JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 8)
If $$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$$$\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right]$$....$$\left[ {\matrix{
1 & {n - 1} \cr
0 & 1 \cr
} } \right] = \left[ {\matrix{
1 & {78} \cr
0 & 1 \cr
} } \right]$$,
then the inverse of $$\left[ {\matrix{ 1 & n \cr 0 & 1 \cr } } \right]$$ is
then the inverse of $$\left[ {\matrix{ 1 & n \cr 0 & 1 \cr } } \right]$$ is
$$\left[ {\matrix{
1 & { 0} \cr
{12} & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & { 0} \cr
{13} & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & { - 13} \cr
0 & 1 \cr
} } \right]$$
$$\left[ {\matrix{
1 & { - 12} \cr
0 & 1 \cr
} } \right]$$
Explanation
Given
$$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$$....$$\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & 6 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & {1 + 2 + 3} \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
.
.
.
.
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & {1 + 2 + 3 + .... + \left( {n - 1} \right)} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
By comparing both sides we get,
1 + 2 + 3 + ........+ (n - 1) = 78
$$ \Rightarrow $$ $${{n\left( {n - 1} \right)} \over 2}$$ = 78
$$ \Rightarrow $$ n = 13, - 12(not possible)
$$ \therefore $$ The inverse of $$\left[ {\matrix{ 1 & 13 \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & -13 \cr 0 & 1 \cr } } \right]$$
$$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$$....$$\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & 6 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & {1 + 2 + 3} \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
.
.
.
.
$$ \Rightarrow $$ $$\left[ {\matrix{ 1 & {1 + 2 + 3 + .... + \left( {n - 1} \right)} \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$
By comparing both sides we get,
1 + 2 + 3 + ........+ (n - 1) = 78
$$ \Rightarrow $$ $${{n\left( {n - 1} \right)} \over 2}$$ = 78
$$ \Rightarrow $$ n = 13, - 12(not possible)
$$ \therefore $$ The inverse of $$\left[ {\matrix{ 1 & 13 \cr 0 & 1 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & -13 \cr 0 & 1 \cr } } \right]$$
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