Exercise 3.3 answers: Matrices

Open NCERT p. 66Matches NCERT’s answer

(i) Find the transpose of each

1. The matrix is a single column with three entries: 5, 1/2, −1.
2. Transposing turns the one column into one row, keeping the entries in the same order.

Answer[5, 1/2, −1] (a 1×3 row matrix).

(ii) Find the transpose of each

1. Write down the first column of the matrix, [1, 2], as the first row of the transpose.
2. Write down the second column, [−1, 3], as the second row of the transpose.

Answer[[1, 2], [−1, 3]].

(iii) Find the transpose of each

1. The first column, [−1, √3, 2], becomes the first row of the transpose.
2. The second column, [5, 5, 3], becomes the second row.
3. The third column, [6, 6, −1], becomes the third row.

Answer[[−1, √3, 2], [5, 5, 3], [6, 6, −1]].

Watch this explained “The definition, and its two symbols”, 2:04 into Flipping rows into columns, and why the transpose of a product reverses it

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(i) then verify that

1. Add A and B entry by entry: A + B = [[−5, 3, −2], [6, 9, 9], [−1, 4, 2]].
2. Transpose this sum: (A + B)′ = [[−5, 6, −1], [3, 9, 4], [−2, 9, 2]].
3. Now transpose A and B separately: A′ = [[−1, 5, −2], [2, 7, 1], [3, 9, 1]] and B′ = [[−4, 1, 1], [1, 2, 3], [−5, 0, 1]].
4. Add A′ and B′ entry by entry: A′ + B′ = [[−5, 6, −1], [3, 9, 4], [−2, 9, 2]].
5. This matches (A + B)′ exactly, so the two sides are equal.

In short(A + B)′ = A′ + B′ = [[−5, 6, −1], [3, 9, 4], [−2, 9, 2]].

(ii) then verify that

1. Subtract B from A entry by entry: A − B = [[3, 1, 8], [4, 5, 9], [−3, −2, 0]].
2. Transpose this difference: (A − B)′ = [[3, 4, −3], [1, 5, −2], [8, 9, 0]].
3. Subtract B′ from A′ (using the transposes found above): A′ − B′ = [[3, 4, −3], [1, 5, −2], [8, 9, 0]].
4. This matches (A − B)′ exactly, so the two sides are equal.

In short(A − B)′ = A′ − B′ = [[3, 4, −3], [1, 5, −2], [8, 9, 0]].

Watch this explained “Three: adding passes through”, 5:09 into Flipping rows into columns, and why the transpose of a product reverses it

Open NCERT p. 66One way to think about it

(i) then verify that

1. Find A by transposing A′: A = [[3, −1, 0], [4, 2, 1]].
2. Add A and B entry by entry: A + B = [[2, 1, 1], [5, 4, 4]].
3. Transpose this sum: (A + B)′ = [[2, 5], [1, 4], [1, 4]].
4. Transpose B: B′ = [[−1, 1], [2, 2], [1, 3]], and add it to A′: A′ + B′ = [[2, 5], [1, 4], [1, 4]].
5. This matches (A + B)′ exactly, so the two sides are equal.

In short(A + B)′ = A′ + B′ = [[2, 5], [1, 4], [1, 4]].

(ii) then verify that

1. Subtract B from A entry by entry: A − B = [[4, −3, −1], [3, 0, −2]].
2. Transpose this difference: (A − B)′ = [[4, 3], [−3, 0], [−1, −2]].
3. Subtract B′ from A′: A′ − B′ = [[4, 3], [−3, 0], [−1, −2]].
4. This matches (A − B)′ exactly, so the two sides are equal.

In short(A − B)′ = A′ − B′ = [[4, 3], [−3, 0], [−1, −2]].

Watch this explained “Three: adding passes through”, 5:09 into Flipping rows into columns, and why the transpose of a product reverses it

Open NCERT p. 67Matches NCERT’s answer

1. Find A by transposing A′: A = [[−2, 1], [3, 2]].
2. Double B entry by entry: 2B = [[−2, 0], [2, 4]].
3. Add A and 2B: A + 2B = [[−4, 1], [5, 6]].
4. Transpose this sum: (A + 2B)′ = [[−4, 5], [1, 6]].

Answer(A + 2B)′ = [[−4, 5], [1, 6]].

Watch this explained “Three: adding passes through”, 5:09 into Flipping rows into columns, and why the transpose of a product reverses it

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(i) verify that (AB)′ = B′A′

1. Multiply the 3×1 column A by the 1×3 row B: AB = [[−1, 2, 1], [4, −8, −4], [−3, 6, 3]].
2. Transpose this product: (AB)′ = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]].
3. Now transpose B and A: B′ = [[−1], [2], [1]] and A′ = [1, −4, 3].
4. Multiply B′A′: the result is [[−1, 4, −3], [2, −8, 6], [1, −4, 3]].
5. This matches (AB)′ exactly, so the identity holds.

In short(AB)′ = B′A′ = [[−1, 4, −3], [2, −8, 6], [1, −4, 3]].

(ii) verify that (AB)′ = B′A′

1. Multiply the 3×1 column A by the 1×3 row B: AB = [[0, 0, 0], [1, 5, 7], [2, 10, 14]].
2. Transpose this product: (AB)′ = [[0, 1, 2], [0, 5, 10], [0, 7, 14]].
3. Transpose B and A: B′ = [[1], [5], [7]] and A′ = [0, 1, 2].
4. Multiply B′A′: the result is [[0, 1, 2], [0, 5, 10], [0, 7, 14]].
5. This matches (AB)′ exactly, so the identity holds.

In short(AB)′ = B′A′ = [[0, 1, 2], [0, 5, 10], [0, 7, 14]].

Watch this explained “A column times a row, transposed both ways”, 9:12 into Flipping rows into columns, and why the transpose of a product reverses it

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(i) then verify that A′ A = I

1. Transpose A: A′ = [[cos α, −sin α], [sin α, cos α]].
2. Multiply A′ by A: the top-left entry is cos²α + sin²α, and the top-right entry is cos α sin α − sin α cos α.
3. The bottom-left entry is sin α cos α − cos α sin α, and the bottom-right entry is sin²α + cos²α.
4. Using cos²α + sin²α = 1, the diagonal entries become 1, and the off-diagonal entries cancel to 0.
5. So A′A = [[1, 0], [0, 1]] = I.

In shortA′A = I.

(ii) then verify that A′ A = I

1. Transpose A: A′ = [[sin α, −cos α], [cos α, sin α]].
2. Multiply A′ by A: the top-left entry is sin²α + cos²α, and the top-right entry is sin α cos α − cos α sin α.
3. The bottom-left entry is cos α sin α − sin α cos α, and the bottom-right entry is cos²α + sin²α.
4. Using cos²α + sin²α = 1, the diagonal entries become 1, and the off-diagonal entries cancel to 0.
5. So A′A = [[1, 0], [0, 1]] = I.

In shortA′A = I.

Watch this explained “A transpose times its own matrix”, 13:54 into Flipping rows into columns, and why the transpose of a product reverses it

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(i) Show that the matrix A … is a symmetric matrix

1. Write down A′ by turning every row of A into a column.
2. Row 1 of A is 1, −1, 5 — this becomes column 1 of A′.
3. Doing this for all three rows gives A′ = [1 −1 5; −1 2 1; 5 1 3].
4. Compare A′ with A entry by entry — every entry matches.

In shortA′ = A, so A is a symmetric matrix.

(ii) Show that the matrix A … is a skew symmetric matrix

1. Write down A′ by turning every row of A into a column.
2. Row 1 of A is 0, 1, −1 — this becomes column 1 of A′, giving A′ = [0 −1 1; 1 0 −1; −1 1 0].
3. Multiply every entry of A′ by −1: −A′ = [0 1 −1; −1 0 1; 1 −1 0].
4. Compare −A′ with A — every entry matches.

In shortA′ = −A, so A is a skew symmetric matrix.

Watch this explained “Two conditions, one minus sign apart”, 0:00 into Symmetric and skew symmetric, and splitting any square matrix into one of each

Open NCERT p. 67One way to think about it

(i) (A + A′) is a symmetric matrix

1. Write A′ by swapping rows and columns of A: A′ = [1 6; 5 7].
2. Add: A + A′ = [1+1 5+6; 6+5 7+7] = [2 11; 11 14].
3. Take the transpose of this sum: [2 11; 11 14]′ = [2 11; 11 14].
4. The sum equals its own transpose, so it does not change under transposing.

In short(A + A′) = [2 11; 11 14], which is a symmetric matrix.

(ii) (A − A′) is a skew symmetric matrix

1. Subtract: A − A′ = [1−1 5−6; 6−5 7−7] = [0 −1; 1 0].
2. Take the transpose of this difference: [0 −1; 1 0]′ = [0 1; −1 0].
3. This transpose is the negative of the original difference.

In short(A − A′) = [0 −1; 1 0], which is a skew symmetric matrix.

Watch this explained “A matrix plus its transpose is symmetric”, 6:11 into Symmetric and skew symmetric, and splitting any square matrix into one of each

Open NCERT p. 67Matches NCERT’s answer

1. A is already built so that every entry above the diagonal is the negative of the matching entry below it — that is the skew symmetric pattern.
2. Write A′ by swapping rows and columns: A′ = [0 −a −b; a 0 −c; b c 0].
3. This A′ is exactly −A, so A + A′ is the zero matrix, and half of zero is still zero.
4. A − A′ = A − (−A) = 2A, so half of it is just A back again.

Answer(1/2)(A + A′) is the zero matrix, and (1/2)(A − A′) = [0 a b; −a 0 c; −b −c 0], which is A itself.

Watch this explained “The two boundary cases”, 11:59 into Symmetric and skew symmetric, and splitting any square matrix into one of each

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(i) 3 5; 1 −1

1. A′ = [3 1; 5 −1].
2. Symmetric part = (1/2)(A + A′) = (1/2)[6 6; 6 −2] = [3 3; 3 −1].
3. Skew part = (1/2)(A − A′) = (1/2)[0 4; −4 0] = [0 2; −2 0].

AnswerA = [3 3; 3 −1] + [0 2; −2 0].

(ii) 6 −2 2; −2 3 −1; 2 −1 3

1. This matrix is already symmetric, since it equals its own transpose.
2. Symmetric part = (1/2)(A + A′) = A itself.
3. Skew part = (1/2)(A − A′) = the zero matrix.

AnswerA = [6 −2 2; −2 3 −1; 2 −1 3] + zero matrix.

(iii) 3 3 −1; −2 −2 1; −4 −5 2

1. A′ = [3 −2 −4; 3 −2 −5; −1 1 2].
2. Symmetric part = (1/2)(A + A′) = [3 1/2 −5/2; 1/2 −2 −2; −5/2 −2 2].
3. Skew part = (1/2)(A − A′) = [0 5/2 3/2; −5/2 0 3; −3/2 −3 0].

AnswerA = [3 1/2 −5/2; 1/2 −2 −2; −5/2 −2 2] + [0 5/2 3/2; −5/2 0 3; −3/2 −3 0].

(iv) 1 5; −1 2

1. A′ = [1 −1; 5 2].
2. Symmetric part = (1/2)(A + A′) = (1/2)[2 4; 4 4] = [1 2; 2 2].
3. Skew part = (1/2)(A − A′) = (1/2)[0 6; −6 0] = [0 3; −3 0].

AnswerA = [1 2; 2 2] + [0 3; −3 0].

Watch this explained “The split carried out, and added back”, 10:23 into Symmetric and skew symmetric, and splitting any square matrix into one of each

Open NCERT p. 68Matches NCERT’s answer

1. Since A, B are symmetric, A′ = A and B′ = B.
2. Transpose AB − BA: (AB − BA)′ = (AB)′ − (BA)′ = B′A′ − A′B′ (the order of factors reverses under a transpose).
3. Put A′ = A and B′ = B into this: (AB − BA)′ = BA − AB.
4. That is exactly −(AB − BA).

Answer(AB − BA)′ = −(AB − BA), so AB − BA is a skew symmetric matrix — option (A).

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. 68Matches NCERT’s answer

1. Write A′ by swapping rows and columns: A′ = [cos α sin α; −sin α cos α].
2. Add: A + A′ = [2 cos α 0; 0 2 cos α].
3. Set this equal to I = [1 0; 0 1]: 2 cos α = 1.
4. So cos α = 1/2, and matching this to the given options gives α = π/3.

Answerα = π/3 — 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