JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 4)
Let $$A$$ be a matrix such that $$A.\left[ {\matrix{
1 & 2 \cr
0 & 3 \cr
} } \right]$$ is a scalar matrix and |3A| = 108.
Then A2 equals :
Then A2 equals :
$$\left[ {\matrix{
4 & { - 32} \cr
0 & {36} \cr
} } \right]$$
$$\left[ {\matrix{
{36} & 0 \cr
{ - 32} & 4 \cr
} } \right]$$
$$\left[ {\matrix{
4 & 0 \cr
{ - 32} & {36} \cr
} } \right]$$
$$\left[ {\matrix{
{36} & { - 32} \cr
0 & 4 \cr
} } \right]$$
Explanation
According to questions,
A. $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]$$ = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]^{-1}$$
$$ \Rightarrow $$ A = $$1 \over 3$$$$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 3 & {-2} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 1 & { - {2 \over 3}} \cr 0 & {{1 \over 3}} \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & { - {2 \over 3}\lambda } \cr 0 & {{\lambda \over 3}} \cr } } \right]$$
As $$\left| {3A} \right|$$ = 108
$$ \Rightarrow $$ 108 = $$\left| {\matrix{ {3\lambda } & { - 2\lambda } \cr 0 & \lambda \cr } } \right|$$
$$ \Rightarrow $$ 3$$\lambda $$2 = 108
$$ \Rightarrow $$ $$\lambda $$2 = 36
$$ \Rightarrow $$ $$\lambda $$ = $$ \pm $$6
When $$\lambda $$ = +6
then A = $$\left[ {\matrix{ 6 & { - 4} \cr 0 & 2 \cr } } \right]$$
$$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$$
For $$\lambda $$ = -6
A = $$\left[ {\matrix{ { - 6} & 4 \cr 0 & { - 2} \cr } } \right]$$
$$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$$
A. $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]$$ = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]^{-1}$$
$$ \Rightarrow $$ A = $$1 \over 3$$$$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 3 & {-2} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 1 & { - {2 \over 3}} \cr 0 & {{1 \over 3}} \cr } } \right]$$
$$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & { - {2 \over 3}\lambda } \cr 0 & {{\lambda \over 3}} \cr } } \right]$$
As $$\left| {3A} \right|$$ = 108
$$ \Rightarrow $$ 108 = $$\left| {\matrix{ {3\lambda } & { - 2\lambda } \cr 0 & \lambda \cr } } \right|$$
$$ \Rightarrow $$ 3$$\lambda $$2 = 108
$$ \Rightarrow $$ $$\lambda $$2 = 36
$$ \Rightarrow $$ $$\lambda $$ = $$ \pm $$6
When $$\lambda $$ = +6
then A = $$\left[ {\matrix{ 6 & { - 4} \cr 0 & 2 \cr } } \right]$$
$$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$$
For $$\lambda $$ = -6
A = $$\left[ {\matrix{ { - 6} & 4 \cr 0 & { - 2} \cr } } \right]$$
$$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$$
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