NIMCET Determinants Previous Year Question Paper and Solution with PDF

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NIMCET Determinants PYQ

Qus : 1 NIMCET PYQ

The number of values of $k$ for which the system of equations $(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is

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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ

Solution

$ (k+1)x + 8y = 4k $

$ kx + (k+3)y = 3k - 1 $

Since the system has infinitely many solutions, $ \dfrac{k+1}{k} = \dfrac{8}{k+3} = \dfrac{4k}{3k-1} $

Taking 1st and 3rd ratio: $ (k+1)(3k-1) = 4k^2 $

$ 3k^2 + 2k - 1 = 4k^2 $

$ k^2 - 2k + 1 = 0 $

$ (k-1)^2 = 0 $

$ \therefore k = 1 $

Qus : 2 NIMCET PYQ

If $\omega$ is a cube root of unity, then find the value of determinant $\begin{vmatrix} 1+\omega &\omega^{2} &-\omega \\ 1+\omega^{2}&\omega &-\omega^{2} \\ \omega^{2}+\omega&\omega &-\omega^{2} \end{vmatrix}$

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NIMCET Previous Year PYQ NIMCET NIMCET 2013 PYQ

Solution

Qus : 3 NIMCET PYQ

If $x, y, z$ are distinct real numbers, then $$ \begin{vmatrix} x & x^{2} & 2 + x^{3} \\ y & y^{2} & 2 + y^{3} \\ z & z^{2} & 2 + z^{3} \end{vmatrix} = 0 $$ Then find $xyz$.

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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution

Qus : 4 NIMCET PYQ

The system of equations $x+2y+2z=5$, $x+2y+3z=6$, $x+2y+\lambda z=\mu$ has infinitely many solutions if

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

Given System of Equations:

$x + 2y + 2z = 5$
$x + 2y + 3z = 6$
$x + 2y + \lambda z = \mu$

Goal: Find values of $\lambda$ and $\mu$ such that the system has infinitely many solutions

Step 1: Write Augmented Matrix

$ [A|B] = \begin{bmatrix} 1 & 2 & 2 & 5 \\ 1 & 2 & 3 & 6 \\ 1 & 2 & \lambda & \mu \end{bmatrix} $

Step 2: Row operations: Subtract $R_1$ from $R_2$ and $R_3$

$ \Rightarrow \begin{bmatrix} 1 & 2 & 2 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & \lambda - 2 & \mu - 5 \end{bmatrix} $

Step 3: For infinitely many solutions, rank of coefficient matrix = rank of augmented matrix < number of variables (3)

This happens when the third row becomes all zeros:

$ \lambda - 2 = 0 \quad \text{and} \quad \mu - 5 = 0 $

$\Rightarrow \lambda = 2,\quad \mu = 5$

✅ Final Answer: $\boxed{\lambda = 2,\ \mu = 5}$

Qus : 5 NIMCET PYQ

For an invertible matrix A, which of the following is not always true:

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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

Solution

Qus : 6 NIMCET PYQ

If $D={\begin{vmatrix}{1} & 1 & {1} \\ 1 & {2+x} & {1} \\ {1} & {1} & {2+y}\end{vmatrix}}\, for\, x\ne0,\, y\ne0$ then D is

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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

Solution

Qus : 7 NIMCET PYQ

If $a \ne p$, $b \ne q$, $c \ne r$ and $\left|\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}\right| = 0$, then the value of $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ

Solution

Solution: Given $\left|\begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix}\right| = 0,$ the rows are linearly dependent. Using the determinant identity, we get $\frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c} = 1.$

Qus : 8 NIMCET PYQ

The value of the determinant of the following matrix at $x=2026$ is: $\left|\begin{array}{ccc} x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10 \end{array}\right|$

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

Let

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10\end{array}\right|$

Apply row operations:

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

Then,

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ 1 & 2 & 3\\ 3 & 5 & 7\end{array}\right|$

Now expanding along the first row,

$D=x(2\cdot 7-3\cdot 5)-(x+1)(1\cdot 7-3\cdot 3)+(x+3)(1\cdot 5-2\cdot 3)$

$D=x(14-15)-(x+1)(7-9)+(x+3)(5-6)$

$D=-x+2(x+1)-(x+3)$

$D=-x+2x+2-x-3$

$D=-1$

So, at $x=2026$, the value of the determinant is $-1$.

Qus : 9 NIMCET PYQ

The number of values of $\theta$ in the interval $[0,2\pi]$ for which the following homogeneous system of equations has a non-trivial solution is:$x+(\sin\theta)y+(\cos\theta)z=0$$x+(\cos\theta)y+(\sin\theta)z=0$$x-(\sin\theta)y-(\cos\theta)z=0$

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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

So,

$D=\left|\begin{array}{ccc}1 & \sin\theta & \cos\theta\ 1 & \cos\theta & \sin\theta\ 1 & -\sin\theta & -\cos\theta\end{array}\right|$

Let $\sin\theta=s$ and $\cos\theta=c$.

Then,

$D=\left|\begin{array}{ccc}1 & s & c\ 1 & c & s\ 1 & -s & -c\end{array}\right|$

On simplifying,

$D=2(s-c)(s+c)$

For non-trivial solution,

$D=0$

So,

$2(s-c)(s+c)=0$

Hence,

$s-c=0$ or $s+c=0$

So,

$\sin\theta=\cos\theta$ or $\sin\theta=-\cos\theta$

Case 1:

$\sin\theta=\cos\theta$

$\tan\theta=1$

In $[0,2\pi]$,

$\theta=\frac{\pi}{4},\frac{5\pi}{4}$

Case 2:

$\sin\theta=-\cos\theta$

$\tan\theta=-1$

In $[0,2\pi]$,

$\theta=\frac{3\pi}{4},\frac{7\pi}{4}$

Total number of values of $\theta$ is $4$.

Qus : 10 NIMCET PYQ

If $a, b, c$ are the roots of the equation $x^3 - 3x^2 + 3x + 7 = 0$, then the value of\[\begin{vmatrix}2bc - a^2 & c^2 & b^2 \\c^2 & 2ac - b^2 & a^2 \\b^2 & a^2 & 2ab - c^2\end{vmatrix}\]is

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NIMCET Previous Year PYQ NIMCET NIMCET 2018 PYQ

Solution

Qus : 11 NIMCET PYQ

If the system of equations $3x-y+4z=3$ , $x+2y-3z=-2$ , $6x+5y+λz=-3 $ has atleast one solution, then $λ=$

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NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ

Solution

Qus : 12 NIMCET PYQ

If $a+b+c=\pi$ , then the value of $\begin{vmatrix} sin(A+B+C) &sinB &cosC \\ -sinB & 0 &tanA \\ cos(A+B)&-tanA &0 \end{vmatrix}$ is

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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ

Solution

Qus : 13 NIMCET PYQ

Suppose, the system of linear equations

-2x + y + z = l

x - 2y + z = m

x + y - 2z = n

is such that l + m + n = 0, then the system has:

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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution

Qus : 14 NIMCET PYQ

The number of values of $k$ for which the linear equations

4x + ky + 2z = 0

kx + 4y + z = 0

2x + 2y + z = 0

posses a non-zero solution is

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NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

A homogeneous system has a non-trivial solution $\iff$ the determinant of its coefficient matrix is $0$.
Coefficient matrix $A=\begin{bmatrix}4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1\end{bmatrix}$. Hence, $$ \det(A)= \begin{vmatrix} 4 & k & 2\\ k & 4 & 1\\ 2 & 2 & 1 \end{vmatrix} =-(k-4)(k-2). $$ Setting $\det(A)=0 \Rightarrow -(k-4)(k-2)=0 \Rightarrow k=2 \text{ or } k=4.$
Therefore, the number of values of $k$ is $\boxed{2}$.

Qus : 15 NIMCET PYQ

Let A = (a_ij) and B = (b_ij) be two square matricesof order n and det(A) denotes the determinant of A. Then, which of the following is not correct.

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NIMCET Previous Year PYQ NIMCET NIMCET 2020 PYQ

Solution

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