If $A, B$ are two non singular matrices of order $3,|B|=k$, a positive integer, then match the items of list-I with the items of list-II.
| $$ \text { List-I } $$ |
$$ \text { List-II } $$ |
||
|---|---|---|---|
| A. | $\quad\left|k^{-1} A^{-1}\right|$ | I. | $$ B A^k+A^k B $$ |
| B. | $\left|\operatorname{Adj}\left(A^{-1}\right)\right|$ | II. | $$ \frac{B \operatorname{Adj}(B)}{|B|} $$ |
| C. | $B A B^{-1}=I, \Rightarrow B A^k B^{-1}=$ | III. | $$ \frac{1}{|B|^3|A|} $$ |
| D. | $\quad \operatorname{Adj}\left(\operatorname{Adj}\left(A^{-1}\right)\right)=$ | IV. | $$ \frac{1}{|A|}\left(A^{-1}\right) $$ |
| V. | $$ \frac{1}{|A|^2} $$ |
||
$$ \text { The correct match is } $$
All the real values of $p, q$ so that the system of equations
$$ 2 x+p y+6 z=8, x+2 y+q z=5 $$
and $\quad x+y+3 z=4$
may have no solution are
If $p$ and $q$ are two distinct real values of $\lambda$ for which the system of equations
$$ \begin{array}{r} (\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0 \\ (\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0 \\ 2 x+(3 \lambda+1) y+3(\lambda-1) z=0 \end{array} $$
has non-zero solution, then $p^2+q^2-p q=$
Let $z=x+i y$ be a complex number, $A=\{z /|z| \leq 2\}$ and $B=\{z /(1-i) z+(1+i) \bar{z} \geq 4\}$ Then which one of the following options belongs to $A \cap B$ ?
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