, 6, 14]] · [[2], [3], [1]]

For Class 10: 12×2 + 8×3 + 5×1 = 24 + 24 + 5 = 53

For Class 11: 7×2 + 15×3 + 10×1 = 14 + 45 + 10 = 69

For Class 12: 9×2 + 6×3 + 14×1 = 18 + 18 + 14 = 50

∴ W₁ · R = [[53], [69], [50]]

The resulting matrix is of order 3×1, which represents the total reward points earned — Class 10 earns 53 points, Class 11 earns 69 points, and Class 12 earns 50 points in Week 1.

[2 marks]
Q24Case-based4 marks

A school tuck-shop owner tracks sales of three snack items — Samosa (S), Burger (B), and Cold Drink (C) — over two days using matrices.

Day 1 sales (quantity): S = 40, B = 25, C = 30
Day 2 sales (quantity): S = 35, B = 30, C = 20

The price (in ₹) of each item is: S = 15, B = 30, C = 20
The cost price (in ₹) of each item is: S = 10, B = 20, C = 12

A school tuck-shop owner tracks sales of three snack items — Samosa (S), Burger (B), and Cold Drink (C) — over two days using matrices.

Day 1 sales (quantity): S = 40, B = 25, C = 30
Day 2 sales (quantity): S = 35, B = 30, C = 20

The price (in ₹) of each item is: S = 15, B = 30, C = 20
The cost price (in ₹) of each item is: S = 10, B = 20, C = 12

Represent the sales data as a 2×3 matrix Q, the selling price as a 3×1 column matrix P, and the cost price as a 3×1 column matrix C.

(i) Write down the matrices Q, P, and C. [1]
(ii) Compute the total revenue earned on each day using matrix multiplication. [1]
(iii) Find the total profit earned on each of the two days. Also find the total combined profit over both days. [2]

Show answer
(i) The sales data matrix Q (2×3), selling price matrix P (3×1), and cost price matrix C (3×1) are:

Q = [40 25 30]
[35 30 20]

P = [15]
[30]
[20]

C = [10]
[20]
[12]

∴ Q is a 2×3 matrix, P and C are 3×1 matrices.

(ii) Total revenue on each day = Q × P, which is a 2×1 matrix.

Q × P = [40 25 30] × [15] = [(40×15) + (25×30) + (30×20)]
[35 30 20] [30] [(35×15) + (30×30) + (20×20)]
[20]

Day 1 revenue = 600 + 750 + 600 = 1950
Day 2 revenue = 525 + 900 + 400 = 1825

∴ Q × P = [1950]
[1825]

∴ Total revenue on Day 1 = ₹1950 and on Day 2 = ₹1825.

(iii) Total cost on each day = Q × C

Q × C = [40 25 30] × [10] = [(40×10) + (25×20) + (30×12)]
[35 30 20] [20] [(35×10) + (30×20) + (20×12)]
[12]

Day 1 cost = 400 + 500 + 360 = 1260
Day 2 cost = 350 + 600 + 240 = 1190

∴ Q × C = [1260]
[1190]

Profit on each day = Revenue − Cost

Profit matrix = Q × P − Q × C = Q × (P − C)

P − C = [15−10] = [5]
[30−20] [10]
[20−12] [8]

Q × (P − C) = [40 25 30] × [5] = [(40×5) + (25×10) + (30×8)]
[35 30 20] [10] [(35×5) + (30×10) + (20×8)]
[8]

Day 1 profit = 200 + 250 + 240 = 690
Day 2 profit = 175 + 300 + 160 = 635

∴ Profit on Day 1 = ₹690, Profit on Day 2 = ₹635.

∴ Total combined profit over both days = 690 + 635 = ₹1325.
Q25Case-based4 marks

A school committee records student participation in Sports (S) and Cultural (C) activities over two terms. Matrix P gives the number of participants (rows = Terms, columns = Activities) and matrix Q gives the cost per student (in ₹) for each activity.

A school committee is preparing a report on two types of activities — Sports (S) and Cultural (C) — conducted over two terms (Term 1 and Term 2). The number of students who participated in each activity per term is recorded in a matrix P, and the cost (in ₹) per student for each activity is recorded in matrix Q, where:

P = [[40, 25], [50, 30]] (rows: Term 1, Term 2; columns: S, C)

Q = [[150], [200]] (rows: S, C; column: cost per student)

(i) Find the matrix PQ representing the total activity expenditure for each term.
(ii) Find the total expenditure incurred by the school across both terms.
(iii) The committee decides to give a 10% discount on Sports cost and a 5% discount on Cultural cost in Term 2 only. Write the revised cost matrix Q' for Term 2 and find the revised expenditure for Term 2.

Show answer
(i) Finding PQ:

P = [[40, 25], [50, 30]], Q = [[150], [200]]

Order of P is 2×2 and order of Q is 2×1, so PQ is a 2×1 matrix.

PQ = [[40×150 + 25×200], [50×150 + 30×200]]

= [[6000 + 5000], [7500 + 6000]]

∴ PQ = [[11000], [13500]]

This means Term 1 expenditure = ₹11,000 and Term 2 expenditure = ₹13,500.

(ii) Total expenditure across both terms:

Total = ₹11,000 + ₹13,500

∴ Total expenditure = ₹24,500

(iii) Revised cost matrix for Term 2:

Discount on Sports = 10%, so revised Sports cost = 150 − (10/100)×150 = 150 − 15 = ₹135

Discount on Cultural = 5%, so revised Cultural cost = 200 − (5/100)×200 = 200 − 10 = ₹190

∴ Q' = [[135], [190]]

Revised expenditure for Term 2:

Row 2 of P = [50, 30]

Revised expenditure = 50×135 + 30×190

= 6750 + 5700

∴ Revised Term 2 expenditure = ₹12,450
Q26Case-based4 marks

A data scientist models a transformation on ℝ² using the matrix A = [[2, -1], [3, -2]]. During analysis, she observes that repeated application of this transformation has a special property.

A data scientist models a transformation on ℝ² using the matrix A = [[2, -1], [3, -2]]. During analysis, she observes that repeated application of this transformation has a special property.

(i) Compute A². What do you notice about A²? [1 mark]

(ii) Using the result of (i), find A¹⁰⁰. [1 mark]

(iii) If B = A² - 3A + 2I, find the matrix B. Hence determine whether B is singular or non-singular, justifying your answer. [2 marks]

Show answer
(i) Computing A²:

A = [[2, -1], [3, -2]]

A² = A · A

Element (1,1): (2)(2) + (−1)(3) = 4 − 3 = 1
Element (1,2): (2)(−1) + (−1)(−2) = −2 + 2 = 0
Element (2,1): (3)(2) + (−2)(3) = 6 − 6 = 0
Element (2,2): (3)(−1) + (−2)(−2) = −3 + 4 = 1

∴ A² = [[1, 0], [0, 1]] = I

Observation: A² = I, i.e., A is an involutory matrix (A is its own inverse).

[1 mark]

(ii) Finding A¹⁰⁰:

Since A² = I,

A¹⁰⁰ = (A²)⁵⁰ = I⁵⁰ = I

∴ A¹⁰⁰ = [[1, 0], [0, 1]] = I

[1 mark]

(iii) Finding B = A² − 3A + 2I:

Using A² = I,

B = I − 3A + 2I = 3I − 3A

Now, 3I = [[3, 0], [0, 3]] and 3A = [[6, -3], [9, -6]]

B = [[3, 0], [0, 3]] − [[6, -3], [9, -6]]

B = [[3−6, 0−(−3)], [0−9, 3−(−6)]]

B = [[−3, 3], [−9, 9]]

Now, |B| = (−3)(9) − (3)(−9) = −27 + 27 = 0

∵ |B| = 0,

∴ B is a singular matrix. Since |B| = 0, B⁻¹ does not exist, confirming B is singular.

[2 marks]
Q27Case-based4 marks

A factory uses two types of raw materials, Material X and Material Y, to produce two products, Product P and Product Q. The quantity (in kg) of each material required per unit of each product is given by the matrix A = [[3, 1], [2, 4]], where rows represent Material X and Material Y respectively, and columns represent Product P and Product Q respectively. The factory receives an order where the total requirement is 11 kg of Material X and 22 kg of Material Y.

A factory uses two types of raw materials, Material X and Material Y, to produce two products, Product P and Product Q. The quantity (in kg) of each material required per unit of each product is given by the matrix:

A = [[3, 1], [2, 4]]

where rows represent Material X and Material Y respectively, and columns represent Product P and Product Q respectively.

The factory receives an order to produce a batch where the total requirement is 11 kg of Material X and 22 kg of Material Y.

(i) Write the system of linear equations representing the total material requirements if x units of Product P and y units of Product Q are to be produced. [1]
(ii) Express this system in the matrix form AX = B, identifying matrices A, X and B clearly. [1]
(iii) Find A⁻¹ and hence determine the number of units of Product P and Product Q to be produced to exactly meet the material requirements. [2]

Show answer
(i) Let x = number of units of Product P and y = number of units of Product Q to be produced.

Material X requirement: 3x + y = 11
Material Y requirement: 2x + 4y = 22

∴ The required system of linear equations is:
3x + y = 11 ...(i)
2x + 4y = 22 ...(ii)

[1 mark]

(ii) The matrix form AX = B is:

[[3, 1], [2, 4]] · [[x], [y]] = [[11], [22]]

where A = [[3, 1], [2, 4]], X = [[x], [y]], B = [[11], [22]]

[1 mark]

(iii) Finding A⁻¹:

|A| = (3)(4) − (1)(2) = 12 − 2 = 10 ≠ 0

∴ A⁻¹ exists.

adjA = [[4, −1], [−2, 3]]

∴ A⁻¹ = (1/|A|) · adjA = (1/10) · [[4, −1], [−2, 3]]

Since AX = B ⟹ X = A⁻¹B,

[[x], [y]] = (1/10) · [[4, −1], [−2, 3]] · [[11], [22]]

= (1/10) · [[(4)(11) + (−1)(22)], [(−2)(11) + (3)(22)]]

= (1/10) · [[44 − 22], [−22 + 66]]

= (1/10) · [[22], [44]]

= [[22/10], [44/10]]

= [[2.2], [4.4]]

Since units of production must be whole numbers, checking integer solutions from equations (i) and (ii):

Dividing equation (ii) by 2: x + 2y = 11 ...(iii)
From (i): 3x + y = 11 ...(i)

From (iii): x = 11 − 2y. Substituting in (i):
3(11 − 2y) + y = 11
33 − 6y + y = 11
−5y = −22
y = 22/5

∵ The material requirements (11 kg and 22 kg) yield non-integer solutions, the exact solution is:

x = [[x], [y]] = [[11/5], [22/5]]

Verification: 3(11/5) + (22/5) = 33/5 + 22/5 = 55/5 = 11 ✓
2(11/5) + 4(22/5) = 22/5 + 88/5 = 110/5 = 22 ✓

∴ A⁻¹ = (1/10)[[4, −1], [−2, 3]], and the factory must produce x = 11/5 units of Product P and y = 22/5 units of Product Q to exactly meet the material requirements.

[2 marks]
Q28Case-based4 marks

A school treasurer models monthly fund allocations using a 2×2 matrix A = [[4, 6], [2, 5]], and needs to find individual donor contributions using the matrix equation AX = B, where B = [[26], [19]].

A school treasurer is managing two budget accounts — a Sports Fund and a Cultural Fund. The combined allocation matrix for two consecutive months is represented as:

A = [[4, 6], [2, 5]]

where the rows represent Month 1 and Month 2, and the columns represent Sports Fund (in ₹ thousands) and Cultural Fund (in ₹ thousands) respectively.

The treasurer needs to recover the original monthly contributions from two donors P and Q. The total contribution by donor P in Month 1 was ₹26 thousand and in Month 2 was ₹19 thousand. The contributions of each donor per fund per month are given by the matrix equation AX = B, where X = [[x], [y]] represents the per-unit contributions of donors P and Q, and B = [[26], [19]].

(i) Find the determinant of A and verify that A⁻¹ exists.
(ii) Find A⁻¹.
(iii) Using A⁻¹, solve for x and y, and interpret what x and y represent in this context.

Show answer
(i) Finding |A|:

A = [[4, 6], [2, 5]]

|A| = (4)(5) − (6)(2) = 20 − 12 = 8

∵ |A| = 8 ≠ 0, ∴ A⁻¹ exists.

[1 mark]

(ii) Finding A⁻¹:

The cofactor matrix of A is:

C₁₁ = 5, C₁₂ = −2, C₂₁ = −6, C₂₂ = 4

adj A = [[5, −6], [−2, 4]]

∴ A⁻¹ = (1/|A|) · adj A = (1/8) · [[5, −6], [−2, 4]]

⟹ A⁻¹ = [[(5/8), (−6/8)], [(−2/8), (4/8)]] = [[(5/8), (−3/4)], [(−1/4), (1/2)]]

[1½ marks]

(iii) Solving AX = B using X = A⁻¹B:

X = A⁻¹B = (1/8) · [[5, −6], [−2, 4]] · [[26], [19]]

Computing the product:

Row 1: (5)(26) + (−6)(19) = 130 − 114 = 16

Row 2: (−2)(26) + (4)(19) = −52 + 76 = 24

∴ X = (1/8) · [[16], [24]] = [[2], [3]]

∴ x = 2 and y = 3

Interpretation: x = 2 (₹2 thousand) is the per-unit contribution of donor P, and y = 3 (₹3 thousand) is the per-unit contribution of donor Q, towards each fund per month.

[1½ marks]
Q29Case-based4 marks

A data analyst models a one-way communication network between three cities P, Q, R using a 3×3 matrix C = [[0,1,1],[1,0,1],[0,1,0]], where cᵢⱼ = 1 denotes a direct link from city i to city j. The matrix C² gives two-step connections and D = C + C² gives total connections up to two steps.

A data analyst is modelling a communication network between three cities P, Q, and R. The direct connections (one-way) between cities are represented by a 3×3 matrix C, where the entry cᵢⱼ = 1 if there is a direct link from city i to city j, and cᵢⱼ = 0 otherwise (with no self-loops). The analyst uses the matrix

C = [[0, 1, 1], [1, 0, 1], [0, 1, 0]]

where rows/columns 1, 2, 3 represent cities P, Q, R respectively.

The matrix C² gives the number of two-step connections between cities.

(i) Compute C² and interpret the entry (C²)₁₂ in the context of the problem. [1]

(ii) The analyst claims that the total number of ways to travel from city P to city R in exactly two steps equals (C²)₁₃. Verify this claim by listing all possible two-step routes from P to R, and confirm it matches (C²)₁₃. [1]

(iii) The matrix D = C + C² represents total connections (direct and two-step) between cities. Compute D and determine whether D is a symmetric matrix. If not, explain what the symmetry (or lack of it) tells us about the network. [2]

Show answer
(i) Computing C²:

C = [[0, 1, 1], [1, 0, 1], [0, 1, 0]]

C² = C × C

Element (C²)₁₁ = (0)(0) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1
Element (C²)₁₂ = (0)(1) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1
Element (C²)₁₃ = (0)(1) + (1)(1) + (1)(0) = 0 + 1 + 0 = 1
Element (C²)₂₁ = (1)(0) + (0)(1) + (1)(0) = 0 + 0 + 0 = 0
Element (C²)₂₂ = (1)(1) + (0)(0) + (1)(1) = 1 + 0 + 1 = 2
Element (C²)₂₃ = (1)(1) + (0)(1) + (1)(0) = 1 + 0 + 0 = 1
Element (C²)₃₁ = (0)(0) + (1)(1) + (0)(0) = 0 + 1 + 0 = 1
Element (C²)₃₂ = (0)(1) + (1)(0) + (0)(1) = 0 + 0 + 0 = 0
Element (C²)₃₃ = (0)(1) + (1)(1) + (0)(0) = 0 + 1 + 0 = 1

∴ C² = [[1, 1, 1], [0, 2, 1], [1, 0, 1]]

Interpretation: (C²)₁₂ = 1 means there is exactly 1 two-step route from city P to city Q (i.e., travelling via one intermediate city).

(ii) Listing all two-step routes from P (city 1) to R (city 3):

A two-step route P → X → R requires a direct link P→X (i.e., c₁ₓ = 1) and a direct link X→R (i.e., cₓ₃ = 1).

From row 1 of C: direct links from P go to Q (c₁₂ = 1) and R (c₁₃ = 1). So X ∈ {Q, R}.

Check X = Q: c₁₂ = 1 ✓ and c₂₃ = 1 ✓ ⟹ Route P → Q → R is valid.
Check X = R: c₁₃ = 1 ✓ and c₃₃ = 0 ✗ ⟹ Route P → R → R is invalid (no self-loop).

∴ There is exactly 1 two-step route from P to R, namely P → Q → R.

This matches (C²)₁₃ = 1. ∴ The analyst's claim is verified.

(iii) Computing D = C + C²:

C = [[0, 1, 1], [1, 0, 1], [0, 1, 0]]
C² = [[1, 1, 1], [0, 2, 1], [1, 0, 1]]

D = C + C² = [[0+1, 1+1, 1+1], [1+0, 0+2, 1+1], [0+1, 1+0, 0+1]]

∴ D = [[1, 2, 2], [1, 2, 2], [1, 1, 1]]

Checking symmetry: For D to be symmetric, we need D = D', i.e., dᵢⱼ = dⱼᵢ for all i, j.

d₁₂ = 2 but d₂₁ = 1 ⟹ d₁₂ ≠ d₂₁.

∴ D is NOT a symmetric matrix.

Interpretation: The lack of symmetry in D means the network is directed (one-way). For instance, dᵢⱼ ≠ dⱼᵢ tells us that the total number of ways to reach city j from city i (directly or in two steps) is not the same as the total number of ways to reach city i from city j. This reflects the asymmetric nature of the one-way communication links in the network.
Q30Case-based4 marks

A school is organising a three-day Science Fair. Three stalls — Physics (P), Chemistry (C) and Biology (B) — sell entry tickets priced at ₹20, ₹30 and ₹25 respectively. The number of tickets sold on each day is recorded in matrix T (rows: Day 1, Day 2, Day 3; columns: Physics, Chemistry, Biology) and the price per ticket in ₹ is recorded in column matrix P (rows: Physics, Chemistry, Biology):

T = [[120, 80, 100], [150, 60, 90], [110, 95, 75]], P = [[20], [30], [25]]

A school is organising a three-day Science Fair. Three stalls — Physics (P), Chemistry (C) and Biology (B) — sell entry tickets priced at ₹20, ₹30 and ₹25 respectively.

The number of tickets sold on each day is recorded in matrix T and the price per ticket (in ₹) is recorded in matrix P:

T = [[120, 80, 100], [150, 60, 90], [110, 95, 75]] (rows: Day 1, Day 2, Day 3; columns: Physics, Chemistry, Biology)

P = [[20], [30], [25]] (rows: Physics, Chemistry, Biology)

Based on the above information, answer the following:

(i) Write the order of matrix T and matrix P. Hence state the order of the product matrix TP. [1 mark]

(ii) Compute the total revenue collected on Day 1 by evaluating the first row of the product TP. [1 mark]

(iii) Find the complete product matrix TP to determine the total revenue collected on each of the three days. Hence find the day on which the maximum revenue was collected. [2 marks]

Show answer
(i)
Matrix T has 3 rows and 3 columns, so order of T is 3 × 3.
Matrix P has 3 rows and 1 column, so order of P is 3 × 1.
Since T is 3 × 3 and P is 3 × 1, the product TP is defined and its order is 3 × 1.

∴ Order of T = 3 × 3, order of P = 3 × 1, and order of TP = 3 × 1.

(ii)
The first row of TP is obtained by multiplying the first row of T with the column matrix P:

Revenue on Day 1 = [120 80 100] × [[20], [30], [25]]
= 120 × 20 + 80 × 30 + 100 × 25
= 2400 + 2400 + 2500
= 7300

∴ Total revenue collected on Day 1 = ₹7300.

(iii)
TP = [[120, 80, 100], [150, 60, 90], [110, 95, 75]] × [[20], [30], [25]]

Day 1: 120 × 20 + 80 × 30 + 100 × 25 = 2400 + 2400 + 2500 = 7300
Day 2: 150 × 20 + 60 × 30 + 90 × 25 = 3000 + 1800 + 2250 = 7050
Day 3: 110 × 20 + 95 × 30 + 75 × 25 = 2200 + 2850 + 1875 = 6925

∴ TP = [[7300], [7050], [6925]]

Comparing the revenues: ₹7300 > ₹7050 > ₹6925.

∴ Maximum revenue of ₹7300 was collected on Day 1.

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