NIMCET 2027 Determinants Previous Year Questions (PYQs) – Page 2 of 2
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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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Solution
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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Solution
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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Solution
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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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 $\boxed{2}$.
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 $\boxed{2}$.
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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Solution
NIMCET 2027
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and More.
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