JEE MAIN Matrices Previous Year Questions (PYQs) – Page 10 of 15

For two $3 \times 3$ matrices $A$ and $B$, let $A + B = 2B^T$ and $3A + 2B = I_3$, where $B^T$ is the transpose of $B$ and $I_3$ is $3 \times 3$ identity matrix. Then:

Let $B=\left[\begin{array}{lll}1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4\end{array}\right], \alpha > 2$ be the adjoint of a matrix $A$ and $|A|=2$. Then $\left[\begin{array}{ccc}\alpha & -2 \alpha & \alpha\end{array}\right] B\left[\begin{array}{c}\alpha \\ -2 \alpha \\ \alpha\end{array}\right]$$ is equal to :

If $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 3 \\ 0 & 1\end{bmatrix} \cdots \begin{bmatrix}1 & n-1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 78 \\ 0 & 1\end{bmatrix}$, then the inverse of $\begin{bmatrix}1 & n \\ 0 & 1\end{bmatrix}$ is:

The system of linear equations
3x - 2y - kz = 10
2x - 4y - 2z = 6
x+2y - z = 5m
is inconsistent if :

Let $A=\begin{bmatrix}\dfrac{1}{\sqrt{10}} & \dfrac{3}{\sqrt{10}}\\[4pt]-\dfrac{3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}}\end{bmatrix}$ and $B=\begin{bmatrix}1 & -i\\[2pt] 0 & 1\end{bmatrix}$, where $i=\sqrt{-1}$.

Let $\mathrm{A}=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(\mathrm{A})=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(I+A)^8$ is :

$\text{The number of symmetric matrices of order }3\text{, with all entries from the set }{0,1,2,3,4,5,6,7,8,9}\text{ is:}$

Let $A + 2B = \left[ {\matrix{ 1 & 2 & 0 \cr 6 & { - 3} & 3 \cr { - 5} & 3 & 1 \cr } } \right]$ and $2A - B = \left[ {\matrix{ 2 & { - 1} & 5 \cr 2 & { - 1} & 6 \cr 0 & 1 & 2 \cr } } \right]$. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) $-$ Tr(B) has value equal to

Let $\alpha$ and $\beta$ be real numbers. Consider a $3\times 3$ matrix $A$ such that $A^{2}=3A+\alpha I$. If $A^{4}=21A+\beta I$, then

If the system of linear equations

$\begin{aligned} & x-2 y+z=-4 \\ & 2 x+\alpha y+3 z=5 \\ & 3 x-y+\beta z=3 \end{aligned}$

has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to