NIMCET 2027 Determinants Previous Year Questions (PYQs) – Page 1 of 2
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Solution
Solution
Solution
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: $\boxed{\lambda = 2,\ \mu = 5}$
Solution
Solution
Solution
Solution
$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10\end{array}\right|$
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 $-1$.
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 $4$.
Solution
NIMCET 2027
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NIMCET 2027
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