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CUET Previous Year Questions (PYQs)

CUET Determinants PYQ


CUET PYQ
If $x,y,z$ are all distinct and $\left| \begin{array}{ccc} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{array} \right|=0$ then the value of $xyz$ is.





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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Expand the determinant by using column operation: $C_3 \rightarrow C_3 - C_1^3$ Then the determinant becomes a Vandermonde determinant multiplied by $(xyz+1)$. Since $x,y,z$ are distinct, the Vandermonde determinant is non-zero. Hence, $xyz+1=0$ $\Rightarrow xyz=-1$

CUET PYQ
If $|\begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{vmatrix} = k$, then the value of k is:





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CUET Previous Year PYQ CUET CUET 2025 PYQ

Solution



CUET PYQ
Consider the system of linear equations as 2x + 2y + z = 1, 4x + ky + 2z = 2 and kx + 4y + z = 1 then choosethe correct statement(s) from blow 
(A) The system of equation has a unique solution if k≠4 and k≠2
(B) The system of equations is inconsistent for every real number k
(C) The system of equations have infinite number of solutions if k = 4
(D) The system of equations have infinite number of solutions if k = 2
Choose the correct answer from the options given below





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CUET Previous Year PYQ CUET CUET 2024 PYQ

Solution

The system of equations is:

2x + 2y + z = 1
4x + ky + 2z = 2
kx + 4y + z = 1

The coefficient matrix is

\(A = \begin{bmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{bmatrix}\).

The determinant is

\(\Delta = \begin{vmatrix} 2 & 2 & 1 \\ 4 & k & 2 \\ k & 4 & 1 \end{vmatrix}\).

Expanding:

\(\Delta = 2\begin{vmatrix} k & 2 \\ 4 & 1 \end{vmatrix} - 2\begin{vmatrix} 4 & 2 \\ k & 1 \end{vmatrix} + 1\begin{vmatrix} 4 & k \\ k & 4 \end{vmatrix}\).

\(\Delta = 2(k-8) - 2(4-2k) + (16-k^2)\).

\(\Delta = -k^2 + 6k - 8 = -(k-2)(k-4)\).

  • If \(k \neq 2,4\), then \(\Delta \neq 0\) and the system has a unique solution.
  • If \(k=4\): equations (1) and (2) are dependent, equation (3) reduces to the same relation, hence infinitely many solutions.
  • If \(k=2\): substituting gives \(y=0\) and \(2x+z=1\), equation (3) is the same, hence infinitely many solutions.

Correct Statements: (A), (C), (D)


CUET PYQ
Arrange the following in decreasing order (based on determinant value).
A. $ \begin{vmatrix} 1 & 3 & 5\\ 2 & 6 & 10\\ 31 & 11 & 38 \end{vmatrix} $

B.$ \begin{vmatrix} 67 & 19 & 21\\ 39 & 13 & 14\\ 81 & 24 & 26 \end{vmatrix} $

C. $ \begin{vmatrix} 1 & -3 & 2\\ 4 & -1 & 2\\ 3 & 5 & 2 \end{vmatrix} $

D. $ \begin{vmatrix} 1 & 4 & 9\\ 4 & 9 & 16\\ 9 & 16 & 25 \end{vmatrix} $





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Solution



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