GATE 2026 Engineering Mathematics Questions and Solutions

1 mark, multiple choice. Let \(A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\) and let \(I\) denote the \(2\times 2\) identity matrix. The inverse \(A^{-1}\) is

1. \(\dfrac{1}{10}\,(7I - A)\)
2. \(\dfrac{1}{10}\,(A - 7I)\)
3. \(10I - 7A\)
4. \(\dfrac{1}{7}\,(A - 10I)\)

Solution

The characteristic polynomial of \(A\) is built from its trace and determinant:

By the Cayley-Hamilton theorem the matrix satisfies its own characteristic equation:

Since \(\det A = 10 \neq 0\), the inverse exists. Multiplying throughout by \(A^{-1}\) and rearranging,

Checking directly,

which is precisely the adjugate formula for the inverse of a \(2\times 2\) matrix. Background: Matrices.

2 marks, numerical answer. Let \(A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{bmatrix}\). The sum of the eigenvalues of \(A^{-1}\), rounded off to one decimal place, is _____.

Solution

Write \(A\) as a multiple of the identity plus the all-ones matrix \(J\):

The rank of \(J\) is 1, so \(J\) has eigenvalue 0 with multiplicity 2. Its remaining eigenvalue equals its trace, namely 3, with eigenvector \((1,1,1)^{T}\). Adding \(2I\) shifts every eigenvalue by 2:

As a check, \(\operatorname{tr}A = 9 = 5 + 2 + 2\) and \(\det A = 5\times 2\times 2 = 20\). Since \(A\) is invertible, the eigenvalues of \(A^{-1}\) are the reciprocals of those of \(A\):

The same number follows from \(\sum\lambda(A^{-1}) = \operatorname{tr}(A^{-1}) = \operatorname{tr}(\operatorname{adj}A)/\det A\). Each diagonal cofactor of \(A\) is \(\det\begin{bmatrix} 3 & 1 \\ 1 & 3\end{bmatrix} = 8\), so \(\operatorname{tr}(\operatorname{adj}A) = 24\) and \(24/20 = 1.2\). Background: Determinants.

1 mark, numerical answer. The value of the limit

is _____.

Solution

Both numerator and denominator vanish at \(x = 0\), so the limit is of the \(0/0\) form. Expanding the exponential as a Maclaurin series,

The first two terms cancel against \(-1-2x\), leaving

The same answer follows from two applications of L'Hopital's rule:

Background: Functions, limits and continuity.

2 marks, numerical answer. A scalar field is defined by \(f(x,y,z) = x^{2}y + y z^{2}\). The maximum rate of increase of \(f\) at the point \((1,\,2,\,-1)\), taken over all possible unit directions, is _____.

Solution

The directional derivative of \(f\) along a unit vector \(\hat{u}\) is \(\nabla f \cdot \hat{u} = |\nabla f|\cos\theta\), where \(\theta\) is the angle between \(\hat{u}\) and the gradient. This is largest when \(\theta = 0\), so the maximum rate of increase is the magnitude of the gradient.

Evaluating at \(x = 1\), \(y = 2\), \(z = -1\):

The direction in which this maximum is attained is the unit vector \((4,2,-4)/6 = (2/3,\,1/3,\,-2/3)\). Background: Differentiability and differentiation.

1 mark, multiple choice. The solution of the initial value problem

on the interval \(-\pi/2 \lt x \lt \pi/2\) is

1. \(y = \tan x\)
2. \(y = 1 - \cos x\)
3. \(y = \sin x\)
4. \(y = x\cos x\)

Solution

The equation is linear and first order, of the form \(y' + P(x)y = Q(x)\) with \(P = \tan x\). Its integrating factor is

Multiplying through by \(\sec x\) makes the left side an exact derivative:

Applying \(y(0) = 0\), and noting \(\sec 0 = 1\) and \(\tan 0 = 0\), gives \(C = 0\). Hence

Verification: with \(y = \sin x\), the left side is \(\cos x + \sin x\tan x = \cos x + \sin^{2}x/\cos x = (\cos^{2}x + \sin^{2}x)/\cos x = \sec x\), which matches the right side. Background: Differential equations.

2 marks, numerical answer. The function \(y(t)\) satisfies

The value of \(y\) at \(t = 0.5\), rounded off to four decimal places, is _____.

Solution

The auxiliary equation is

The root is repeated, so the general solution carries an extra factor of \(t\) on the second term:

Applying the first condition, \(y(0) = A = 1\). Differentiating,

At \(t = 0.5\), with \(e^{-1} = 0.3678794\),

The response is critically damped: it rises to a single maximum at \(t = 1/6\) and then decays without oscillation. Background: Differential equations.

1 mark, multiple choice. Let \(C\) be the circle \(|z| = 2\) traversed once in the counter-clockwise sense. The value of the contour integral

is

1. \(2\pi i\)
2. \(-\pi i\)
3. \(\pi i\)
4. \(0\)

Solution

The integrand has simple poles at \(z = 1\) and \(z = 3\). Only the pole at \(z = 1\) lies inside \(C\), since \(|1| = 1\) is less than 2 while \(|3| = 3\) is greater than 2. The residue theorem therefore involves a single residue.

Equivalently, writing \(f(z) = z/(z-3)\), which is analytic inside and on \(C\), Cauchy's integral formula gives \(2\pi i\, f(1) = 2\pi i \times \frac{1}{-2} = -\pi i\). Background: Complex numbers.

2 marks, numerical answer. Box A contains 3 red and 2 white balls. Box B contains 2 red and 4 white balls. One box is selected at random, each box being equally likely, and a single ball is drawn at random from it. The drawn ball turns out to be red. The probability that it came from box A, rounded off to two decimal places, is _____.

Solution

Let \(A\) and \(B\) denote the events that box A and box B are selected, and let \(R\) denote the event that the drawn ball is red. The prior probabilities and the conditional probabilities are

By the law of total probability,

Bayes' theorem then reverses the conditioning:

In exact fractions this is \(\dfrac{1/2 \times 3/5}{1/2 \times 3/5 + 1/2 \times 1/3} = \dfrac{3/10}{3/10 + 1/6} = \dfrac{9/30}{14/30} = \dfrac{9}{14} = 0.64\). Observing a red ball raises the probability of box A from 0.50 to 0.64, because box A is the richer source of red balls. Background: Probability.

2 marks, numerical answer. The Newton-Raphson method is applied to \(f(x) = x^{3} - 2x - 5\) starting from \(x_{0} = 2\). The value of the second iterate \(x_{2}\), rounded off to four decimal places, is _____.

Solution

The Newton-Raphson iteration is

First iteration. At \(x_{0} = 2\),

Second iteration. At \(x_{1} = 2.1\),

The residual has fallen from \(-1\) to \(0.061\) and the correction from \(0.1\) to \(0.0054\), the roughly quadratic contraction expected of Newton-Raphson near a simple root. The exact root of \(x^{3} - 2x - 5 = 0\) is 2.0945515 to seven places, so two iterations already give four correct significant figures.

2 marks, multiple choice. The inverse Laplace transform of

for \(t \ge 0\) is

1. \(e^{-2t}\left(\cos 3t + 3\sin 3t\right)\)
2. \(e^{-2t}\left(\cos 3t - \tfrac{1}{3}\sin 3t\right)\)
3. \(e^{-3t}\left(\cos 2t + \tfrac{1}{2}\sin 2t\right)\)
4. \(e^{-2t}\left(\cos 3t + \tfrac{1}{3}\sin 3t\right)\)

Solution

Complete the square in the denominator. Since \(s^{2} + 4s + 13 = (s+2)^{2} + 9\), the poles sit at \(s = -2 \pm j3\), so the answer must decay as \(e^{-2t}\) and oscillate at 3 rad/s. Options (A), (B) and (D) share that envelope; option (C) has the damping and frequency interchanged and can be discarded at once.

Split the numerator so that each piece matches a standard shifted transform:

Using the first shifting theorem together with the pairs \(\mathcal{L}\{\cos\omega t\} = s/(s^{2}+\omega^{2})\) and \(\mathcal{L}\{\sin\omega t\} = \omega/(s^{2}+\omega^{2})\),

A quick check with the initial value theorem: \(f(0^{+}) = \lim_{s\to\infty} sF(s) = \lim_{s\to\infty} \frac{s^{2}+3s}{s^{2}+4s+13} = 1\), and the proposed \(f(t)\) gives \(f(0) = 1(1 + 0) = 1\). Background: Chapter 3: Laplace transform and transfer functions.