JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 11)
Let $$M = \left[ {\matrix{
0 & 1 & a \cr
1 & 2 & 3 \cr
3 & b & 1 \cr
} } \right]$$ and
adj $$M = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
where a and b are real numbers. Which of the following options is/are correct?
adj $$M = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
where a and b are real numbers. Which of the following options is/are correct?
det(adj M2) = 81
If $$M\left[ {\matrix{
\alpha \cr
\beta \cr
\gamma \cr
} } \right] = \left[ {\matrix{
1 \cr
2 \cr
3 \cr
} } \right]$$, then $$\alpha - \beta + \gamma = 3$$
$${(adj\,M)^{ - 1}} + adj\,{M^{ - 1}} = - M$$
a + b = 3
Explanation
Given square matrix
$$M = \left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right]$$
and adj $$(M) = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
$$ \because $$ |adj (M)| = |M|2 = $$\left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
$$ \Rightarrow $$ $$|M{|^2} = - 1(6 - 6) - 1( - 8 + 10) - 1(24 - 30) = - 2 + 6 = 4$$
$$ \Rightarrow $$ $$|M| = \pm 2$$
$$ \therefore $$ det (adj M2) = $$|{M^2}{|^2} = |M{|^4} = 16$$
As we know A(adj A) = |A| I
$$ \Rightarrow $$ M = |M| (adj M)-1 ....(i)
$$ \because $$ (adj M)-1 = $${1 \over {|adj\,M|}}{\left[ {\matrix{ 0 & { - 2} & { - 6} \cr { - 2} & { - 4} & { - 2} \cr { - 4} & { - 6} & { - 2} \cr } } \right]^T}$$
= $${1 \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$$
So$$\left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right] = {{|M|} \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$$
$$ \Rightarrow $$ M = $$\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]$$
Now, If M$$\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$$
$$ \Rightarrow $$ $$\beta $$ + 2$$\gamma $$ = 1, $$\alpha $$ + 2$$\beta $$ + 3$$\gamma $$ = 2 and 3$$\alpha $$ + $$\beta $$ + $$\gamma $$ = 3
$$ \Rightarrow $$ $$\alpha $$ = 1, $$\beta $$ = -1 and $$\gamma $$ = 1
$$ \therefore $$ $$\alpha $$ - $$\beta $$ + $$\gamma $$ = 3
And $${(adj\,M)^{ - 1}} + adj\,({M^{ - 1}}) = 2{(adj\,M)^{ - 1}}$$
[$$ \because $$ adj (M-1) = (adj M)-1]
= $$2\left( { - {M \over 2}} \right) = - M$$
[$$ \because $$ (adj M)-1 = $${M \over {|M|}}$$ from Eq. (i) ]
and $$ \because $$ a = 2 and b = 1, so a + b = 3
Hence, options (b), (c) and (d) are correct.
$$M = \left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right]$$
and adj $$(M) = \left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
$$ \because $$ |adj (M)| = |M|2 = $$\left[ {\matrix{ { - 1} & 1 & { - 1} \cr 8 & { - 6} & 2 \cr { - 5} & 3 & { - 1} \cr } } \right]$$
$$ \Rightarrow $$ $$|M{|^2} = - 1(6 - 6) - 1( - 8 + 10) - 1(24 - 30) = - 2 + 6 = 4$$
$$ \Rightarrow $$ $$|M| = \pm 2$$
$$ \therefore $$ det (adj M2) = $$|{M^2}{|^2} = |M{|^4} = 16$$
As we know A(adj A) = |A| I
$$ \Rightarrow $$ M = |M| (adj M)-1 ....(i)
$$ \because $$ (adj M)-1 = $${1 \over {|adj\,M|}}{\left[ {\matrix{ 0 & { - 2} & { - 6} \cr { - 2} & { - 4} & { - 2} \cr { - 4} & { - 6} & { - 2} \cr } } \right]^T}$$
= $${1 \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$$
So$$\left[ {\matrix{ 0 & 1 & a \cr 1 & 2 & 3 \cr 3 & b & 1 \cr } } \right] = {{|M|} \over 4}\left[ {\matrix{ 0 & { - 2} & { - 4} \cr { - 2} & { - 4} & { - 6} \cr { - 6} & { - 2} & { - 2} \cr } } \right]$$
$$ \Rightarrow $$ M = $$\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]$$
Now, If M$$\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$$
$$ \Rightarrow $$ $$\left[ {\matrix{ 0 & 1 & 2 \cr 1 & 2 & 3 \cr 3 & 1 & 1 \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr \gamma \cr } } \right] = \left[ {\matrix{ 1 \cr 2 \cr 3 \cr } } \right]$$
$$ \Rightarrow $$ $$\beta $$ + 2$$\gamma $$ = 1, $$\alpha $$ + 2$$\beta $$ + 3$$\gamma $$ = 2 and 3$$\alpha $$ + $$\beta $$ + $$\gamma $$ = 3
$$ \Rightarrow $$ $$\alpha $$ = 1, $$\beta $$ = -1 and $$\gamma $$ = 1
$$ \therefore $$ $$\alpha $$ - $$\beta $$ + $$\gamma $$ = 3
And $${(adj\,M)^{ - 1}} + adj\,({M^{ - 1}}) = 2{(adj\,M)^{ - 1}}$$
[$$ \because $$ adj (M-1) = (adj M)-1]
= $$2\left( { - {M \over 2}} \right) = - M$$
[$$ \because $$ (adj M)-1 = $${M \over {|M|}}$$ from Eq. (i) ]
and $$ \because $$ a = 2 and b = 1, so a + b = 3
Hence, options (b), (c) and (d) are correct.
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