CBSE Class 12 Mathematics Matrices MCQs Set 09

Answer: (b) \( |C| = |A|^{n-1} \)

Explanation:
1. C is the matrix of cofactors, and adj A = Cᵀ, so |C| = |adj A|.
2. From A(adj A) = |A|I, taking determinants: |A||adj A| = |A|ⁿ.
3. So |adj A| = |A|ⁿ⁻¹, and hence |C| = |A|ⁿ⁻¹.

Teacher's Note:
1. A matrix and its transpose have the same determinant, which links the cofactor matrix to adj A.
2. Check with n = 2: for A = [[a,b],[c,d]], C = [[d,−c],[−b,a]] and |C| = ad − bc = |A|¹.

Answer: (b) A = O

Explanation:
1. B is non-singular, so B⁻¹ exists.
2. Multiplying AB = O on the right by B⁻¹: ABB⁻¹ = OB⁻¹.
3. A I = O, so A = O.

Teacher's Note:
1. Cancellation of a matrix is allowed only when it is invertible, and only from the correct side.
2. Without the non-singular condition, AB = O does not imply A = O.

Answer: (c) A + B

Explanation:
1. Using A = BA and B = AB: A² = A(BA) = (AB)A = BA = A.
2. Similarly B² = B(AB) = (BA)B = AB = B.
3. So A² + B² = A + B.

Teacher's Note:
1. Substitute A = BA or B = AB at the right moment and use associativity to regroup.
2. Such matrices A and B are idempotent (A² = A), which is a useful name to recognise.

Answer: (b) A and B are square matrices of the same order

Explanation:
1. A + B is defined only if A and B have the same order, say m × n.
2. AB is defined only if columns of A = rows of B, so n = m.
3. So A and B are square matrices of the same order.

Teacher's Note:
1. Combining two order conditions is a typical way such questions are built.
2. Option (c) is needed but not enough; the question asks for what both conditions together imply.

Answer: (d) 0

Explanation:
1. \( A^2 = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^2 + 1 & x \\ x & 1 \end{bmatrix} \).
2. For A² = I we need x² + 1 = 1 and x = 0.
3. So x = 0.

Teacher's Note:
1. The off-diagonal entry x must be 0, which settles it immediately.
2. Then A = [[0,1],[1,0]], the row-swap matrix, and swapping twice gives I.

Answer: (a) \( A^2 + B^2 \)

Explanation:
1. \( AB = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \) and \( BA = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \), so BA = −AB.
2. (A + B)² = A² + AB + BA + B² = A² + B² + (AB + BA).
3. Since AB + BA = O, (A + B)² = A² + B².

Teacher's Note:
1. For anticommuting matrices (AB = −BA), the cross terms cancel when squaring a sum.
2. Here A² = B² = I, so (A + B)² = 2I, which you can also check by direct multiplication.

Answer: (a) \( A^2 - B^2 \)

Explanation:
1. A = iI is a scalar matrix, so it commutes with B: AB = BA = iB.
2. (A + B)(A − B) = A² − AB + BA − B².
3. Since AB = BA, the middle terms cancel and (A + B)(A − B) = A² − B².

Teacher's Note:
1. A scalar matrix kI commutes with every square matrix of the same order.
2. Option (c) would add 2AB = 2iB ≠ O, so it is not equal to the product.

Answer: (a) 3 × 4

Explanation:
1. A is 3 × 4, so A′ is 4 × 3. Let B be p × q.
2. A′B is (4 × 3)(p × q): defined only if p = 3.
3. BA′ is (p × q)(4 × 3): defined only if q = 4. So B is 3 × 4.

Teacher's Note:
1. Work with the transpose's order (4 × 3), not with A's order.
2. B has the same order as A here, which makes both A′B (4 × 4) and BA′ (3 × 3) square.

Answer: (a) Every skew-symmetric matrix of odd order is non-singular

Explanation:
1. For a skew-symmetric A of odd order n, |A| = |Aᵀ| = |−A| = (−1)ⁿ|A| = −|A|.
2. So |A| = 0: every skew-symmetric matrix of odd order is singular, and statement (a) is false.
3. (b) is the definition of non-singular, and (c), (d) are standard properties of the adjoint, so they are true.

Teacher's Note:
1. The trick |−A| = (−1)ⁿ|A| is what makes odd-order skew-symmetric determinants vanish.
2. For example, [[0,1,2],[−1,0,3],[−2,−3,0]] has determinant 0.

Answer: (c) If A′ = A, then A is a square matrix

Explanation:
1. A non-singular matrix has a unique inverse, so (a) is false; its determinant is non-zero, so (b) is false.
2. If A′ = A, then A and A′ have the same order; if A is m × n, A′ is n × m, so m = n and A is square. (c) is true.
3. For (d), |A · adj A| = ||A|I| = |A|ⁿ, not |A|ⁿ⁻¹, so (d) is false.

Teacher's Note:
1. Only a square matrix can equal its own transpose, because transposing swaps the order m × n to n × m.
2. Remember |adj A| = |A|ⁿ⁻¹ but |A · adj A| = |A|ⁿ; mixing the two is the trap in (d).

Answer: (c) \( A.\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} = 2I \)

Explanation:
1. \( A' = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \), so (a) is false. |A| = 1 + 1 = 2, so \( A^{-1} = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \) and (b) is false.
2. \( A\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1-1 \\ 1-1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I \), so (c) is true.
3. λA = [[λ, −λ],[λ, λ]], so (d) is false. Answer (c).

Teacher's Note:
1. The matrix in (c) is adj A, and A(adj A) = |A|I = 2I, which confirms (c) without multiplying.
2. Do not forget the factor 1/|A| when writing an inverse; (b) misses it.

Answer: (c) AB ≠ BA

Explanation:
1. \( A^2 = \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix}^2 = \begin{bmatrix} -5 & 2 \\ -3 & -6 \end{bmatrix} \neq A \) and \( B^2 = \begin{bmatrix} 1 & 0 \\ 4 & 9 \end{bmatrix} \neq B \).
2. \( AB = \begin{bmatrix} -1+4 & 6 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 3 & 0 \end{bmatrix} \).
3. \( BA = \begin{bmatrix} -1 & -2 \\ 2-9 & 4 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -7 & 4 \end{bmatrix} \), so AB ≠ BA.

Teacher's Note:
1. Comparing a single entry, such as the (1, 1) entries 3 and −1, already shows AB ≠ BA.
2. Most pairs of matrices do not commute, so (c) is the usual answer unless one is a scalar matrix.

Answer: (b) Skew-symmetric matrix of odd order is singular

Explanation:
1. For a skew-symmetric matrix A of order n, Aᵀ = −A, so |A| = |Aᵀ| = |−A| = (−1)ⁿ|A|.
2. If n is odd, |A| = −|A|, so |A| = 0 and A is singular. So (b) is correct and (a) is false.
3. For even order this argument gives nothing; e.g. \( \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) has determinant 1, so (c) is false.

Teacher's Note:
1. Odd order forces a zero determinant; even order allows any non-negative value.
2. Keep [[0,1],[−1,0]] in mind as a non-singular skew-symmetric example.

Answer: (a) Every identity matrix is a scalar matrix

Explanation:
1. An identity matrix I is diagonal with all diagonal entries equal (to 1), so it is a scalar matrix: (a) is true.
2. A scalar matrix kI with k ≠ 1, such as 2I, is not an identity matrix, so (b) is false; diag(1, 2) shows (c) is false.
3. A matrix of all 1s has non-zero off-diagonal entries, so (d) is false.

Teacher's Note:
1. The chain is: identity ⊂ scalar ⊂ diagonal; each is a special case of the next.
2. Reversing such an inclusion, as in (b) and (c), is the usual trap.

Answer: (b) \( A^2 + B^2 \)

Explanation:
1. B = −A⁻¹BA gives AB = −BA after multiplying on the left by A.
2. (A + B)² = A² + AB + BA + B².
3. Since AB + BA = O, (A + B)² = A² + B².

Teacher's Note:
1. Turn conditions involving A⁻¹ into product relations by multiplying by A on the correct side.
2. AB = −BA means the matrices anticommute, so the cross terms of (A + B)² cancel.

Answer: (c) p = r, q = s

Explanation:
1. Subtraction, like addition, is defined only for matrices of the same order.
2. p × q = r × s means p = r and q = s.

Teacher's Note:
1. p = q would only make the first matrix square, which has nothing to do with subtraction.
2. Compare rows with rows and columns with columns.

Answer: (c) It is not necessary that either A = O or B = O

Explanation:
1. \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \) are both non-zero, yet AB = O.
2. So AB = O does not force A = O or B = O; (a) and (b) are false.
3. (d) is false too, since A = O also gives AB = O. So it is not necessary that either A or B is O.

Teacher's Note:
1. Matrices have zero divisors, so the zero-product rule of real numbers fails.
2. AB = O does force A = O when B is invertible; that is the exception to remember.

Answer: (b) \( \begin{bmatrix} 1/2 & -1/2 \\ 2 & 1 \end{bmatrix} \)

Explanation:
1. \( a_{11} = \frac{1}{2}(3 - 2) = \frac{1}{2} \), \( a_{12} = \frac{1}{2}(3 - 4) = -\frac{1}{2} \).
2. \( a_{21} = \frac{1}{2}(6 - 2) = 2 \), \( a_{22} = \frac{1}{2}(6 - 4) = 1 \).
3. \( A = \begin{bmatrix} 1/2 & -1/2 \\ 2 & 1 \end{bmatrix} \).

Teacher's Note:
1. In aᵢⱼ, i is the row number and j the column number; swapping them produces the transpose, which is option (a).
2. Write the four (i, j) pairs before substituting to stay organised.

Answer: (c) AB = O implies A = O or B = O

Explanation:
1. (a) and (d) are standard true properties, and (b) is true because AB = AC need not give B = C (A may be singular).
2. (c) is false: \( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = O \) with both factors non-zero.
3. So (c) is not true for matrices.

Teacher's Note:
1. Cancellation and the zero-product rule both fail for matrices unless the matrix is invertible.
2. Read 'not true' carefully: (b) itself states a failure of cancellation, which is a true statement.

Answer: (b) \( (AB)^{-1} = B^{-1}A^{-1} \)

Explanation:
1. (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I.
2. So (AB)⁻¹ = B⁻¹A⁻¹ (reversal law), and (a) is false in general.
3. AB = BA is not always true, so (c) and (d) are false. Answer (b).

Teacher's Note:
1. Inverses and transposes of products both reverse the order: (AB)⁻¹ = B⁻¹A⁻¹ and (AB)′ = B′A′.
2. Think of 'socks and shoes': undo the last step first.

Answer: (c) \( A^2 = A \) and \( B^2 = B \)

Explanation:
1. A² = A·A = (AB)A = A(BA) = AB = A.
2. B² = B·B = (BA)B = B(AB) = BA = B.
3. So A² = A and B² = B.

Teacher's Note:
1. Replace A by AB (or B by BA) at the right place, then regroup using associativity.
2. Both A and B turn out idempotent; the two parts are mirror images of each other.

Answer: (b) A + B is symmetric

Explanation:
1. (A + B)ᵀ = Aᵀ + Bᵀ = A + B, so A + B is symmetric.
2. A + B need not be skew-symmetric or diagonal, and A − B ≠ O since A ≠ B.

Teacher's Note:
1. The sum of two symmetric matrices is symmetric; the same holds for differences and scalar multiples.
2. Products are different: AB is symmetric only when AB = BA.

Answer: (b) 8

Explanation:
1. An m × n matrix has mn elements, so we need mn = 24.
2. Factor pairs: 1 × 24, 2 × 12, 3 × 8, 4 × 6 and their reverses 24 × 1, 12 × 2, 8 × 3, 6 × 4.
3. So there are 8 possible orders.

Teacher's Note:
1. The number of orders equals the number of divisors of the element count; 24 = 2³ × 3 has 4 × 2 = 8 divisors.
2. m × n and n × m are different orders, so count both.

Answer: (d) AB = I ⇒ AB = BA

Explanation:
1. (a) AB = AC is not true in general; (b) (AB)² = ABAB, which equals A²B² only if AB = BA.
2. (c) is false, since non-zero matrices can have AB = O.
3. (d): if AB = I for square matrices, then |A||B| = 1, so A is invertible with B = A⁻¹; hence BA = A⁻¹A = I = AB. So (d) is true.

Teacher's Note:
1. For square matrices, a one-sided inverse is automatically a two-sided inverse.
2. Taking determinants is the quick way to show A is invertible from AB = I.

Answer: (a) 1 × 13, 13 × 1

Explanation:
1. An m × n matrix has mn elements, so mn = 13.
2. 13 is prime, so the only possibilities are 1 × 13 and 13 × 1.

Teacher's Note:
1. A prime number of elements forces a row or a column matrix.
2. Options (b) and (c) have 26 elements, not 13.

Answer: (b) C′B′A′

Explanation:
1. By the reversal law, (XY)′ = Y′X′.
2. (ABC)′ = ((AB)C)′ = C′(AB)′ = C′B′A′.

Teacher's Note:
1. The transpose of a product is the product of the transposes in reverse order.
2. The same reversal holds for inverses: (ABC)⁻¹ = C⁻¹B⁻¹A⁻¹.

Answer: (c) \( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \)

Explanation:
1. A is 3 × 1 and A′ = [1  2  3] is 1 × 3, so AA′ is 3 × 3.
2. Entry (i, j) is aᵢaⱼ: \( AA' = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \).

Teacher's Note:
1. AA′ (column times row) is 3 × 3, while A′A (row times column) is the 1 × 1 matrix [14].
2. Option (a), 14, comes from computing A′A instead.

Answer: (a) \( \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} \)

Explanation:
1. \( A^T = \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix} \).
2. \( A + A^T = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix} \).

Teacher's Note:
1. A + Aᵀ is always symmetric, so options that are not symmetric can be rejected at once.
2. Only (a) is symmetric among the given matrices.

Answer: (b) \( \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix} \)

Explanation:
1. A is 2 × 3 and B is 3 × 2, so AB is 2 × 2. Row 1 of A (1, −2, 1) gives (2 − 6 + 1, 1 − 4 + 1) = (−3, −2).
2. Row 2 (2, 1, 3) gives (4 + 3 + 3, 2 + 2 + 3) = (10, 7). So \( AB = \begin{bmatrix} -3 & -2 \\ 10 & 7 \end{bmatrix} \).
3. \( (AB)^T = \begin{bmatrix} -3 & 10 \\ -2 & 7 \end{bmatrix} \).

Teacher's Note:
1. You can also use (AB)ᵀ = BᵀAᵀ, but finding AB and transposing is usually quicker.
2. Option (a) is AB itself; remember the final transpose.

Answer: (d) None of these

Explanation:
1. \( A' = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & 5 \\ -2 & 5 & 0 \end{bmatrix} \).
2. A′ ≠ A, since a₁₂ = 1 but a₂₁ = −1. A′ ≠ −A, since a₂₃ = a₃₂ = 5 (for −A we would need a₃₂ = −5). A′ ≠ 2A clearly.
3. So none of the given relations holds: None of these.

Teacher's Note:
1. For skew-symmetry, every pair aᵢⱼ and aⱼᵢ must be negatives of each other; one equal pair like a₂₃ = a₃₂ = 5 breaks it.
2. Check all three off-diagonal pairs, not just the first one.

Answer: (b) Column matrix

Explanation:
1. A row matrix of order 1 × n has a transpose of order n × 1.
2. An n × 1 matrix is a column matrix.

Teacher's Note:
1. Transposing turns rows into columns, so row and column matrices are transposes of each other.
2. It is a square matrix only when n = 1.

Answer: (d) None of these

Explanation:
1. \( AB = \begin{bmatrix} -5 & 8 \\ -5 & 6 \end{bmatrix} \) and \( BA = \begin{bmatrix} -1 & 6 \\ -2 & 2 \end{bmatrix} \), so AB ≠ BA.
2. \( AA^T = \begin{bmatrix} 13 & 11 \\ 11 & 17 \end{bmatrix} \) but \( A^2 = \begin{bmatrix} 11 & 14 \\ 7 & 18 \end{bmatrix} \), so AAᵀ ≠ A².
3. \( B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \neq AB \). So none of the statements is correct.

Teacher's Note:
1. AAᵀ is always symmetric, while A² is symmetric only for special A, so (b) fails at a glance.
2. When 'None of these' is an option, each statement must be checked and rejected.

Answer: (d) Order of \( B^T A^T \)

Explanation:
1. AB is m × p, so (AB)ᵀ is p × m.
2. Bᵀ is p × n and Aᵀ is n × m, so BᵀAᵀ is p × m, the same order.
3. AB is m × p and BA, AᵀBᵀ need not even be defined. So the order of (AB)ᵀ equals the order of BᵀAᵀ.

Teacher's Note:
1. This order check is consistent with the reversal law (AB)ᵀ = BᵀAᵀ.
2. AᵀBᵀ is (n × m)(p × n), which is defined only if m = p.

Answer: (a) \( \begin{pmatrix} 69 & 80 \\ 80 & 100 \end{pmatrix} \)

Explanation:
1. \( A^T = \begin{pmatrix} 4 & 6 \\ 2 & 0 \\ 7 & 8 \end{pmatrix} \), so AAᵀ is (2 × 3)(3 × 2) = 2 × 2.
2. (1, 1): 16 + 4 + 49 = 69; (1, 2) = (2, 1): 24 + 0 + 56 = 80; (2, 2): 36 + 0 + 64 = 100.
3. \( AA^T = \begin{pmatrix} 69 & 80 \\ 80 & 100 \end{pmatrix} \).

Teacher's Note:
1. The diagonal entries of AAᵀ are the sums of squares of the rows of A, a quick check.
2. AAᵀ is always symmetric, which rules out option (b).

Answer: (b) [0]

Explanation:
1. Let C be n × 1, so CᵀAC is 1 × 1, a number k, and it equals its own transpose.
2. (CᵀAC)ᵀ = CᵀAᵀC = Cᵀ(−A)C = −CᵀAC.
3. So k = −k, which gives k = 0: CᵀAC = [0].

Teacher's Note:
1. A 1 × 1 matrix is always symmetric, so if it is also skew-symmetric it must be zero.
2. This is why xᵀAx = 0 for every vector x when A is skew-symmetric.

Answer: (b) \( (A + B)^T = A^T + B^T \)

Explanation:
1. (A + B)ᵀ = Aᵀ + Bᵀ is a basic property of transpose, so (b) is correct.
2. (a) and (c) reverse the order wrongly: (AB)ᵀ = BᵀAᵀ and (AB)⁻¹ = B⁻¹A⁻¹.
3. (d) is false, since (kA)ᵀ = kAᵀ.

Teacher's Note:
1. Transpose distributes over sums directly but reverses order over products.
2. Scalars come straight out of a transpose: (kA)ᵀ = kAᵀ.

Answer: (a) \( (AB)^T = A^T B \)

Explanation:
1. The correct rule is (AB)ᵀ = BᵀAᵀ, so (AB)ᵀ = AᵀB is wrong in general.
2. (b) (Aᵀ)ᵀ = A, (c) (A − B)ᵀ = Aᵀ − Bᵀ and (d) (Aᵀ)⁻¹ = (A⁻¹)ᵀ are standard true results.
3. So the wrong statement is (a).

Teacher's Note:
1. (Aᵀ)⁻¹ = (A⁻¹)ᵀ follows from transposing AA⁻¹ = I.
2. In 'wrong statement' questions, a single false rule is the answer; check the product rule first.

Answer: (d) 2

Explanation:
1. A square matrix and its transpose have the same determinant: |A′| = |A|.
2. So |A′| = 2.

Teacher's Note:
1. Expanding |A| along a row gives the same result as expanding |A′| along the matching column.
2. Do not confuse this with |A⁻¹| = 1/|A|, which would give 1/2.

Answer: (c) Skew-symmetric

Explanation:
1. Check aⱼᵢ = −aᵢⱼ: a₁₂ = −4, a₂₁ = 4; a₁₃ = 1, a₃₁ = −1; a₂₃ = −5, a₃₂ = 5.
2. All diagonal entries are 0.
3. So Aᵀ = −A and the matrix is skew-symmetric.

Teacher's Note:
1. A skew-symmetric matrix must have every diagonal entry zero, since aᵢᵢ = −aᵢᵢ.
2. Zero diagonal alone is not enough; every off-diagonal pair must also be opposite.

Answer: (a) Symmetric

Explanation:
1. Let Aᵀ = A with A invertible.
2. (A⁻¹)ᵀ = (Aᵀ)⁻¹ = A⁻¹.
3. So A⁻¹ is symmetric.

Teacher's Note:
1. The key fact is that transpose and inverse can be taken in either order.
2. Likewise, the inverse of an invertible skew-symmetric matrix is skew-symmetric.

Answer: (a) Symmetric

Explanation:
1. (Aⁿ)ᵀ = (A·A·…·A)ᵀ = Aᵀ·Aᵀ·…·Aᵀ = (Aᵀ)ⁿ.
2. Since Aᵀ = A, (Aⁿ)ᵀ = Aⁿ.
3. So Aⁿ is symmetric.

Teacher's Note:
1. The reversal law does no harm here because all factors are the same matrix.
2. Contrast with skew-symmetric A, where Aⁿ is symmetric or skew-symmetric depending on whether n is even or odd.

Answer: (d) None of these

Explanation:
1. (Aⁿ)ᵀ = (Aᵀ)ⁿ = (−A)ⁿ = (−1)ⁿAⁿ.
2. If n is even, Aⁿ is symmetric; if n is odd, Aⁿ is skew-symmetric.
3. Since n is an arbitrary positive integer, no single option always holds: None of these.

Teacher's Note:
1. Check with A = [[0,1],[−1,0]]: A² = −I is symmetric, while A itself is skew-symmetric.
2. The sign factor (−1)ⁿ decides the type, so the parity of n matters.

Answer: (b) 5

Explanation:
1. A symmetric matrix needs a₁₂ = a₂₁: x + 2 = 2x − 3.
2. So x = 5.

Teacher's Note:
1. Only the off-diagonal pair gives a condition; diagonal entries can be anything.
2. Check: with x = 5 both off-diagonal entries are 7.

Answer: (b) Symmetric matrix

Explanation:
1. (A + Aᵀ)ᵀ = Aᵀ + (Aᵀ)ᵀ = Aᵀ + A.
2. So A + Aᵀ equals its own transpose and is symmetric.

Teacher's Note:
1. Similarly A − Aᵀ is always skew-symmetric; together they give A = ½(A + Aᵀ) + ½(A − Aᵀ).
2. A + Aᵀ can be singular, e.g. for skew-symmetric A it is O, so (a) is not always true.