1

### JEE Main 2019 (Online) 11th January Morning Slot

Let A = $\left( {\matrix{ 0 & {2q} & r \cr p & q & { - r} \cr p & { - q} & r \cr } } \right).$   If  AAT = I3,   then   $\left| p \right|$ is
A
${1 \over {\sqrt 2 }}$
B
${1 \over {\sqrt 5 }}$
C
${1 \over {\sqrt 6 }}$
D
${1 \over {\sqrt 3 }}$

## Explanation

A is orthogonal matrix

$\Rightarrow$  02 + p2 + p2 = 1

$\Rightarrow$  $\left| p \right| = {1 \over {\sqrt 2 }}$
2

### JEE Main 2019 (Online) 11th January Evening Slot

If  $\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

= (a + b + c) (x + a + b + c)2, x $\ne$ 0,

then x is equal to :
A
–2(a + b + c)
B
2(a + b + c)
C
abc
D
–(a + b + c)

## Explanation

$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

R1 $\to$ R1 + R2 + R3

$= \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$= \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr } } \right|$

$=$ (a + b + c) (a + b + c)2

$\Rightarrow$  x $=$ $-$ 2(a + b + c)
3

### JEE Main 2019 (Online) 11th January Evening Slot

Let A and B be two invertible matrices of order 3 $\times$ 3. If det(ABAT) = 8 and det(AB–1) = 8,
then det (BA–1 BT) is equal to :
A
${1 \over 4}$
B
16
C
${1 \over {16}}$
D
1

## Explanation

${\left| A \right|^2}.\left| B \right| = 8$

and ${{\left| A \right|} \over {\left| B \right|}} = 8 \Rightarrow \left| A \right| = 4$

and $\left| B \right| = {1 \over 2}$

$\therefore$  det(BA$-$1. BT) $= {1 \over 4} \times {1 \over 4} = {1 \over {16}}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

An ordered pair ($\alpha$, $\beta$) for which the system of linear equations
(1 + $\alpha$) x + $\beta$y + z = 2
$\alpha$x + (1 + $\beta$)y + z = 3
$\alpha$x + $\beta$y + 2z = 2
has a unique solution, is :
A
(–3, 1)
B
(1, –3)
C
(–4, 2)
D
(2, 4)

## Explanation

For unique solution

$\Delta$ $\ne$ 0 $\Rightarrow$ $\left| {\matrix{ {1 + \alpha } & \beta & 1 \cr \alpha & {1 + \beta } & 1 \cr \alpha & \beta & 2 \cr } } \right| \ne 0$

$\left| {\matrix{ 1 & { - 1} & 0 \cr 0 & 1 & { - 1} \cr \alpha & \beta & 2 \cr } } \right| \ne 0 \Rightarrow \alpha + \beta \ne - 2$