Which statements are true for n × n real matrices A and B?

A rotation matrix R satisfies R⁻¹ = Rᵀ

The integral ∫_{−1}^{1} dx/x²:

equals 0, since the integrand is even

Matrices, Taylor Series, Improper Integrals and the Vector Identities: the Part A.1 Topics the Shared Chapters Leave Out

Section A.1 of the GATE Robotics and Automation paper is its Engineering Mathematics, and most of it is taught in the shared chapters: rank, systems and eigenvalues; limits, derivatives and definite integrals; partial derivatives, maxima and minima and multiple integrals. This chapter teaches the A.1 sub-items those chapters do not: the transpose, determinant, adjoint and trace rules that every rotation-matrix question leans on; Taylor series in one and two variables; improper integrals; the vector identities beyond the two the shared vector-calculus chapter proves (the product rule for divergence, curl of curl, the position vector), with gradient, divergence and curl restated for the worked cases; and the total derivative of a vector function, worked on the forward-kinematics map of a two-link arm, whose determinant is the number a manipulator question most often wants.

1. Transpose, determinant, adjoint, inverse and trace

The transpose reverses a product: (AB)ᵀ = BᵀAᵀ, and (A⁻¹)ᵀ = (Aᵀ)⁻¹. A matrix with Aᵀ = A is symmetric; with Aᵀ = −A skew-symmetric, and its diagonal is then zero. The determinant survives transposition, |Aᵀ| = |A|, multiplies over products, |AB| = |A||B|, and scales with the order: for an n × n matrix |kA| = kⁿ|A|, not k|A|, because every one of the n rows is multiplied by k.

The adjoint (adjugate) of A is the transpose of its cofactor matrix, and it satisfies A · adj A = adj A · A = |A| I. So when |A| ≠ 0, A⁻¹ = adj A / |A|. Taking determinants of that identity gives |adj A| = |A|ⁿ⁻¹ and, applied twice, adj(adj A) = |A|ⁿ⁻² A. For A = [[2, 1], [1, 3]]: |A| = 5, adj A = [[3, −1], [−1, 2]] (swap the diagonal, negate the off-diagonal — the 2 × 2 shortcut), so A⁻¹ = (1/5)[[3, −1], [−1, 2]].

The rules, with a 3 × 3 matrix of determinant 4
Quantity	Rule	Value
\|2A\|	2³\|A\|	32
\|A⁻¹\|	1/\|A\|	1/4
\|adj A\|	\|A\|³⁻¹	16
\|Aᵀ A\|	\|A\|²	16

The trace is the sum of the diagonal entries. It is linear, tr(A + B) = tr A + tr B, it ignores the order of a product, tr(AB) = tr(BA), and it equals the sum of the eigenvalues, as the determinant equals their product. A rotation matrix is the case this paper cares about most: it is orthogonal, Rᵀ R = I, so R⁻¹ = Rᵀ at no cost, and |R| = +1. The 2D rotation [[cos θ, −sin θ], [sin θ, cos θ]] has trace 2 cos θ, which is how the angle is read back from a given matrix.

⚠️ |kA| is not k|A|

Scaling a 3 × 3 matrix by 2 multiplies its determinant by 8. The same slip appears inside adjoint questions: adj(kA) = kⁿ⁻¹ adj A, so |adj(2A)| for a 3 × 3 A with |A| = 4 is |2A|² = 32² = 1024, not 2 × 16.

2. Taylor series in one and two variables

About x = a, f(x) = f(a) + f′(a)(x − a) + f″(a)(x − a)²/2! + … + f⁽ⁿ⁾(a)(x − a)ⁿ/n! + …; about a = 0 it is the Maclaurin series. The ones to know outright: eˣ = 1 + x + x²/2! + x³/3! + …; sin x = x − x³/3! + x⁵/5! − …; cos x = 1 − x²/2! + x⁴/4! − …; ln(1 + x) = x − x²/2 + x³/3 − … for −1 < x ≤ 1. A coefficient question is a derivative question: the coefficient of (x − a)ⁿ is f⁽ⁿ⁾(a)/n!. For ln x about x = 1, f″(x) = −1/x², so the coefficient of (x − 1)² is −1/2.

In two variables, with h = x − a and k = y − b: f(x, y) = f + (h fₓ + k f_y) + (1/2!)(h² fₓₓ + 2hk fₓ_y + k² f_yy) + …, every derivative evaluated at (a, b). For f = eˣ sin y about (0, 0): f = 0, fₓ = 0, f_y = 1, fₓₓ = 0, fₓ_y = 1, f_yy = 0, so up to second order f ≈ y + xy. The first-order part is the linearisation, the tangent plane, and it is what "small-angle" reasoning is: sin θ ≈ θ and cos θ ≈ 1 − θ²/2 are the first terms of the series.

🧠 The 2hk term

The mixed term carries a 2 because fₓ_y and f_yₓ are equal and both appear. Dropping it is the commonest error in two-variable expansions; writing the quadratic part as ½[h k] H [h k]ᵀ with the Hessian H keeps it automatically.

3. Improper integrals

An integral is improper when a limit is infinite or the integrand is unbounded inside the interval, and it is defined as a limit: ∫ₐ^∞ f dx = lim_R→∞ ∫ₐᴿ f dx. It converges if the limit is finite. The two p-tests decide most questions: ∫₁^∞ dx/xᵖ converges iff p > 1, to 1/(p − 1); ∫₀¹ dx/xᵖ converges iff p < 1, to 1/(1 − p). So ∫₁^∞ dx/x³ = 1/2, while ∫₁^∞ dx/x and ∫₀¹ dx/x both diverge.

∫₀^∞ e^−ax dx = 1/a for a > 0, and ∫₀^∞ x e^−x dx = 1 (integration by parts; it is Γ(2) = 1!).
∫_−∞^∞ dx/(1 + x²) = π, since tan⁻¹ x runs from −π/2 to π/2.
Comparison: if 0 ≤ f ≤ g and ∫g converges, so does ∫f; if ∫f diverges, so does ∫g. e^−x² ≤ e^−x for x ≥ 1, so ∫₁^∞ e^−x² dx converges without being evaluated.

⚠️ A hidden singularity

∫₋₁¹ dx/x² looks like an ordinary definite integral, and the antiderivative −1/x gives −2 — a negative number for a positive integrand, which is the warning. The integrand is unbounded at x = 0, and ∫₀¹ dx/x² diverges (p = 2 ≥ 1), so the whole integral diverges.

4. Gradient, divergence, curl and the vector identities

With ∇ = î ∂/∂x + ĵ ∂/∂y + k̂ ∂/∂z: the gradient of a scalar f is the vector ∇f = (fₓ, f_y, f_z), normal to the level surface f = c and pointing in the direction of fastest increase, with magnitude equal to that rate; the rate in the direction of a unit vector û is ∇f · û. The divergence of F = (P, Q, R) is the scalar ∇ · F = Pₓ + Q_y + R_z, the net outflow per unit volume. The curl is the vector ∇ × F = (R_y − Q_z, P_z − Rₓ, Qₓ − P_y), the local rotation. F is solenoidal if ∇ · F = 0 and irrotational if ∇ × F = 0.

Worked: F = (x²y, yz, xz). ∇ · F = 2xy + z + x, which is 4 at (1, 1, 1). ∇ × F = (∂(xz)/∂y − ∂(yz)/∂z, ∂(x²y)/∂z − ∂(xz)/∂x, ∂(yz)/∂x − ∂(x²y)/∂y) = (−y, −z, −x²). For f = x² + y² + z², ∇f = (2x, 2y, 2z), so at (1, 2, 2) |∇f| = 2 × 3 = 6.

The identities, for smooth f and F
Identity	What it says
∇ × (∇f) = 0	A gradient field is irrotational.
∇ · (∇ × F) = 0	A curl field is solenoidal.
∇ · (∇f) = ∇²f	Divergence of a gradient is the Laplacian, not zero in general.
∇ · (fF) = f ∇ · F + F · ∇f	The product rule for divergence.
∇ × (∇ × F) = ∇(∇ · F) − ∇²F	Curl of curl.
r = (x, y, z): ∇ · r = 3, ∇ × r = 0, ∇\|r\| = r/\|r\|	The position vector: divergence 3, not 1.

5. The total derivative of a vector function: the two-link arm

A map from (θ₁, θ₂) to (x, y) has as its total derivative the Jacobian matrix of partial derivatives, and a small change of the inputs gives, to first order, [dx, dy]ᵀ = J [dθ₁, dθ₂]ᵀ. For the planar two-link arm of Section A.3, x = l₁ cos θ₁ + l₂ cos(θ₁ + θ₂) and y = l₁ sin θ₁ + l₂ sin(θ₁ + θ₂), so J = [[−l₁ sin θ₁ − l₂ sin(θ₁ + θ₂), −l₂ sin(θ₁ + θ₂)], [l₁ cos θ₁ + l₂ cos(θ₁ + θ₂), l₂ cos(θ₁ + θ₂)]].

Expanding the determinant, every θ₁ term cancels and det J = l₁ l₂ sin θ₂. It depends only on the elbow angle: with l₁ = 1 m, l₂ = 0.5 m and θ₂ = 30°, det J = 0.5 × 0.5 = 0.25 m². It vanishes at θ₂ = 0° and 180° — the arm fully stretched or fully folded — where the two columns of J are parallel and the tip cannot be moved in the direction along the arm, however the joints turn. Those configurations lie on the boundary of the workspace, which is how the calculus and the geometry of the next section meet.

ℹ️ Why this sits in the mathematics

Section A.3 names forward kinematics and no velocity kinematics; what A.1 names is partial derivatives and the total derivative. The Jacobian here is that calculus applied to the forward-kinematics map, and its determinant is an ordinary 2 × 2 determinant.

Key takeaways

|kA| = kⁿ|A|, A · adj A = |A| I, A⁻¹ = adj A/|A| and |adj A| = |A|ⁿ⁻¹; tr(AB) = tr(BA).
A rotation matrix is orthogonal with determinant +1, so its inverse is its transpose.
The coefficient of (x − a)ⁿ in a Taylor series is f⁽ⁿ⁾(a)/n!; the two-variable second-order term carries 2hk fₓ_y.
∫₁^∞ dx/xᵖ converges iff p > 1 and ∫₀¹ dx/xᵖ iff p < 1; check for a singularity inside the interval before integrating.
curl grad = 0 and div curl = 0; div grad is the Laplacian; ∇ · r = 3. For the two-link arm, det J = l₁ l₂ sin θ₂.

Practice questions (13)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

For conformable matrices A and B, (AB)ᵀ equals:
1. BᵀAᵀ
2. AᵀBᵀ
3. BA
4. (BA)⁻¹
Show answer
Answer: A — BᵀAᵀ
The transpose reverses the order of a product: the (i, j) entry of (AB)ᵀ is row j of A times column i of B, which is row i of Bᵀ times column j of Aᵀ. AᵀBᵀ is generally not even conformable when A and B are not square.
A is a 3 × 3 matrix with |A| = 4. The value of |adj A| is ____.
Numerical answer — type the value.
Show answer
Answer: 16
From A · adj A = |A| I, |A| |adj A| = |A|³, so |adj A| = |A|² = 16. Answering 4 treats the adjoint as if it had the determinant of A; answering 1/4 confuses it with the inverse.
A is a 3 × 3 matrix with |A| = 4. The value of |adj(2A)| is ____.
Numerical answer — type the value.
Show answer
Answer: 1024
|2A| = 2³ × 4 = 32, and for any 3 × 3 matrix M, |adj M| = |M|², so |adj(2A)| = 32² = 1024. Equivalently adj(2A) = 2² adj A, whose determinant is (2²)³ × 16 = 1024. Using |2A| = 8 (that is, 2|A|) gives 64.
The inverse of A = [[2, 1], [1, 3]] is:
1. (1/5)[[3, −1], [−1, 2]]
2. (1/5)[[2, −1], [−1, 3]]
3. (1/7)[[3, −1], [−1, 2]]
4. (1/5)[[3, 1], [1, 2]]
Show answer
Answer: A — (1/5)[[3, −1], [−1, 2]]
|A| = 6 − 1 = 5. For a 2 × 2 matrix the adjoint swaps the diagonal entries and negates the off-diagonal ones: adj A = [[3, −1], [−1, 2]]. Check: A × adj A = [[5, 0], [0, 5]] = 5I. Keeping the diagonal in place or the signs positive fails this check; 7 is the trace-like sum 6 + 1.
Which statements are true for n × n real matrices A and B?
1. tr(AB) = tr(BA)
2. |A + B| = |A| + |B|
3. A · adj A = |A| I
4. A rotation matrix R satisfies R⁻¹ = Rᵀ
Show answer
Answer: A — tr(AB) = tr(BA); C — A · adj A = |A| I; D — A rotation matrix R satisfies R⁻¹ = Rᵀ
(A) Both traces equal Σᵢ Σⱼ aᵢⱼ bⱼᵢ. (B) False: I + I has determinant 2ⁿ, not 2. (C) The defining property of the adjoint. (D) A rotation matrix is orthogonal, RᵀR = I.
In the Taylor series of ln x about x = 1, the coefficient of (x − 1)² is ____ (type a negative value with a minus sign).
Numerical answer — type the value.
Show answer
Answer: -0.5
f′(x) = 1/x and f″(x) = −1/x², so f″(1) = −1 and the coefficient is f″(1)/2! = −0.5. The series is (x − 1) − (x − 1)²/2 + (x − 1)³/3 − …, the ln(1 + u) series with u = x − 1. Forgetting the 2! gives −1.
The Taylor expansion of f(x, y) = eˣ sin y about (0, 0), up to and including second-order terms, is:
1. y + xy
2. x + y + xy
3. y + xy − y²/2
4. 1 + x + y
Show answer
Answer: A — y + xy
At (0, 0): f = 0, fₓ = eˣ sin y = 0, f_y = eˣ cos y = 1, fₓₓ = 0, fₓ_y = eˣ cos y = 1, f_yy = −eˣ sin y = 0. So f ≈ 0 + (0·x + 1·y) + ½(0 + 2·1·xy + 0) = y + xy. A −y²/2 term would need f_yy = −1, which is the cosine’s behaviour, not the sine’s; the constant 1 would need f(0, 0) = 1.
The value of ∫₁^∞ dx/x³ is ____.
Numerical answer — type the value.
Show answer
Answer: 0.5
p = 3 > 1, so it converges to 1/(p − 1) = 1/2: the antiderivative −1/(2x²) is 0 at infinity and −1/2 at x = 1, and 0 − (−1/2) = 0.5.
The integral ∫₋₁¹ dx/x²:
1. diverges
2. equals −2
3. equals 2
4. equals 0, since the integrand is even
Show answer
Answer: A — diverges
The integrand is unbounded at x = 0, inside the interval, so the integral is improper and splits at 0; ∫₀¹ dx/x² has p = 2 ≥ 1 and diverges, so the whole integral diverges. −2 comes from applying −1/x across the singularity. An even integrand doubles the half-interval integral; it does not make it zero.
For f(x, y, z) = x² + y² + z², the magnitude of ∇f at the point (1, 2, 2) is ____.
Numerical answer — type the value.
Show answer
Answer: 6
∇f = (2x, 2y, 2z) = (2, 4, 4) at (1, 2, 2), and |∇f| = √(4 + 16 + 16) = √36 = 6. Reporting f(1, 2, 2) = 9 gives the value of the function, not of its gradient.
For F = (x²y, yz, xz), the divergence ∇ · F at the point (1, 1, 1) is ____.
Numerical answer — type the value.
Show answer
Answer: 4
∇ · F = ∂(x²y)/∂x + ∂(yz)/∂y + ∂(xz)/∂z = 2xy + z + x = 2 + 1 + 1 = 4 at (1, 1, 1). Differentiating each component with respect to the wrong variable — for instance x²y by y — is the usual slip.
For smooth scalar f and vector F, which of these are identities?
1. ∇ × (∇f) = 0
2. ∇ · (∇ × F) = 0
3. ∇ · (∇f) = 0 for every f
4. ∇ · r = 1 for r = (x, y, z)
Show answer
Answer: A — ∇ × (∇f) = 0; B — ∇ · (∇ × F) = 0
(A) and (B) follow from the equality of mixed partial derivatives. (C) ∇ · ∇f = ∇²f, zero only for harmonic f — for f = x², it is 2. (D) ∇ · r = 1 + 1 + 1 = 3.
A planar two-link arm has link lengths l₁ = 1 m and l₂ = 0.5 m, with tip position x = l₁ cos θ₁ + l₂ cos(θ₁ + θ₂), y = l₁ sin θ₁ + l₂ sin(θ₁ + θ₂). The determinant of the Jacobian ∂(x, y)/∂(θ₁, θ₂) at θ₁ = 45°, θ₂ = 30° is ____ m².
Numerical answer — type the value.
Show answer
Answer: 0.25
Expanding the 2 × 2 determinant of partials, the θ₁ terms cancel and det J = l₁ l₂ sin θ₂ = 1 × 0.5 × sin 30° = 0.25. The value of θ₁ is a distractor: rotating the whole arm about its base cannot change how the elbow maps joint motion to tip motion.