1
BITSAT 2021
+3
-1

Matrix $$A = \left| {\matrix{ x & 3 & 2 \cr 1 & y & 4 \cr 2 & 2 & z \cr } } \right|$$, if xyz = 60 and 8x + 4y + 3z = 20, then A(adj A) is equal to

A
$$\left[ {\matrix{ {64} & 0 & 0 \cr 0 & {64} & 0 \cr 0 & 0 & {64} \cr } } \right]$$
B
$$\left[ {\matrix{ {88} & 0 & 0 \cr 0 & {88} & 0 \cr 0 & 0 & {88} \cr } } \right]$$
C
$$\left[ {\matrix{ {68} & 0 & 0 \cr 0 & {68} & 0 \cr 0 & 0 & {68} \cr } } \right]$$
D
$$\left[ {\matrix{ {34} & 0 & 0 \cr 0 & {34} & 0 \cr 0 & 0 & {34} \cr } } \right]$$
2
BITSAT 2020
+3
-1

An ordered pair ($$\alpha$$, $$\beta$$) for which the system of linear $$(1 + \alpha )x + \beta y + z = 2$$, $$\alpha x + (1 + \beta )y + z = 3$$, $$\alpha x + \beta y + 2z = 2$$ has a unique solution.

A
(1, $$-$$3)
B
($$-$$3, 1)
C
(2, 4)
D
($$-$$4, 2)
3
BITSAT 2020
+3
-1

Consider matrix $$A = \left[ {\matrix{ 2 & 1 \cr 1 & 2 \cr } } \right]$$, if $${A^{ - 1}} = \alpha I + \beta A$$, where $$\alpha$$, $$\beta$$ $$\notin$$ R, then ($$\alpha$$ + $$\beta$$) is equal to (where A$$-$$1 denotes the inverse of matrix A)

A
1
B
$${4 \over 3}$$
C
$${5 \over 3}$$
D
$${1 \over 3}$$
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