Exercise 3.2 answers: Matrices

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(i) A + B

1. Add matching entries of A and B.

Answer[[3, 7], [1, 7]]

(ii) A – B

1. Subtract matching entries of B from A.

Answer[[1, 1], [5, −3]]

(iii) 3A – C

1. Multiply every entry of A by 3 to get 3A = [[6, 12], [9, 6]].
2. Subtract matching entries of C from 3A.

Answer[[8, 7], [6, 2]]

(iv) AB

1. For AB, each entry is (row of A) matched with (column of B), multiplied and added.
2. Top-left: 2×1 + 4×(−2) = 2 − 8 = −6. Top-right: 2×3 + 4×5 = 6 + 20 = 26.
3. Bottom-left: 3×1 + 2×(−2) = 3 − 4 = −1. Bottom-right: 3×3 + 2×5 = 9 + 10 = 19.

Answer[[−6, 26], [−1, 19]]

(v) BA

1. For BA, each entry is (row of B) matched with (column of A), multiplied and added.
2. Top-left: 1×2 + 3×3 = 2 + 9 = 11. Top-right: 1×4 + 3×2 = 4 + 6 = 10.
3. Bottom-left: −2×2 + 5×3 = −4 + 15 = 11. Bottom-right: −2×4 + 5×2 = −8 + 10 = 2.
4. Notice AB and BA are different, so matrix multiplication does not give the same answer both ways round.

Answer[[11, 10], [11, 2]]

Watch this explained “Scale, negate, add”, 14:32 into Addition and scalar multiplication done entry by entry, and why orders must agree

Open NCERT p. 58Matches NCERT’s answer

(i)

1. Compute: [[a, b], [−b, a]] + [[a, b], [b, a]].
2. Add matching entries: a+a = 2a, b+b = 2b, −b+b = 0, a+a = 2a.

Answer[[2a, 2b], [0, 2a]]

(ii)

1. Compute: [[a² + b², b² + c²], [a² + c², a² + b²]] + [[2ab, 2bc], [−2ac, −2ab]].
2. Add matching entries and group as perfect squares.
3. a²+b²+2ab = (a+b)². b²+c²+2bc = (b+c)². a²+c²−2ac = (a−c)². a²+b²−2ab = (a−b)².

Answer[[(a+b)², (b+c)²], [(a−c)², (a−b)²]]

(iii)

1. Compute: [[−1, 4, −6], [8, 5, 16], [2, 8, 5]] + [[12, 7, 6], [8, 0, 5], [3, 2, 4]].
2. Add matching entries position by position.
3. Row 1: −1+12=11, 4+7=11, −6+6=0. Row 2: 8+8=16, 5+0=5, 16+5=21. Row 3: 2+3=5, 8+2=10, 5+4=9.

Answer[[11, 11, 0], [16, 5, 21], [5, 10, 9]]

(iv)

1. Compute: [[cos² x, sin² x], [sin² x, cos² x]] + [[sin² x, cos² x], [cos² x, sin² x]].
2. Add matching entries: cos²x+sin²x = 1 at every position, using the identity sin²x+cos²x = 1.

Answer[[1, 1], [1, 1]]

Watch this explained “Entries that collapse”, 16:29 into Addition and scalar multiplication done entry by entry, and why orders must agree

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(i)

1. Compute: [[a, b], [−b, a]] [[a, −b], [b, a]]
2. Multiply row 1 of the first matrix by each column of the second: a×a+b×b = a²+b², a×(−b)+b×a = 0.
3. Multiply row 2 of the first matrix by each column of the second: −b×a+a×b = 0, −b×(−b)+a×a = a²+b².

Answer[[a²+b², 0], [0, a²+b²]]

(ii)

1. Compute: [[1], [2], [3]] [2 3 4]
2. A 3×1 column times a 1×3 row gives a 3×3 matrix.
3. Each entry is (column entry) × (row entry): row 1 uses 1, row 2 uses 2, row 3 uses 3, each times 2, 3, 4.

Answer[[2, 3, 4], [4, 6, 8], [6, 9, 12]]

(iii)

1. Compute: [[1, −2], [2, 3]] [[1, 2, 3], [2, 3, 1]]
2. The first matrix is 2×2, the second is 2×3, so the product is 2×3.
3. Row 1: 1×1+(−2)×2=−3, 1×2+(−2)×3=−4, 1×3+(−2)×1=1.
4. Row 2: 2×1+3×2=8, 2×2+3×3=13, 2×3+3×1=9.

Answer[[−3, −4, 1], [8, 13, 9]]

(iv)

1. Compute: [[2, 3, 4], [3, 4, 5], [4, 5, 6]] [[1, −3, 5], [0, 2, 4], [3, 0, 5]]
2. Both matrices are 3×3, so the product is 3×3.
3. Row 1: 2×1+3×0+4×3=14, 2×(−3)+3×2+4×0=0, 2×5+3×4+4×5=42.
4. Row 2: 3×1+4×0+5×3=18, 3×(−3)+4×2+5×0=−1, 3×5+4×4+5×5=56.
5. Row 3: 4×1+5×0+6×3=22, 4×(−3)+5×2+6×0=−2, 4×5+5×4+6×5=70.

Answer[[14, 0, 42], [18, −1, 56], [22, −2, 70]]

(v)

1. Compute: [[2, 1], [3, 2], [−1, 1]] [[1, 0, 1], [−1, 2, 1]]
2. The first matrix is 3×2, the second is 2×3, so the product is 3×3.
3. Row 1: 2×1+1×(−1)=1, 2×0+1×2=2, 2×1+1×1=3.
4. Row 2: 3×1+2×(−1)=1, 3×0+2×2=4, 3×1+2×1=5.
5. Row 3: −1×1+1×(−1)=−2, −1×0+1×2=2, −1×1+1×1=0.

Answer[[1, 2, 3], [1, 4, 5], [−2, 2, 0]]

(vi)

1. Compute: [[3, −1, 3], [−1, 0, 2]] [[2, −3], [1, 0], [3, 1]]
2. The first matrix is 2×3, the second is 3×2, so the product is 2×2.
3. Row 1: 3×2+(−1)×1+3×3=14, 3×(−3)+(−1)×0+3×1=−6.
4. Row 2: −1×2+0×1+2×3=4, −1×(−3)+0×0+2×1=5.

Answer[[14, −6], [4, 5]]

Watch this explained “One row, one column”, 4:58 into Row into column: why multiplication needs the inner orders to match

Open NCERT p. 59Matches NCERT’s answer

1. Add A and B entry by entry. Row 1: 1 + 3 = 4, 2 + (−1) = 1, −3 + 2 = −1. Row 2: 5 + 4 = 9, 0 + 2 = 2, 2 + 5 = 7. Row 3: 1 + 2 = 3, −1 + 0 = −1, 1 + 3 = 4.
2. So A + B = [[4, 1, −1], [9, 2, 7], [3, −1, 4]].
3. Subtract C from B entry by entry. Row 1: 3 − 4 = −1, −1 − 1 = −2, 2 − 2 = 0. Row 2: 4 − 0 = 4, 2 − 3 = −1, 5 − 2 = 3. Row 3: 2 − 1 = 1, 0 − (−2) = 2, 3 − 3 = 0.
4. So B − C = [[−1, −2, 0], [4, −1, 3], [1, 2, 0]].
5. Left side, A + (B − C): add A to the B − C just found. Row 1: 1 + (−1) = 0, 2 + (−2) = 0, −3 + 0 = −3. Row 2: 5 + 4 = 9, 0 + (−1) = −1, 2 + 3 = 5. Row 3: 1 + 1 = 2, −1 + 2 = 1, 1 + 0 = 1. This gives [[0, 0, −3], [9, −1, 5], [2, 1, 1]].
6. Right side, (A + B) − C: subtract C from the A + B just found. Row 1: 4 − 4 = 0, 1 − 1 = 0, −1 − 2 = −3. Row 2: 9 − 0 = 9, 2 − 3 = −1, 7 − 2 = 5. Row 3: 3 − 1 = 2, −1 − (−2) = 1, 4 − 3 = 1. This gives [[0, 0, −3], [9, −1, 5], [2, 1, 1]].
7. Both sides give the same matrix, so A + (B − C) = (A + B) − C.

AnswerA + B = [[4, 1, −1], [9, 2, 7], [3, −1, 4]], B − C = [[−1, −2, 0], [4, −1, 3], [1, 2, 0]], and A + (B − C) and (A + B) − C both equal [[0, 0, −3], [9, −1, 5], [2, 1, 1]], so they are equal.

Watch this explained “Grouping does not matter”, 3:33 into The laws addition and scaling obey, and what they buy you

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1. Multiply every entry of A by 3 to get 3A.
2. Multiply every entry of B by 5 to get 5B.
3. Subtract 5B from 3A entry by entry.
4. 3A works out to [[2, 3, 5], [1, 2, 4], [7, 6, 2]] and 5B works out to the same matrix, [[2, 3, 5], [1, 2, 4], [7, 6, 2]].
5. So every entry of 3A − 5B is 0.

Answer[[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Watch this explained “Scale, negate, add”, 14:32 into Addition and scalar multiplication done entry by entry, and why orders must agree

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1. Simplify: cos θ [[cos θ, sin θ], [−sin θ, cos θ]] + sin θ [[sin θ, −cos θ], [cos θ, sin θ]].
2. Multiply the first matrix by cosθ: [[cos²θ, sinθcosθ], [−sinθcosθ, cos²θ]].
3. Multiply the second matrix by sinθ: [[sin²θ, −sinθcosθ], [sinθcosθ, sin²θ]].
4. Add the two matrices entry by entry.
5. Top-left and bottom-right both become cos²θ+sin²θ, which is 1 by the Pythagorean identity.
6. Top-right becomes sinθcosθ−sinθcosθ = 0, and bottom-left becomes −sinθcosθ+sinθcosθ = 0.

Answer[[1, 0], [0, 1]]

Watch this explained “Entries that collapse”, 16:29 into Addition and scalar multiplication done entry by entry, and why orders must agree

Open NCERT p. 59Matches NCERT’s answer

(i)

1. Given: X + Y = [[7, 0], [2, 5]] and X – Y = [[3, 0], [0, 3]].
2. Add the two given equations: (X+Y) + (X−Y) = 2X, so 2X = [[10, 0], [2, 8]], giving X = [[5, 0], [1, 4]].
3. Subtract the second equation from the first: (X+Y) − (X−Y) = 2Y, so 2Y = [[4, 0], [2, 2]], giving Y = [[2, 0], [1, 1]].

AnswerX = [[5, 0], [1, 4]], Y = [[2, 0], [1, 1]]

(ii)

1. Given: 2X + 3Y = [[2, 3], [4, 0]] and 3X + 2Y = [[2, −2], [−1, 5]].
2. Multiply the first equation by 3 and the second by 2, then subtract to eliminate X: 9Y − 4Y = 3×[[2,3],[4,0]] − 2×[[2,−2],[−1,5]], so 5Y = [[2, 13], [14, −10]], giving Y = [[2/5, 13/5], [14/5, −2]].
3. Put Y back into 2X + 3Y = [[2, 3], [4, 0]] and solve for X.
4. 2X = [[2, 3], [4, 0]] − 3×[[2/5, 13/5], [14/5, −2]] = [[4/5, −24/5], [−22/5, 6]], so X = [[2/5, −12/5], [−11/5, 3]].

AnswerX = [[2/5, −12/5], [−11/5, 3]], Y = [[2/5, 13/5], [14/5, −2]]

Watch this explained “Two unknowns at once”, 14:06 into The laws addition and scaling obey, and what they buy you

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1. Subtract Y from both sides: 2X = [[1, 0], [−3, 2]] − [[3, 2], [1, 4]] = [[−2, −2], [−4, −2]].
2. Divide every entry by 2 to get X.

Answer[[−1, −1], [−2, −1]]

Watch this explained “Solving for an unknown, one law per line”, 11:57 into The laws addition and scaling obey, and what they buy you

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1. Multiply the first matrix by 2: [[2, 6], [0, 2x]].
2. Add the second matrix: [[2+y, 6], [1, 2x+2]].
3. Match this with [[5, 6], [1, 8]]: 2+y = 5, so y = 3, and 2x+2 = 8, so x = 3.
4. The other two positions, 6 = 6 and 1 = 1, are already true and confirm the working.

Answerx = 3, y = 3

Watch this explained “Unknowns inside an arrangement”, 16:10 into The laws addition and scaling obey, and what they buy you

Open NCERT p. 59Matches NCERT’s answer

1. Multiply the right side out: 3×[[3,5],[4,6]] = [[9,15],[12,18]].
2. Multiply the known matrix on the left by 3: 3×[[1,−1],[0,2]] = [[3,−3],[0,6]].
3. Move that matrix to the right side by subtracting it: 2[[x,z],[y,t]] = [[9,15],[12,18]] − [[3,−3],[0,6]] = [[6,18],[12,12]].
4. Divide every entry by 2: [[x,z],[y,t]] = [[3,9],[6,6]].
5. Two matrices are equal only when the matching positions are equal, so read off x = 3, z = 9, y = 6, t = 6.

Answerx = 3, y = 6, z = 9, t = 6.

Watch this explained “Unknowns inside an arrangement”, 16:10 into The laws addition and scaling obey, and what they buy you

Open NCERT p. 59Matches NCERT’s answer

1. Multiply out the left side entry by entry: [[2x − y], [3x + y]] = [[10], [5]].
2. Matching the top entries gives 2x − y = 10, and matching the bottom entries gives 3x + y = 5.
3. Add the two equations so that y cancels: 5x = 15, so x = 3.
4. Put x = 3 back into 3x + y = 5 to get y = 5 − 9 = −4.

Answerx = 3, y = −4.

Watch this explained “When the equations split”, 12:20 into Equality as two demands, and why the orders have to agree before the entries are looked at

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1. Add the two matrices on the right first: [[x + 4, x + y + 6], [z + w − 1, 2w + 3]].
2. Compare the top-left entries: 3x = x + 4, so 2x = 4 and x = 2.
3. Compare the top-right entries: 3y = x + y + 6, so 2y = x + 6 = 8 and y = 4.
4. Compare the bottom-right entries: 3w = 2w + 3, so w = 3.
5. Compare the bottom-left entries: 3z = z + w − 1, so 2z = w − 1 = 2 and z = 1.

Answerx = 2, y = 4, z = 1, w = 3.

Watch this explained “Unknowns inside an arrangement”, 16:10 into The laws addition and scaling obey, and what they buy you

Open NCERT p. 60One way to think about it

1. Write F(x) and F(y) with the same pattern, one using angle x and the other using angle y.
2. Multiply row 1 of F(x) by column 1 of F(y): cos x cos y − sin x sin y, which equals cos(x + y) by the compound-angle formula.
3. Multiply row 1 by column 2: −cos x sin y − sin x cos y, which equals −sin(x + y).
4. Multiply row 2 by column 1: sin x cos y + cos x sin y, which equals sin(x + y).
5. Multiply row 2 by column 2: −sin x sin y + cos x cos y, which equals cos(x + y).
6. The third row and third column only touch the entry 1, since every other entry there is 0, so they reproduce the third row/column of F(x + y) exactly.
7. Putting these nine entries together gives exactly F(x + y), so F(x) F(y) = F(x + y).

In shortF(x) F(y) = [[cos(x+y), −sin(x+y), 0], [sin(x+y), cos(x+y), 0], [0, 0, 1]] = F(x + y).

Watch this explained “A turn, raised to the nth power”, 14:30 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60One way to think about it

(i) Show that

1. Multiply [[5,−1],[6,7]] by [[2,1],[3,4]] in that order: the product is [[10−3, 5−4],[12+21, 6+28]] = [[7, 1], [33, 34]].
2. Now multiply in the other order, [[2,1],[3,4]] by [[5,−1],[6,7]]: the product is [[10+6, −2+7],[15+24, −3+28]] = [[16, 5], [39, 25]].
3. Compare the two results entry by entry: [[7,1],[33,34]] is not the same as [[16,5],[39,25]], so the two products are different.

In short[[5,−1],[6,7]][[2,1],[3,4]] = [[7,1],[33,34]], but [[2,1],[3,4]][[5,−1],[6,7]] = [[16,5],[39,25]]; these are not equal.

(ii) Show that

1. Multiply the first 3×3 matrix by the second in the order given; working row by row gives [[5, 8, 14], [0, −1, 1], [−1, 0, 1]].
2. Now multiply them in the reverse order; working row by row gives [[−1, −1, −3], [1, 0, 0], [6, 11, 6]].
3. The two results do not match position by position, so the two products are different.

In shortThe two products come out to [[5,8,14],[0,−1,1],[−1,0,1]] and [[−1,−1,−3],[1,0,0],[6,11,6]], which are not equal.

Watch this explained “The hard half: same size, still different”, 2:09 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60Matches NCERT’s answer

1. Multiply A by itself, row into column, to get A² = [[5, −1, 2], [9, −2, 5], [0, −1, −2]].
2. Multiply every entry of A by 5 to get 5A = [[10, 0, 5], [10, 5, 15], [5, −5, 0]].
3. 6I is 6 on the diagonal and 0 elsewhere: [[6, 0, 0], [0, 6, 0], [0, 0, 6]].
4. Add A² and 6I, then subtract 5A, entry by entry, to get [[1, −1, −3], [−1, −1, −10], [−5, 4, 4]].

AnswerA² − 5A + 6I = [[1, −1, −3], [−1, −1, −10], [−5, 4, 4]].

Watch this explained “Powers, and a polynomial in a matrix”, 11:53 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60One way to think about it

1. Multiply A by itself to get A² = [[5, 0, 8], [2, 4, 5], [8, 0, 13]].
2. Multiply A² by A once more to get A³ = [[21, 0, 34], [12, 8, 23], [34, 0, 55]].
3. Multiply A² by 6: 6A² = [[30, 0, 48], [12, 24, 30], [48, 0, 78]].
4. Multiply A by 7: 7A = [[7, 0, 14], [0, 14, 7], [14, 0, 21]].
5. 2I is [[2, 0, 0], [0, 2, 0], [0, 0, 2]].
6. Add A³ + 7A + 2I and subtract 6A², entry by entry: every position comes out to 0.
7. So A³ − 6A² + 7A + 2I is the zero matrix, which is what had to be proved.

In shortA³ − 6A² + 7A + 2I = 0 (the 3×3 zero matrix).

Watch this explained “Powers, and a polynomial in a matrix”, 11:53 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60Matches NCERT’s answer

1. Multiply A by itself: A² = [[9−8, −6+4], [12−8, −8+4]] = [[1, −2], [4, −4]].
2. Write kA − 2I as [[3k − 2, −2k], [4k, −2k − 2]].
3. Compare the top-left entries: 3k − 2 = 1, so k = 1.
4. Check the other three positions with k = 1: −2k = −2 matches −2; 4k = 4 matches 4; −2k − 2 = −4 matches −4.
5. All four positions agree, so k = 1 works.

Answerk = 1.

Watch this explained “Powers, and a polynomial in a matrix”, 11:53 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60One way to think about it

1. Write t for tan(α/2) to keep the working short. Then I + A = [[1, −t], [t, 1]] and I − A = [[1, t], [−t, 1]].
2. Multiply (I − A) by [[cos α, −sin α], [sin α, cos α]]: the top-left entry is cos α + t sin α, and the top-right entry is −sin α + t cos α.
3. The bottom-left entry is −t cos α + sin α, and the bottom-right entry is t sin α + cos α.
4. Replace t by tan(α/2) and use cos α = (1 − t²)/(1 + t²) and sin α = 2t/(1 + t²), the half-angle formulas.
5. The top-left entry simplifies to (1 − t² + 2t²)/(1 + t²) = 1, and the top-right entry simplifies to (−2t + t(1 − t²))/(1 + t²) = −t.
6. The bottom-left entry simplifies to t, and the bottom-right entry simplifies to 1, using the same two formulas.
7. So the product comes out to [[1, −t], [t, 1]], which is exactly I + A, as required.

In short(I − A)[[cos α, −sin α],[sin α, cos α]] = [[1, −tan(α/2)], [tan(α/2), 1]] = I + A.

Watch this explained “A turn, raised to the nth power”, 14:30 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 60Matches NCERT’s answer

(a) an annual total interest of

1. Let ₹x be invested at 5%, so ₹(30000 − x) is invested at 7%.
2. Write the interest as a matrix product: [x, 30000 − x] [[0.05], [0.07]] = 1800.
3. This gives 0.05x + 0.07(30000 − x) = 1800, which simplifies to 2100 − 0.02x = 1800.
4. Solve: 0.02x = 300, so x = 15000.
5. So ₹15000 is invested at 5% and the remaining ₹15000 at 7%.

Answer₹15,000 at 5% and ₹15,000 at 7%.

(b) an annual total interest of

1. Use the same set-up with the new total interest: 0.05x + 0.07(30000 − x) = 2000.
2. This simplifies to 2100 − 0.02x = 2000, so 0.02x = 100 and x = 5000.
3. So ₹5000 is invested at 5% and the remaining ₹25000 at 7%.

Answer₹5,000 at 5% and ₹25,000 at 7%.

Watch this explained “Every entry means something”, 14:53 into Row into column: why multiplication needs the inner orders to match

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1. Convert every dozen to books before building the matrix: 10 dozen = 120, 8 dozen = 96, 10 dozen = 120 books.
2. Write the number of books as a row matrix and the prices as a column matrix: [120, 96, 120] [[80], [60], [40]].
3. Multiply and add: 120×80 + 96×60 + 120×40 = 9600 + 5760 + 4800.
4. Add the three amounts to get ₹20,160.

Answer₹20,160.

Watch this explained “A trap, and what to keep”, 16:29 into Row into column: why multiplication needs the inner orders to match

Open NCERT p. 61Matches NCERT’s answer

1. PY multiplies a p×k matrix by a 3×k matrix, so the inner numbers must match: k must equal 3.
2. WY multiplies an n×3 matrix by a 3×k matrix; the inner numbers (3 and 3) already match, so WY always exists, with order n×k.
3. With k = 3, PY has order p×k = p×3, and WY has order n×k = n×3.
4. To add PY and WY, their orders must be the same, so p must equal n.
5. So the restriction is k = 3 and p = n, which is option (A).

Answer(A) k = 3, p = n.

Watch this explained “Orders with no numbers”, 13:18 into Row into column: why multiplication needs the inner orders to match

Open NCERT p. 61Matches NCERT’s answer

1. X has order 2×n and Z has order 2×p.
2. Subtraction 7X − 5Z needs both matrices to have the same order, which is why the question sets n = p.
3. With n = p, both matrices are 2×n, so the difference is also 2×n.
4. That matches option (B).

Answer(B) 2 × n.

Watch this explained “Orders with no numbers”, 13:18 into Row into column: why multiplication needs the inner orders to match