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nimcet Previous Year Questions (PYQs)

nimcet Determinants PYQ


nimcet PYQ
If x, y, z are distinct real numbers then  = 0, then xyz=





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nimcet Previous Year PYQnimcet NIMCET 2019 PYQ

Solution


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 PYQnimcet 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: $\boxed{\lambda = 2,\ \mu = 5}$


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





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nimcet Previous Year PYQnimcet NIMCET 2024 PYQ

Solution


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 PYQnimcet NIMCET 2022 PYQ

Solution


nimcet PYQ
If a, b, c are the roots of the equation , then the value of  is





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nimcet Previous Year PYQnimcet NIMCET 2018 PYQ

Solution


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 PYQnimcet NIMCET 2021 PYQ

Solution


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 PYQnimcet NIMCET 2015 PYQ

Solution


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 PYQnimcet NIMCET 2014 PYQ

Solution


nimcet PYQ
The number of values of k for which the linear equations
4x + ky + z = 0
kx + 4y + z = 0
2x + 2y + z = 0
posses a non-zero solution is





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nimcet Previous Year PYQnimcet NIMCET 2020 PYQ

Solution

Since, equation has non-zero solution.
Δ = 0

nimcet PYQ
Let A = (aij) and B = (bij) 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 PYQnimcet NIMCET 2020 PYQ

Solution



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