$$\because$$ I $$-$$ EF = G^$$-$$1 $$\Rightarrow$$ G $$-$$ GEF = I ..... (i)

and G $$-$$ EFG = I ..... (ii)

Clearly, GEF = EFG $$\to$$ option (c) is correct.

Also, (I $$-$$ FE) (I + FGE)

= I $$-$$ FE + FGE $$-$$ FEFGE

= I $$-$$ FE + FGE $$-$$ F(G $$-$$ I) E

= I $$-$$ FE + FGE $$-$$ FGE + FE

= I $$\to$$ option (b) is correct but option (d) is incorrect.

$$\because$$ (I $$-$$ FE) (I $$-$$ FGE) = I $$-$$ FE $$-$$ FGE + F(G $$-$$ I) E

= I $$-$$ 2FE

Now, (I $$-$$ FE) ($$-$$ FGE) = $$-$$ FE

$$\Rightarrow$$ | I $$-$$ FE | | FGE | = | FE |

$$\to$$ option (a) is correct.

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.

It is given that matrix M be a 3 $$ \times $$ 3 invertible matrix, such that

M^$$-$$1 = adj(adj M) $$ \Rightarrow $$ M^$$-$$1 = |M| M

($$ \because $$ for a matrix A of order 'n' adj(adjA) = |A|^n$$-$$2 A}

$$ \Rightarrow $$ M^$$-$$1 M = |M|M² $$ \Rightarrow $$ M²|M| = I .....(i)

$$ \because $$ det(M² |M|) = det(I) = 1

$$ \Rightarrow $$ |M|³|M|² = 1 $$ \Rightarrow $$ |M| = 1 .....(ii)

from Eqs. (i) and (ii), we get

M² = I

As, adj M = |M|M^$$-$$1 = M

$$ \Rightarrow $$ (adj M)² = M² (adj M)² = I

$${P_1} = I = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_2} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right],\,{P_3} = \left[ {\matrix{ 0 & 1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 \cr } } \right],\,{P_4} = \left[ {\matrix{ 0 & 1 & 0 \cr 0 & 0 & 1 \cr 1 & 0 & 0 \cr } } \right],\,{P_5} = \left[ {\matrix{ 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \cr } } \right],\,{P_6} = \left[ {\matrix{ 0 & 0 & 1 \cr 0 & 1 & 0 \cr 1 & 0 & 0 \cr } } \right]$$ and $$X = \sum\limits_{k = 1}^6 {{P_k}} \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]P_k^T$$

$$ \because $$ $$P_1^T = {P_1},\,P_2^T = {P_2},\,P_3^T = {P_3},\,P_4^T = {P_5},\,P_5^T = {P_4}$$ and $$P_6^T = {P_6}$$

and Let $$Q = \left[ {\matrix{ 2 & 1 & 3 \cr 1 & 0 & 2 \cr 3 & 2 & 1 \cr } } \right]$$

and $$ \because $$ Q^T = Q

Now,

$$X = ({P_1}QP_1^T) + ({P_2}QP_2^T) + ({P_3}QP_3^T) + ({P_4}QP_4^T) + ({P_5}QP_5^T) + ({P_6}QP_6^T)$$

So, $${X^T} = {({P_1}QP_1^T)^T} + {({P_2}QP_2^T)^T} + {({P_3}QP_3^T)^T} + {({P_4}QP_4^T)^T} + {({P_5}QP_5^T)^T} + {({P_6}QP_6^T)^T}$$

= $${P_1}QP_1^T$$ + $${P_2}QP_2^T$$ + $${P_3}QP_3^T$$ + $${P_4}QP_4^T$$ + $${P_5}QP_5^T$$ + $${P_6}QP_6^T$$

[$$ \because $$ (ABC)^T = C^TB^TA^T and (A^T)^T = A and Q^T = Q]

$$ \Rightarrow $$ X^T = X

$$ \Rightarrow $$ X is a symmetric matrix.

The sum of diagonal entries of X = Tr(X)

$$ = \sum\limits_{i = 1}^6 {{T_r}({P_i}QP_i^T) = \sum\limits_{i = 1}^6 {{T_r}(QP_i^T{P_i})} } $$ [$$ \because $$ T_r(ABC) = T_r(BCA)]

$$ = \sum\limits_{i = 1}^6 {{T_r}(QI)} $$

[$$ \because $$ P_i's are orthogonal matrices]

$$ = \sum\limits_{i = 1}^6 {{T_r}(Q) = 6{T_r}(Q) = 6 \times 3 = 18} $$

Now, Let $$R = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, then

$$XR = \sum\limits_{K = 1}^6 {({P_K}} QP_K^T)R = \sum\limits_{K = 1}^6 {({P_K}QP_K^TR)} $$

$$ = \sum\limits_{K = 1}^6 {({P_K}QR)} $$ [$$ \because $$ $$P_K^T$$R = R]

$$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} $$

$$ = \sum\limits_{K = 1}^6 {{P_K}\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]} = \left[ {\matrix{ 2 & 2 & 2 \cr 2 & 2 & 2 \cr 2 & 2 & 2 \cr } } \right]\left[ {\matrix{ 6 \cr 3 \cr 6 \cr } } \right]$$

$$ \Rightarrow XR = \left[ {\matrix{ {30} \cr {30} \cr {30} \cr } } \right] \Rightarrow XR = 30R$$

$$ \Rightarrow X\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = 30\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$

$$ \Rightarrow $$ (X $$-$$ 30I) R = 0 $$ \Rightarrow $$ |X $$-$$ 30I| = 0

So, (X $$-$$ 30I) is not invertible and value of $$\alpha $$ = 30.

Hence, options (a), (b) and (d) are correct.