Miscellaneous Exercise answers: Matrices

Open NCERT p. 72One way to think about it

1. Since A, B are symmetric, A′ = A and B′ = B.
2. Transpose AB − BA: (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′.
3. Replace A′ by A and B′ by B: this becomes BA − AB.
4. BA − AB is exactly −(AB − BA).

In short(AB − BA)′ = −(AB − BA), so AB − BA is skew symmetric.

Watch this explained “When a product of two symmetric matrices is symmetric”, 14:36 into Symmetric and skew symmetric, and splitting any square matrix into one of each

Open NCERT p. 72One way to think about it

A symmetric

1. Take the transpose of B′AB: (B′AB)′ = B′A′(B′)′ (reversing the order of the three factors).
2. (B′)′ is just B again, so this is B′A′B.
3. Since A is symmetric, A′ = A, so (B′AB)′ = B′AB.

In short(B′AB)′ = B′AB, so B′AB is symmetric.

A skew symmetric

1. As before, (B′AB)′ = B′A′B.
2. Since A is skew symmetric, A′ = −A, so (B′AB)′ = B′(−A)B = −B′AB.

In short(B′AB)′ = −(B′AB), so B′AB is skew symmetric.

Watch this explained “Four: multiplying does not”, 6:19 into Flipping rows into columns, and why the transpose of a product reverses it

Open NCERT p. 72Checked by computerAnswers can differ: one example

1. Write A′ by turning each row of A into a column: A′ = [0 x x; 2y y −y; z −z z].
2. Diagonal entries of A′A: row 1 of A′ times column 1 of A = 0·0 + x·x + x·x = 2x². In the same way the other two are 4y² + y² + y² = 6y² and z² + z² + z² = 3z².
3. Off-diagonal entries cancel whatever x, y, z are. For example, row 1 of A′ times column 2 of A = 0·2y + x·y + x·(−y) = 0; the other five work the same way.
4. So A′A = I needs 2x² = 1, 6y² = 1 and 3z² = 1, that is x² = 1/2, y² = 1/6, z² = 1/3.
5. Each square has a positive and a negative root, and the three signs can be chosen independently, so there are 8 answers. Taking all three positive gives one of them.

Answerx = ±1/√2, y = ±1/√6, z = ±1/√3 (any choice of signs). One answer: x = 1/√2 = √2/2, y = 1/√6 = √6/6, z = 1/√3 = √3/3.

Watch this explained “Three unknowns, and the factors turn around”, 15:24 into Flipping rows into columns, and why the transpose of a product reverses it

Open NCERT p. 72Matches NCERT’s answer

1. First multiply the row [1 2 1] with the square matrix [1 2 0; 2 0 1; 1 0 2].
2. This gives the row [1·1+2·2+1·1, 1·2+2·0+1·0, 1·0+2·1+1·2] = [6, 2, 4].
3. Now multiply this row by the column [0; 2; x]: 6·0 + 2·2 + 4·x = 4 + 4x.
4. Set this equal to 0 (the single entry of O): 4 + 4x = 0, so x = −1.

Answerx = −1.

Watch this explained “One collapses, one expands”, 11:31 into Row into column: why multiplication needs the inner orders to match

Open NCERT p. 72One way to think about it

1. Compute A² = A × A = [3·3+1·(−1) 3·1+1·2; −1·3+2·(−1) −1·1+2·2] = [8 5; −5 3].
2. Compute 5A = [15 5; −5 10].
3. Compute 7I = [7 0; 0 7].
4. Add and subtract entry by entry: A² − 5A + 7I = [8−15+7 5−5+0; −5+5+0 3−10+7] = [0 0; 0 0].

In shortA² − 5A + 7I is the zero matrix, so the equation holds.

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. 72Matches NCERT’s answer

1. Multiply the row [x −5 −1] with the matrix [1 0 2; 0 2 1; 2 0 3].
2. This gives the row [x·1+(−5)·0+(−1)·2, x·0+(−5)·2+(−1)·0, x·2+(−5)·1+(−1)·3] = [x−2, −10, 2x−8].
3. Multiply this row by the column [x; 4; 1]: (x−2)·x + (−10)·4 + (2x−8)·1 = x² − 48.
4. Set this equal to 0: x² − 48 = 0, so x² = 48 and x = ±4√3.

Answerx = 4√3 or x = −4√3.

Watch this explained “One collapses, one expands”, 11:31 into Row into column: why multiplication needs the inner orders to match

Open NCERT p. 72Matches NCERT’s answer

(a) find the total revenue in each market

1. Write the sales as a 2×3 matrix (rows = markets, columns = x, y, z) and the prices as a 3×1 column: [2.50; 1.50; 1.00].
2. Multiply sales by the price column. For Market I: 10000×2.50 + 2000×1.50 + 18000×1.00 = 25000 + 3000 + 18000 = 46000.
3. For Market II: 6000×2.50 + 20000×1.50 + 8000×1.00 = 15000 + 30000 + 8000 = 53000.

AnswerRevenue in Market I = ₹46,000; revenue in Market II = ₹53,000.

(b) Find the gross profit

1. Write the unit costs as a column: [2.00; 1.00; 0.50] (50 paise = ₹0.50).
2. Multiply sales by the cost column to get total cost. Market I: 10000×2 + 2000×1 + 18000×0.50 = 20000 + 2000 + 9000 = 31000.
3. Market II: 6000×2 + 20000×1 + 8000×0.50 = 12000 + 20000 + 4000 = 36000.
4. Gross profit = revenue − cost: Market I: 46000 − 31000 = 15000. Market II: 53000 − 36000 = 17000.

AnswerGross profit in Market I = ₹15,000; gross profit in Market II = ₹17,000.

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

Open NCERT p. 73Matches NCERT’s answer

1. The known matrix is 2×3 and the answer on the right is also 2×3, so for X times a 2×3 to give a 2×3, X must be 2×2.
2. Let X = [a b; c d] and multiply it out: X[1 2 3; 4 5 6] = [a+4b, 2a+5b, 3a+6b; c+4d, 2c+5d, 3c+6d].
3. Match each entry to the right-hand side: a+4b=−7, 2a+5b=−8, 3a+6b=−9 (top row), and c+4d=2, 2c+5d=4, 3c+6d=6 (bottom row).
4. Solving the top row's equations gives a=1, b=−2. Solving the bottom row's equations gives c=2, d=0.

AnswerX = [1 −2; 2 0].

Watch this explained “An unknown, on one side only”, 16:32 into What multiplication keeps from ordinary algebra, and the two things it loses

Open NCERT p. 73Matches NCERT’s answer

1. Multiply A by itself: the off-diagonal entries of A² work out to αβ − βα = 0 and γα − αγ = 0, so they vanish on their own.
2. The diagonal entries both come out as α² + βγ.
3. So A² = [α²+βγ 0; 0 α²+βγ]. For this to equal I, we need α² + βγ = 1.
4. Rearranging α² + βγ = 1 gives 1 − α² − βγ = 0.

Answer1 − α² − βγ = 0 — option (C).

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. 73Matches NCERT’s answer

1. A symmetric matrix satisfies A′ = A; a skew symmetric matrix satisfies A′ = −A. Both definitions are for square matrices.
2. If A is both, then A = A′ = −A, so A = −A.
3. Add A to both sides: 2A = O, so A = O. Every entry of A is 0.
4. The zero matrix is also square and diagonal, so (A) and (C) are true too, but they say less: a matrix like I is square and diagonal without being skew symmetric. Only (B) tells you exactly what A must be.

AnswerA is the zero matrix — option (B).

Watch this explained “Can a matrix be both kinds at once?”, 16:06 into Symmetric and skew symmetric, and splitting any square matrix into one of each

Open NCERT p. 73Matches NCERT’s answer

1. Expand (I + A)³ using the binomial pattern, which is allowed here because I and A commute: (I+A)³ = I + 3A + 3A² + A³.
2. Since A² = A, we also get A³ = A²·A = A·A = A² = A.
3. Substitute A² = A and A³ = A: (I+A)³ = I + 3A + 3A + A = I + 7A.
4. Now subtract 7A: (I + 7A) − 7A = I.

Answer(I + A)³ − 7A = I — option (C).

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