JEE Advance - Mathematics (2021 - Paper 1 Online - No. 11)
For any 3 $$\times$$ 3 matrix M, let | M | denote the determinant of M. Let
$$E = \left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]$$, $$P = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$ and $$F = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right]$$
If Q is a nonsingular matrix of order 3 $$\times$$ 3, then which of the following statements is(are) TRUE?
$$E = \left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]$$, $$P = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$ and $$F = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right]$$
If Q is a nonsingular matrix of order 3 $$\times$$ 3, then which of the following statements is(are) TRUE?
F = PEP and $${P^2} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$
| EQ + PFQ$$-$$1 | = | EQ | + | PFQ$$-$$1 |
| (EF)3 | > | EF |2
Sum of the diagonal entries of P$$-$$1EP + F is equal to the sum of diagonal entries of E + P$$-$$1FP
Explanation
For option (a)
$$PEP = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$
$$ = \left[ {\matrix{ 1 & 2 & 3 \cr 8 & {13} & {18} \cr 2 & 3 & 4 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right] = F$$
and $${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
Hence, option (a) is correct.
For option (b)
$$\left| {EQ + PF{Q^{ - 1}}} \right| = \left| {EQ} \right| + \left| {PF{Q^{ - 1}}} \right|$$ .... (i)
$$\because$$ | E | = 0 and | F | = 0 and | Q | $$\ne$$ 0
$$\therefore$$ $$\left| {EQ} \right| = \left| E \right|\left| Q \right| = 0$$
and $$\left| {PF{Q^{ - 1}}} \right| = {{\left| P \right|\left| F \right|} \over {\left| Q \right|}} = 0$$
Let $$R = EQ + PF{Q^{ - 1}}$$ .... (ii)
$$ \Rightarrow RQ = E{Q^2} + PF = E{Q^2} + {P^2}EP = E{Q^2} + EP$$ [$$\because$$ P2 = I]
$$ = E({Q^2} + P)$$
$$ \Rightarrow \left| {RQ} \right| = \left| {E({Q^2} + P)} \right|$$
$$ \Rightarrow \left| R \right|\left| Q \right| = \left| E \right|\left| {{Q^2} + P} \right| = 0$$ [$$\because$$ | E | = O]
$$ \Rightarrow \left| R \right| = 0$$ (as $$\left| Q \right| \ne 0$$) ..... (iii)
From Eqs. (ii) and (iii), we get Eq. (i) is true.
Hence, option (b) is correct.
For option (c)
$$\left| {{{(EF)}^3}} \right| > {\left| {EF} \right|^2}$$
i.e. 0 > 0 which is false.
For option (d)
$$\because$$ $${P^2} = I \Rightarrow {P^{ - 1}} = P$$
$$\therefore$$ $${P^{ - 1}}FP = PFP = PPEPP = E$$
So, $$E + {P^{ - 1}}FP = E + E = 2E$$
$$ \Rightarrow Tr(E + {P^{ - 1}}FP) = Tr(2E) = 2Tr(E)$$ ..... (iv)
and $${P^{ - 1}}EP + F$$
$$ \Rightarrow PEP + F = 2PEP$$ [$$\because$$ F = PEP]
$$\therefore$$ $$Tr(2PEP) = 2Tr(PEP) = 2Tr(F) = 2Tr(E)$$ ..... (v)
From Eqs. (iv) and (v) option (d) is also correct.
$$PEP = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 2 & 3 \cr 2 & 3 & 4 \cr 8 & {13} & {18} \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]$$
$$ = \left[ {\matrix{ 1 & 2 & 3 \cr 8 & {13} & {18} \cr 2 & 3 & 4 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 3 & 2 \cr 8 & {18} & {13} \cr 2 & 4 & 3 \cr } } \right] = F$$
and $${P^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$
Hence, option (a) is correct.
For option (b)
$$\left| {EQ + PF{Q^{ - 1}}} \right| = \left| {EQ} \right| + \left| {PF{Q^{ - 1}}} \right|$$ .... (i)
$$\because$$ | E | = 0 and | F | = 0 and | Q | $$\ne$$ 0
$$\therefore$$ $$\left| {EQ} \right| = \left| E \right|\left| Q \right| = 0$$
and $$\left| {PF{Q^{ - 1}}} \right| = {{\left| P \right|\left| F \right|} \over {\left| Q \right|}} = 0$$
Let $$R = EQ + PF{Q^{ - 1}}$$ .... (ii)
$$ \Rightarrow RQ = E{Q^2} + PF = E{Q^2} + {P^2}EP = E{Q^2} + EP$$ [$$\because$$ P2 = I]
$$ = E({Q^2} + P)$$
$$ \Rightarrow \left| {RQ} \right| = \left| {E({Q^2} + P)} \right|$$
$$ \Rightarrow \left| R \right|\left| Q \right| = \left| E \right|\left| {{Q^2} + P} \right| = 0$$ [$$\because$$ | E | = O]
$$ \Rightarrow \left| R \right| = 0$$ (as $$\left| Q \right| \ne 0$$) ..... (iii)
From Eqs. (ii) and (iii), we get Eq. (i) is true.
Hence, option (b) is correct.
For option (c)
$$\left| {{{(EF)}^3}} \right| > {\left| {EF} \right|^2}$$
i.e. 0 > 0 which is false.
For option (d)
$$\because$$ $${P^2} = I \Rightarrow {P^{ - 1}} = P$$
$$\therefore$$ $${P^{ - 1}}FP = PFP = PPEPP = E$$
So, $$E + {P^{ - 1}}FP = E + E = 2E$$
$$ \Rightarrow Tr(E + {P^{ - 1}}FP) = Tr(2E) = 2Tr(E)$$ ..... (iv)
and $${P^{ - 1}}EP + F$$
$$ \Rightarrow PEP + F = 2PEP$$ [$$\because$$ F = PEP]
$$\therefore$$ $$Tr(2PEP) = 2Tr(PEP) = 2Tr(F) = 2Tr(E)$$ ..... (v)
From Eqs. (iv) and (v) option (d) is also correct.
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