NIMCET Matrices Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Matrices PYQ

Qus : 1 NIMCET PYQ

If A and B are square matrices such that $B=-A^{-1} BA$, then $(A + B)^2$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

Qus : 2 NIMCET PYQ

If $A$ is a $3\times 3$ matrix with $\det(A)=3$, then $\det(\operatorname{adj}A)$ is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

For an $n \times n$ matrix:

$\det(\operatorname{adj}A) = (\det A)^{,n-1}$

Here: $n = 3$, $\det A = 3$

So, $\det(\operatorname{adj}A) = 3^{3-1} = 3^2 = 9$

Qus : 3 NIMCET PYQ

If $a + b + c \neq 0$, the system of equations: $(b+c)(y+z) - ax = b - c$ $(c+a)(z+x) - by = c - a$ $(a+b)(x+y) - cz = a - b$ has:

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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

This is a symmetric linear system.

Solving gives the unique solution:

$x = y = z = 0$

Since $a+b+c \neq 0$, determinant $\neq 0$ → unique solution.

Qus : 4 NIMCET PYQ

Car travels half distance with $v_1$, half with $v_2$. Average speed is

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

For equal distances: Average velocity $=\dfrac{2v_1v_2}{v_1+v_2}$

Qus : 5 NIMCET PYQ

If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, then $I + A + A^2 + A^3 + \cdots \infty$ equals:

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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

Since $I + A + A^2 + \cdots = (I - A)^{-1}$,

$I - A = \begin{bmatrix} 0 & -2 \\ -3 & -3 \end{bmatrix}$

$(I - A)^{-1} = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{1}{3} \\ -\tfrac{1}{2} & 0 \end{bmatrix}$.

Qus : 6 NIMCET PYQ

If $A=\begin{bmatrix} 1 &0 &0 \\ 0& 1 &1 \\ 0&-2 & 4 \end{bmatrix}$ and $6A^{–1} = A^{2} + cA + dI$, where $A^{–1}$ is A- inverse, I is the identify matrix, then (c, d) is

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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 7 NIMCET PYQ

$ \text{If } B = -A^{-1}BA,\ \text{then } (A+B)^{2} = $

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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

From $ B = -A^{-1}BA $, multiply both sides by $A$: $ BA = -BA $ So $ BA = 0 $. Then $ (A+B)^{2} = A^{2} + AB + BA + B^{2} = A^{2} + AB + 0 + B^{2} $

Qus : 8 NIMCET PYQ

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

Qus : 9 NIMCET PYQ

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

Qus : 10 NIMCET PYQ

If A is an invertible skew-symmetric matrix, then

is a

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 11 NIMCET PYQ

Consider the matrix $$B=\begin{pmatrix}{-1} & {-1} & {2} \\ {0} & {-1} & {-1} \\ {0} & {0} & {-1}\end{pmatrix}$$. The sum of all the entries of the matrix $B^{19}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

The matrix is \[ B=\begin{pmatrix} -1 & -1 & 2\\ 0 & -1 & -1\\ 0 & 0 & -1 \end{pmatrix} \] Since $B$ is upper–triangular with diagonal entries $-1$, write \[ B = -I + N, \] where \[ N=\begin{pmatrix} 0 & -1 & 2\\ 0 & 0 & -1\\ 0 & 0 & 0 \end{pmatrix}, \qquad N^{3}=0. \] Using the binomial expansion: \[ B^{19}=(-I+N)^{19} = (-1)^{19}I + \binom{19}{1}(-1)^{18}N + \binom{19}{2}(-1)^{17}N^{2}. \] Compute signs: \[ (-1)^{19}=-1,\quad (-1)^{18}=1,\quad (-1)^{17}=-1. \] So: \[ B^{19} = -I + 19N - \binom{19}{2}N^{2}. \] Now compute sums: \[ \text{sum}(-I) = -3 \] \[ N^{2} = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \quad\Rightarrow\quad \text{sum}(N^{2}) = 1 \] \[ \Rightarrow -\binom{19}{2}\cdot 1 = -171 \] Final total: \[ -3 -171 = -174. \]

Qus : 12 NIMCET PYQ

The value of ‘a’ for which the system of equations $a^3 x + (a+1)^3 y + (a+2)^3 z = 0$ $ax + (a+1) y + (a+2) z = 0$ $x + y + z = 0$ has a non–zero solution, is

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

For non-zero solution, determinant must be zero. Matrix: $\begin{vmatrix} a^3 & (a+1)^3 & (a+2)^3 \ a & a+1 & a+2 \ 1 & 1 & 1 \end{vmatrix} = 0$ Factor out structure: This determinant becomes zero when columns become linearly dependent → when $a=-1$ or $a=0$ or $a=1$. Checking each value in equations: • $a = -1$ → valid • $a = 0$ → equations collapse but still allow nonzero solution • $a = 1$ → also gives dependence But only one of these matches the options where system definitely has non-zero solution. Correct value = $-1$

Qus : 13 NIMCET PYQ

If $F(\theta)=\begin{bmatrix}{\cos \theta} & {-\sin \theta} & {0} \\ {\sin \theta} & {\cos \theta} & {0} \\ {0} & {0} & {1}\end{bmatrix}$ , then $F(\theta)F(\alpha)$ is equal to

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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution

Qus : 14 NIMCET PYQ

Let A and B be two square matrices of same order satisfying $A^2+5A+5I =0$ and $B^2+3B+I=0$ repectively. Where I is the identity matrix. Then the inverse of the matrix $C= BA+2B+2A+4I$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

Given: $A^2 + 5A + 5I = 0 \quad\Rightarrow\quad A^2 = -5A - 5I$

$B^2 + 3B + I = 0 \quad\Rightarrow\quad B^2 = -3B - I$

Matrix: $C = BA + 2B + 2A + 4I$

We want $C^{-1}$.

Step 1: Try a linear expression as inverse Assume: $C^{-1} = \alpha AB + \beta A + \gamma B + \delta I$

Test using the fact that quadratic equations give inverses as linear polynomials.

Check option (2):

$AB + A + 3B + 3I$ Multiply with $C$:

$C(AB + A + 3B + 3I)$

$= (BA + 2B + 2A + 4I)(AB + A + 3B + 3I)$

Using reductions:

$A^2 = -5A - 5I$ $B^2 = -3B - I$

Everything simplifies to: $I$

Hence: $C^{-1} = AB + A + 3B + 3I$

Qus : 15 NIMCET PYQ

Find a matrix X such that 2A + B + X = 0, whose A =

and B =

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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

Qus : 16 NIMCET PYQ

If x, y and z are three cube roots of 27, then the determinant of the matrix $\begin{bmatrix}{x} & {y} & {z} \\ {y} & {z} & {x} \\ {z} & {x} & {y}\end{bmatrix}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

If $x, y, z$ are three cube roots of $27$, then the determinant of the matrix \[ \begin{pmatrix} x & y & z\\[4pt] y & z & x\\[4pt] z & x & y \end{pmatrix} \] is: The cube roots of $27 = 3^3$ are: \[ x = 3,\qquad y = 3\omega,\qquad z = 3\omega^2, \] where $\omega$ is a cube root of unity satisfying \[ \omega^3 = 1,\qquad 1+\omega+\omega^2 = 0. \] For a circulant matrix, the determinant is: \[ (x+y+z)(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z). \] Now compute the first factor: \[ x+y+z = 3(1+\omega+\omega^2) = 3\cdot 0 = 0. \] Therefore, \[ \det = 0. \]

Qus : 17 NIMCET PYQ

If $A = \begin{bmatrix} 0 & 5 \\ 0 & 0 \end{bmatrix}$ and $f(x) = 1 + x + x^2 + \cdots + x^{16}$, then $f(A) =$

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 18 NIMCET PYQ

If $A=\begin{bmatrix} a &b &c \\ b & c & a\\ c& a &b \end{bmatrix}$ , where $a, b, c$ are real positive numbers such that $abc = 1$ and $A^{T}A=I$ then the equation that not holds true among the following is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 19 NIMCET PYQ

If the matrix $ \begin{bmatrix} -1 & 3 & 2 \\ 1& k &-3 \\ 1 & 4 & 5\\ \end{bmatrix}$ has an inverse matrix, then the value of K is:

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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution

Qus : 20 NIMCET PYQ

If $f(x)=\left\{\begin{matrix} \frac{sin[x]}{[x]} &, [x]\ne0 \\ 0 &, [x]=0 \end{matrix}\right.$ , where [x] is the largest integer but not larger than x, then $\lim_{x\to0}f(x)$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution

Qus : 21 NIMCET PYQ

The matrix $A$ has $x$ rows and $(x+5)$ columns and the matrix $B$ has $y$ rows and $(11-y)$ columns. If both the matrices $AB$ and $BA$ exist, then value of $x$ and $y$ are

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Qus : 22 NIMCET PYQ

A matrix $M_r$ is defined as $M_r=\begin{bmatrix} r &r-1 \\ r-1&r \end{bmatrix} , r \in N$ then the value of $det(M_1) + det(M_2) +...+ det(M_{2015})$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 23 NIMCET PYQ

The system of equations

x + y + 2z = a

x + z = b

2x + y + 3z = c

has a solution if

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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ

Solution

Given system of equations:
$x + y + 2z = a$
$x + z = b$
$2x + y + 3z = c$
Writing in matrix form $[A|B]$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 1 & 0 & 1 & b \\ 2 & 1 & 3 & c \end{array}\right]$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - 2R_1$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & -1 & -1 & c-2a \end{array}\right]$
Applying $R_3 \rightarrow R_3 - R_2$:
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-2a-(b-a) \end{array}\right]$
$\left[\begin{array}{ccc|c} 1 & 1 & 2 & a \\ 0 & -1 & -1 & b-a \\ 0 & 0 & 0 & c-a-b \end{array}\right]$
For system to have a solution:
$\rho(A) = \rho(A|B)$
$\Rightarrow c - a - b = 0$
$\therefore \boxed{a + b = c}$

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