1

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

If the system of linear equations
2x + 2y + 3z = a
3x – y + 5z = b
x – 3y + 2z = c
where a, b, c are non zero real numbers, has more one solution, then :
A
b – c – a = 0
B
a + b + c = 0
C
b – c + a = 0
D
b + c – a = 0

## Explanation

P1 : 2x + 2y + 3z = a

P2 : 3x $-$ y + 5z = b

P3 : x $-$ 3y + 2z = c

We find

P1 + P3 = P2 $\Rightarrow$ a + c = b
2

### 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

$\Rightarrow$  $\left| p \right| = {1 \over {\sqrt 2 }}$
3

### 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)
4

### 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) $= {1 \over 4} \times {1 \over 4} = {1 \over {16}}$