JEE MAIN Matrices Previous Year Questions (PYQs) – Page 15 of 15

JEE MAIN Matrices Previous Year Questions (PYQs) – Page 15 of 15

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For a $3\times3$ matrix $M$, let $\operatorname{trace}(M)$ denote the sum of all the diagonal elements of $M$. Let $A$ be a $3\times3$ matrix such that $|A|=\dfrac{1}{2}$ and $\operatorname{trace}(A)=3$. If $B=\operatorname{adj}(\operatorname{adj}(2A))$, then the value of $|B|+\operatorname{trace}(B)$ equals:

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The number of real values of $\lambda$, such that the system of linear equations

$2x - 3y + 5z = 9$

$x + 3y - z = -18$

$3x - y + (\lambda^2 - |\lambda|)z = 16$

has no solutions, is

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If the system of equations

$x + y + z = 5$
$x + 2y + 3z = 9$
$x + 3y + az = \beta$

has infinitely many solutions, then $\beta - \alpha =$

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Let $\alpha \in (0,\infty)$ and $A=\begin{bmatrix}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{bmatrix}$. If $\det(\operatorname{adj}(2A-A^T)\cdot\operatorname{adj}(A-2A^T))=2^8$, then $(\det(A))^2$ is equal to:

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The values of a and b, for which the system of equations 2x + 3y + 6z = 8, x + 2y + az = 5, 3x + 5y + 9z = b, has no solution, are :

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If $B = \left[ {\matrix{ 5 & {2\alpha } & 1 \cr 0 & 2 & 1 \cr \alpha & 3 & { - 1} \cr } } \right]$ is the inverse of a 3 × 3 matrix A, then the sum of all values of $\alpha $ for which det(A) + 1 = 0, is :

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If $A=\begin{bmatrix} 1 & 2 & 2\\ 2 & 1 & -2\\ a & 2 & b \end{bmatrix}$ is a matrix satisfying the equation $AA^{T}=9I$, where $I$ is $3\times 3$ identity matrix, then the ordered pair $(a,b)$ is equal to :

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If the system of linear equations
x + y + 3z = 0
x + 3y + k2z = 0
3x + y + 3z = 0
has a non-zero solution (x, y, z) for some k $ \in $ R,then x + $\left( {{y \over z}} \right)$ is equal to :

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If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B = $\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right]$, then AB is equal to :

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If the system of equations $(\lambda-1)x+(\lambda-4)y+\lambda z=5$ $\lambda x+(\lambda-1)y+(\lambda-4)z=7$ $(\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9$ has infinitely many solutions, then $\lambda^2+\lambda$ is equal to:

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