Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
A matrix is an ordered rectangular array of numbers (or functions) arranged in rows and columns and enclosed in brackets. The numbers are its elements or entries. Matrices are used to store and combine data such as sales figures, prices and coefficients of equations.
A matrix with m rows and n columns has order m × n (read 'm by n'). It has mn elements. The entry in row i and column j is written aij, and the matrix is A = [aij]m×n.
Two matrices A = [aij] and B = [bij] are equal if they have the same order and aij = bij for every i and j.
Example: x + y41x − y = 6412 gives x + y = 6 and x − y = 2, so x = 4 and y = 2.
Equality of matrices gives one equation for each entry. Matrices of different orders can never be equal.
| Type | Description | Example |
|---|---|---|
| Column matrix | Only one column (order m × 1) | 147 |
| Row matrix | Only one row (order 1 × n) | 2−53 |
| Square matrix | Same number of rows and columns (m = n) | 1234 |
| Rectangular matrix | Rows and columns not equal | 123456 |
| Type | Condition | Example |
|---|---|---|
| Diagonal | aij = 0 whenever i ≠ j | 2000−10005 |
| Scalar | Diagonal, with all diagonal entries equal | 300030003 |
| Identity (I) | Diagonal, with every diagonal entry 1 | 1001 |
| Zero (O) | Every entry is 0 (any order) | 000000 |
You can add or subtract matrices only when they have the same order. Do it entry by entry:
Example: 1234 + 0110 = 1344.
If k is a number, kA is found by multiplying every entry of A by k. The matrix −A = (−1)A, and A − B = A + (−B).
Example: 31234 = 36912.
AB is defined only when the number of columns of A equals the number of rows of B. If A is m × n and B is n × p, then AB is m × p.
The entry in row i and column j of AB is the sum of products of row i of A with column j of B:
12340110 = 1·0 + 2·11·1 + 2·03·0 + 4·13·1 + 4·0 = 2143.
But 01101234 = 3412, which is different.
Before you multiply, write the orders side by side, for example (3 × 2)(2 × 4). The two inner numbers must match, and the outer numbers give the order of the result (3 × 4).
The transpose of A, written A′ (or Aᵀ), is obtained by interchanging its rows and columns. If A is m × n then A′ is n × m, and the (i, j) entry of A′ is the (j, i) entry of A.
Example: if A = 123456, then A′ = 135246.
Check: with A = 123456 and B = 102310, (BA)′ and A′B′ are equal. The product reverses because the orders must still match: A′ is 2 × 3 and B′ is 3 × 2, so only A′B′ can be the transpose of BA.
A square matrix A is symmetric if A′ = A, that is aij = aji for all i, j. It is skew symmetric if A′ = −A, that is aij = −aji for all i, j.
The first part P = ½(A + A′) is symmetric and the second part Q = ½(A − A′) is skew symmetric. The split is unique.
Example: A = 2375 gives P = 2555 and Q = 0−220.
A square matrix A of order n is invertible if there is a square matrix B of the same order with
B is called the inverse of A and is written A⁻¹. If no such B exists, A is not invertible.
Example: A = 1−1−12 and B = 2111 give AB = 1001 = I (and BA = I too), so B = A⁻¹.
To check an inverse you must have AB = I and BA = I. For square matrices of the same order, either one implies the other.
This chapter only tests whether a given B is an inverse. In Chapter 4 you will find A⁻¹ directly with determinants and the adjoint, and in elementary row operations you will use A = IA.
| Result | Condition |
|---|---|
| (AB)ij = Σ aik bkj | columns of A = rows of B |
| (A′)′ = A | always |
| (A + B)′ = A′ + B′ | same order |
| (AB)′ = B′A′ | AB defined |
| A = ½(A + A′) + ½(A − A′) | A square |
| (AB)⁻¹ = B⁻¹A⁻¹ | A, B invertible, same order |
| AI = IA = A | I of the same order |
Try each problem first, then open the solution to check your working.
Construct the 2 × 3 matrix A = [aij] whose entries are given by aij = 2i − 3j.
Use: Substitute i = 1, 2 and j = 1, 2, 3
Row 1 (i = 1): 2 − 3j = −1, −4, −7. Row 2 (i = 2): 4 − 3j = 1, −2, −5.
A = −1−4−71−2−5
Answer: −1−4−71−2−5
How many 2 × 2 matrices can be formed with every entry from {0, 1, 2}?
Use: Each entry is chosen independently
There are 4 entries, each with 3 choices, so 3⁴ = 81.
Answer: 81
Find x, y and z if x + y4z − 12x − y = 5461.
Use: Equate corresponding entries
x + y = 5 and 2x − y = 1. Adding: 3x = 6, so x = 2 and y = 3.
z − 1 = 6 gives z = 7.
Answer: x = 2, y = 3, z = 7
A is 3 × 2, B is 2 × 4 and C is 4 × 1. Which of AB, BA, ABC, CB are defined? Give the order when defined.
Use: Inner orders must match
AB: (3 × 2)(2 × 4) is defined and has order 3 × 4.
BA: (2 × 4)(3 × 2) needs 4 = 3, so it is not defined.
ABC: (3 × 4)(4 × 1) is defined and has order 3 × 1.
CB: (4 × 1)(2 × 4) needs 1 = 2, so it is not defined.
Answer: AB (3 × 4) and ABC (3 × 1) are defined; BA and CB are not
A 3 × 3 matrix has aij = 5 when i = j and aij = 0 otherwise. Another has aij = i when i = j and 0 otherwise. Name the type of each.
The first is 500050005, a scalar matrix (it is 5I).
The second is 100020003. Its diagonal entries are different, so it is diagonal but not scalar.
Answer: scalar (5I); diagonal but not scalar
Find a and b so that 2a345b371 is a symmetric matrix.
Use: aij = aji
a₁₂ = a₂₁ gives a = 4.
a₂₃ = a₃₂ gives b = 7. The pair a₁₃ = a₃₁ = 3 already agrees.
Check: the matrix is 243457371, equal to its transpose.
Answer: a = 4, b = 7
Find x, y and z so that 0x3−20yz−50 is skew symmetric.
Use: aij = −aji
x = −a₂₁ = −(−2) = 2.
3 = −z, so z = −3.
y = −a₃₂ = −(−5) = 5.
The matrix is then 023−205−3−50, and A′ = −A.
Answer: x = 2, y = 5, z = −3
If A = 1234 and B = 0−125, find 2A − 3B.
2A = 2468 and 3B = 0−3615.
2A − 3B = 270−7
Answer: 270−7
For A = 1234 and B = 0110, find AB and BA. Are they equal?
Use: Row times column
AB = 2143.
BA = 3412.
They are different. Multiplying by B on the right swaps the columns of A, and on the left swaps its rows.
Answer: AB = 2143, BA = 3412; not equal
If A = 01−10, find A² and A⁴. Hence find A¹⁰¹.
A² = −100−1 = −I.
A⁴ = (A²)² = (−I)² = I.
101 = 4 × 25 + 1, so A¹⁰¹ = (A⁴)²⁵ A = IA = A.
Answer: A² = −I, A⁴ = I, A¹⁰¹ = A
If A = 1101, find A² and A³, guess Aⁿ, and give A⁵.
A² = 1201 and A³ = 1301.
The top right entry grows by 1 each time, so Aⁿ = 1n01.
A⁵ = 1501.
Answer: Aⁿ = [[1, n], [0, 1]]; A⁵ = 1501
Find the matrix X if 2X + 1234 = 5016.
2X = 5016 − 1234 = 4−2−22.
X = 2−1−11
Answer: 2−1−11
For A = 1234, show that A² − 5A − 2I = O.
A² = 7101522 and 5A = 5101520.
A² − 5A = 2002 = 2I.
So A² − 5A − 2I = 2I − 2I = O.
Answer: shown
Shop 1 sells 30 pens and 12 notebooks; Shop 2 sells 20 pens and 40 notebooks. A pen costs ₹10 and a notebook ₹25. Use a matrix product to find the total sales of each shop.
Sales matrix 30122040 times price matrix 1025.
= 300 + 300200 + 1000 = 6001200.
Answer: Shop 1: ₹600; Shop 2: ₹1,200
For A = 120341 and B = 210101, verify (AB)′ = B′A′.
AB = 412303941, so (AB)′ = 439104231.
B′A′ = 211001104231 = 439104231. The two are equal.
Answer: verified
Express A = 2375 as the sum of a symmetric and a skew symmetric matrix.
Use: A = ½(A + A′) + ½(A − A′)
A′ = 2735.
P = ½(A + A′) = 2555 (symmetric).
Q = ½(A − A′) = 0−220 (skew symmetric).
P + Q = 2375 = A.
Answer: A = 2555 + 0−220
For A = 123456, find AA′ and show that it is symmetric. Why does this happen for every A?
AA′ = 51117112539173961, which equals its own transpose.
In general (AA′)′ = (A′)′A′ = AA′.
Answer: AA′ = 51117112539173961
If A = 1223, show that A² is symmetric.
A² = 58813, which is symmetric.
In general (A²)′ = (A′)² = A² when A′ = A.
Answer: shown
Show that B = 2111 is the inverse of A = 1−1−12.
Use: AB = BA = I
AB = 1001 = I and BA = 1001 = I.
Both products give I, so B = A⁻¹.
Answer: verified
A and B are invertible with A⁻¹ = 1101 and B⁻¹ = 1021. Find (AB)⁻¹.
Use: (AB)⁻¹ = B⁻¹A⁻¹
(AB)⁻¹ = B⁻¹A⁻¹ = 10211101 = 1123.
Answer: 1123
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Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) A has 3 rows and 4 columns, so the order is 3 × 4.
(ii) The number of elements is 3 × 4 = 12.
(iii) a₁₃ is in row 1, column 3: a₁₃ = 19. a₂₁ = 35. a₃₃ = −5. a₂₄ = 12. a₂₃ = 52.
Answer: 3 × 4; 12; a₁₃ = 19, a₂₁ = 35, a₃₃ = −5, a₂₄ = 12, a₂₃ = 5/2
Possible orders are the ordered pairs of natural numbers with product 24: (1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4).
For 13 elements (13 is prime): (1, 13) and (13, 1).
Answer: 1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4; for 13 elements: 1 × 13 and 13 × 1
Pairs with product 18: (1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3).
For 5 elements: (1, 5) and (5, 1).
Answer: 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3; for 5 elements: 1 × 5 and 5 × 1
(i) a₁₁ = 4/2 = 2, a₁₂ = 9/2, a₂₁ = 9/2, a₂₂ = 16/2 = 8. A = 292928
(ii) a₁₁ = 1, a₁₂ = 1/2, a₂₁ = 2, a₂₂ = 1. A = 11221
(iii) a₁₁ = 9/2, a₁₂ = 25/2, a₂₁ = 16/2 = 8, a₂₂ = 36/2 = 18. A = 92252818
Answer: see the three matrices above
(i) For i = 1: |−3 + j| = 2, 1, 0, 1 for j = 1..4. For i = 2: |−6 + j| = 5, 4, 3, 2. For i = 3: |−9 + j| = 8, 7, 6, 5. Halving each: 112012522321472352
(ii) For i = 1: 2 − j = 1, 0, −1, −2. For i = 2: 4 − j = 3, 2, 1, 0. For i = 3: 6 − j = 5, 4, 3, 2. 10−1−232105432
Answer: see the two matrices above
(i) Equal matrices have equal entries: y = 4, z = 3, x = 1.
(ii) x + y = 6, 5 + z = 5, xy = 8. So z = 0, and x, y are the roots of t² − 6t + 8 = 0, i.e. 2 and 4. So (x, y) = (2, 4) or (4, 2).
(iii) x + y + z = 9, x + z = 5, y + z = 7. Subtract the second from the first: y = 4. Then z = 7 − 4 = 3 and x = 5 − 3 = 2.
Answer: (i) x = 1, y = 4, z = 3; (ii) z = 0 and (x, y) = (2, 4) or (4, 2); (iii) x = 2, y = 4, z = 3
a − b = −1, 2a + c = 5, 2a − b = 0, 3c + d = 13.
Subtract the first from the third: a = 1. Then b = a + 1 = 2.
2a + c = 5 gives c = 3, and 3c + d = 13 gives d = 4.
Answer: a = 1, b = 2, c = 3, d = 4
A square matrix has equally many rows and columns.
Answer: (C) m = n
Equating entries: 3x + 7 = 0 gives x = −7/3. 5 = y − 2 gives y = 7, and y + 1 = 8 agrees with y = 7.
The last pair: 2 − 3x = 4 would need x = −2/3, which conflicts with x = −7/3.
So no x and y make all four entries equal.
Answer: (B) Not possible to find
A 3 × 3 matrix has 9 entries, each chosen in 2 ways, so the number is 2⁹ = 512.
Answer: (D) 512
(i) A + B = 3717
(ii) A − B = 115−3
(iii) 3A = 61296, so 3A − C = 8762
(iv) AB = 2·1 + 4·(−2)2·3 + 4·53·1 + 2·(−2)3·3 + 2·5 = −626−119
(v) BA = 1·2 + 3·31·4 + 3·2−2·2 + 5·3−2·4 + 5·2 = 1110112
Answer: see each part; note AB ≠ BA
(i) a + ab + b−b + ba + a = 2a2b02a
(ii) a² + b² + 2abb² + c² + 2bca² + c² − 2aca² + b² − 2ab = (a + b)²(b + c)²(a − c)²(a − b)²
Answer: (i) [[2a, 2b], [0, 2a]]; (ii) [[(a + b)², (b + c)²], [(a − c)², (a − b)²]]
(iii) 11110165215109
(iv) Each entry is cos²x + sin²x = 1: 1111
Answer: (iii) as above; (iv) [[1, 1], [1, 1]]
(i) a·a + b·b−ab + ba−ba + abb² + a² = a² + b²00a² + b²
(ii) 3 × 1 times 1 × 3 gives a 3 × 3 matrix: 2344686912
(iii) 2 × 2 times 2 × 3 gives 2 × 3: 1 − 42 − 63 − 22 + 64 + 96 + 3 = −3−418139
Answer: see each part
(iv) 3 × 3 times 3 × 3 gives 3 × 3. Row 1: 2 + 0 + 12 = 14; (-6) + 6 + 0 = 0; 10 + 12 + 20 = 42. Row 2: 3 + 0 + 15 = 18; (-9) + 8 + 0 = -1; 15 + 16 + 25 = 56. Row 3: 4 + 0 + 18 = 22; (-12) + 10 + 0 = -2; 20 + 20 + 30 = 70.
So the product is 1404218−15622−270.
(v) 3 × 2 times 2 × 3 gives 3 × 3. Row 1: 2 + (-1) = 1; 0 + 2 = 2; 2 + 1 = 3. Row 2: 3 + (-2) = 1; 0 + 4 = 4; 3 + 2 = 5. Row 3: (-1) + (-1) = -2; 0 + 2 = 2; (-1) + 1 = 0.
So the product is 123145−220.
(vi) 2 × 3 times 3 × 2 gives 2 × 2: 6 − 1 + 9−9 + 0 + 3−2 + 0 + 63 + 0 + 2 = 14−645.
Answer: see each part
A + B = 41−19273−14
B − C = −1−204−13120
A + (B − C) = 00−39−15211
(A + B) − C = 00−39−15211
The two results are equal, so A + (B − C) = (A + B) − C.
Answer: verified
3A = 235124762
5B = 235124762
3A − 5B = 000000000
Answer: the zero matrix of order 3 × 3
= cos²θcos θ sin θ−cos θ sin θcos²θ + sin²θ−sin θ cos θsin θ cos θsin²θ
= cos²θ + sin²θ00cos²θ + sin²θ = 1001
Answer: the identity matrix of order 2
(i) Adding: 2X = 10028, so X = 5014. Subtracting: 2Y = 4022, so Y = 2011.
(ii) Adding the two equations: 5(X + Y) = 4135. Subtracting the first from the second: X − Y = 0−5−55.
So X + Y = 4515351. Hence X = ½[(X + Y) + (X − Y)] = 25−125−1153 and Y = ½[(X + Y) − (X − Y)] = 25135145−2.
Answer: (i) X = [[5, 0], [1, 4]], Y = [[2, 0], [1, 1]]; (ii) as above
2X = 10−32 − 3214 = −2−2−4−2, so X = −1−1−2−1
Answer: X = [[−1, −1], [−2, −1]]
The left side is 2 + y612x + 2. Equating: 2 + y = 5 so y = 3, and 2x + 2 = 8 so x = 3.
Answer: x = 3, y = 3
2x + 3 = 9 gives x = 3. 2z − 3 = 15 gives z = 9. 2y + 0 = 12 gives y = 6. 2t + 6 = 18 gives t = 6.
Answer: x = 3, y = 6, z = 9, t = 6
2x − y = 10 and 3x + y = 5. Adding: 5x = 15, so x = 3. Then y = 5 − 9 = −4.
Answer: x = 3, y = −4
The right side is x + 4x + y + 6z + w − 12w + 3. Equating with 3x3y3z3w:
3x = x + 4 gives x = 2. 3w = 2w + 3 gives w = 3. 3y = x + y + 6 gives 2y = 8, so y = 4. 3z = z + w − 1 gives 2z = 2, so z = 1.
Answer: x = 2, y = 4, z = 1, w = 3
F(x)F(y) = cos x cos y − sin x sin y−cos x sin y − sin x cos y0sin x cos y + cos x sin y−sin x sin y + cos x cos y0001
= cos(x + y)−sin(x + y)0sin(x + y)cos(x + y)0001 = F(x + y).
Answer: proved
(i) Left side: 713334. Right side: 1653925. They are different.
(ii) Left side: 58140−11−101. Right side: −1−1−31006116. They are different.
Answer: shown
A² = 5−129−250−1−2
5A = 1005105155−50 and 6I = 600060006
A² − 5A + 6I = 1−1−3−1−1−10−544
Answer: 1−1−3−1−1−10−544
A² = 5082458013
A³ = A²·A = 210341282334055
A³ − 6A² + 7A + 2I = 210341282334055 − 3004812243048078 + 7014014714021 + 200020002 = 000000000.
Answer: proved
A² = 1−24−4. And kA − 2I = 3k − 2−2k4k−2k − 2.
Equating: 3k − 2 = 1 gives k = 1, and the other three entries (−2k = −2, 4k = 4, −2k − 2 = −4) agree.
Answer: k = 1
Let t = tan(α/2). Then cos α = (1 − t²)/(1 + t²) and sin α = 2t/(1 + t²).
(I − A) = 1t−t1. Multiplying: cos α + t sin α−sin α + t cos α−t cos α + sin αt sin α + cos α.
cos α + t sin α = (1 − t² + 2t²)/(1 + t²) = 1. −sin α + t cos α = (−2t + t − t³)/(1 + t²) = −t(1 + t²)/(1 + t²) = −t. Similarly the second row is t and 1.
So the product is 1−tt1 = I + A.
Answer: proved
Let ₹x go to the first bond and ₹(30000 − x) to the second. In matrix form: x30000 − x5/1007/100 = [interest].
The interest is (5x + 7(30000 − x))/100 = (210000 − 2x)/100.
(a) (210000 − 2x)/100 = 1800 gives 210000 − 2x = 180000, so x = 15000. Invest ₹15,000 in each bond.
(b) (210000 − 2x)/100 = 2000 gives 210000 − 2x = 200000, so x = 5000. Invest ₹5,000 in the first bond and ₹25,000 in the second.
Answer: (a) ₹15,000 at 5% and ₹15,000 at 7%; (b) ₹5,000 at 5% and ₹25,000 at 7%
Numbers of books (in dozens, times 12): chemistry 120, physics 96, economics 120. Prices: 80, 60, 40.
Total = 12096120806040 = [120·80 + 96·60 + 120·40] = [9600 + 5760 + 4800] = [20160].
Answer: ₹20,160
PY: P is p × k and Y is 3 × k. The product is defined only if k = 3, and then PY has order p × k = p × 3.
WY: W is n × 3 and Y is 3 × k, so the product is always defined and has order n × k = n × 3.
To add PY and WY they must have the same order: p × 3 and n × 3, so p = n.
Answer: (A) k = 3, p = n
X has order 2 × n and Z has order 2 × p = 2 × n, so 7X − 5Z is defined and has order 2 × n.
Answer: (B) 2 × n
(i) 5122−1
(ii) 12−13
(iii) −1√3255366−1
Answer: rows and columns interchanged, as shown
A + B = −53−2699−142, so (A + B)′ = −56−1394−292. A′ = −15−2271391, B′ = −411123−501, so A′ + B′ = −56−1394−292. Equal.
A − B = 318459−3−20, so (A − B)′ = 34−315−2890. And A′ − B′ = 34−315−2890. Equal.
Answer: verified
A = (A′)′ = 3−10421. A + B = 211544, so (A + B)′ = 251414. A′ + B′ = 251414. Equal.
A − B = 4−3−130−2, so (A − B)′ = 43−30−1−2 and A′ − B′ = 43−30−1−2. Equal.
Answer: verified
(A + 2B)′ = A′ + 2B′. B′ = −1102, so 2B′ = −2204.
(A + 2B)′ = −2312 + −2204 = −4516
Answer: −4516
(i) AB = −1214−8−4−363, so (AB)′ = −14−32−861−43. B′A′ = −1211−43 = −14−32−861−43. Equal.
(ii) AB = 00015721014, so (AB)′ = 01205100714. B′A′ = 157012 = 01205100714. Equal.
Answer: verified
(i) A′ = cos α−sin αsin αcos α. A′A = cos²α + sin²αcos α sin α − sin α cos αsin α cos α − cos α sin αsin²α + cos²α = I.
(ii) A′ = sin α−cos αcos αsin α. A′A = sin²α + cos²αsin α cos α − cos α sin αcos α sin α − sin α cos αcos²α + sin²α = I.
Answer: verified
(i) A′ = 1−15−121513 = A, so A is symmetric.
(ii) A′ = 0−1110−1−110 = −A, so A is skew symmetric (and its diagonal is zero).
Answer: shown
A′ = 1657.
(i) A + A′ = 2111114, which equals its transpose. Symmetric.
(ii) A − A′ = 0−110, whose transpose is 01−10 = −(A − A′). Skew symmetric.
Answer: verified
A′ = 0−a−ba0−cbc0 = −A, so A is skew symmetric.
½(A + A′) = ½ O = O, the zero matrix of order 3.
½(A − A′) = ½(2A) = A.
Answer: ½(A + A′) = O; ½(A − A′) = A
(i) A′ = 315−1.
P = ½(A + A′) = 333−1
Q = ½(A − A′) = 02−20
P is symmetric and Q is skew symmetric, and P + Q = 351−1 = A.
(ii) The matrix is already symmetric (A′ = A).
P = A and Q = 000000000, so A = A + O.
Answer: (i) as above; (ii) A is itself symmetric: A = A + O
(iii) A′ = 3−2−43−2−5−112.
P = ½(A + A′) = 312−5212−2−2−52−22
Q = ½(A − A′) = 05232−5203−32−30
P is symmetric and Q is skew symmetric, and P + Q = 33−1−2−21−4−52 = A.
(iv) A′ = 1−152.
P = ½(A + A′) = 1222
Q = ½(A − A′) = 03−30
P is symmetric and Q is skew symmetric, and P + Q = 15−12 = A.
Answer: as shown
(AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA).
So AB − BA is skew symmetric.
Answer: (A) Skew symmetric matrix
A + A′ = 2cos α002cos α = 1001, so 2cos α = 1 and cos α = 1/2.
Among the options only α = π/3 has cos α = 1/2.
Answer: (B) π/3
By definition, B is the inverse of A when AB = BA = I.
Answer: (D) AB = BA = I
Since A and B are symmetric, A′ = A and B′ = B. Then (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′ = BA − AB = −(AB − BA).
So AB − BA is skew symmetric.
Answer: proved
(B′AB)′ = B′A′(B′)′ = B′A′B.
If A is symmetric (A′ = A), this is B′AB, so B′AB is symmetric.
If A is skew symmetric (A′ = −A), this is −B′AB, so B′AB is skew symmetric.
Answer: proved
A′A = I means the columns of A have length 1 and are perpendicular to each other.
Column 1 is (0, x, x): x² + x² = 1, so 2x² = 1. Column 2 is (2y, y, −y): 4y² + y² + y² = 1, so 6y² = 1. Column 3 is (z, −z, z): 3z² = 1.
The dot products: col 1 · col 2 = 0 + xy − xy = 0, col 1 · col 3 = 0 − xz + xz = 0 and col 2 · col 3 = 2yz − yz − yz = 0, so they hold for all values.
Hence x = ±1/√2, y = ±1/√6, z = ±1/√3.
Answer: x = ±1/√2, y = ±1/√6, z = ±1/√3
First, 121120201102 = 1 + 4 + 12 + 0 + 00 + 2 + 2 = 624.
Then 62402x = [0 + 4 + 4x] = [4 + 4x].
Setting 4 + 4x = 0 gives x = −1.
Answer: x = −1
A² = 85−53
A² − 5A + 7I = 85−53 − 155−510 + 7007 = 0000.
Answer: shown
First, x−5−1102021203 = x − 2−102x − 8 (entries: x + 0 − 2, 0 − 10 + 0, 2x − 5 − 3).
Then x − 2−102x − 8x41 = [x(x − 2) − 40 + 2x − 8] = [x² − 48].
x² − 48 = 0 gives x = ±4√3.
Answer: x = ±4√3
(a) Revenue = 1000020001800060002000080002.501.501.00 = 25000 + 3000 + 1800015000 + 30000 + 8000 = 4600053000.
(b) Cost = 1000020001800060002000080002.001.000.50 = 20000 + 2000 + 900012000 + 20000 + 4000 = 3100036000.
Gross profit = revenue − cost = 1500017000 (₹15,000 in Market I and ₹17,000 in Market II; ₹32,000 in all).
Answer: (a) ₹46,000 and ₹53,000; (b) ₹15,000 and ₹17,000
The product has order 2 × 3 and the second factor is 2 × 3, so X is 2 × 2. Let X = abcd.
Row 1 gives a + 4b = −7, 2a + 5b = −8, 3a + 6b = −9. From the first two, a = −7 − 4b and −14 − 8b + 5b = −8, so b = −2 and a = 1. Check: 3 − 12 = −9.
Row 2 gives c + 4d = 2, 2c + 5d = 4, 3c + 6d = 6. From the first two, c = 2 − 4d and 4 − 8d + 5d = 4, so d = 0 and c = 2. Check: 6 + 0 = 6.
X = 1−220.
Answer: X = [[1, −2], [2, 0]]
A² = α² + βγαβ − βαγα − αγγβ + α² = α² + βγ00α² + βγ.
A² = I requires α² + βγ = 1, i.e. 1 − α² − βγ = 0.
Answer: (C) 1 − α² − βγ = 0
A′ = A (symmetric) and A′ = −A (skew symmetric) give A = −A, so 2A = O and A = O.
A zero matrix is both, and it is also square and diagonal, but the property that must hold is that A is the zero matrix.
Answer: (B) A is a zero matrix
(I + A)³ = I + 3A + 3A² + A³ (I commutes with A).
Since A² = A, A³ = A²·A = A·A = A. So (I + A)³ = I + 3A + 3A + A = I + 7A.
Then (I + A)³ − 7A = I.
Answer: (C) I