BITSAT 2021 | Matrices and Determinants Question 9 | Mathematics

If p$$\ne$$ q $$\ne$$ r and $$\left| {\matrix{ 0 & {x - p} & {x - q} \cr {x + p} & 0 & {x - r} \cr {x + q} & {x - r} & 0 \cr } } \right| = 0$$, then the value of x which satisfy the equation is

x = p

x = q

x = r

x = 0

BITSAT 2021

MCQ (Single Correct Answer)

-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

$$\left[ {\matrix{ {64} & 0 & 0 \cr 0 & {64} & 0 \cr 0 & 0 & {64} \cr } } \right]$$

$$\left[ {\matrix{ {88} & 0 & 0 \cr 0 & {88} & 0 \cr 0 & 0 & {88} \cr } } \right]$$

$$\left[ {\matrix{ {68} & 0 & 0 \cr 0 & {68} & 0 \cr 0 & 0 & {68} \cr } } \right]$$

$$\left[ {\matrix{ {34} & 0 & 0 \cr 0 & {34} & 0 \cr 0 & 0 & {34} \cr } } \right]$$

BITSAT 2020

MCQ (Single Correct Answer)

-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.

(1, $$-$$3)

($$-$$3, 1)

(2, 4)

($$-$$4, 2)

BITSAT 2020

MCQ (Single Correct Answer)

-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)

$${4 \over 3}$$

$${5 \over 3}$$

$${1 \over 3}$$

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