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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
MCQ
For any $3 \times 3$ matrix $M$, let $| M |$ denote the determinant of $M$. Let

$E=\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18\end{array}\right], P=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right]$ and $F=\left[\begin{array}{ccc}1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3\end{array}\right]$

If $Q$ is a nonsingular matrix of order $3 \times 3$, then which of the following statements is (are) $TRUE$?

$(A)$F $=P E P$ and $P^2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$(B)$ $\left| EQ + PFQ ^{-1}\right|=| EQ |+\left| PFQ ^{-1}\right|$

$(C)$ $\left|( EF )^3\right|>| EF |^2$

$(D)$ Sum of the diagonal entries of $P ^{-1} EP + F$ is equal to the sum of diagonal entries of $E + P ^{-1} FP$

  • A
    $A,B,C$
  • B
    $A,B$
  • ✓
    $A,B,D$
  • D
    $A,C$
✓

Answer

Correct option: C.
$A,B,D$
c
$\begin{aligned} PEP & =\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 3 & 4 \\ 8 & 13 & 18\end{array}\right)\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right) \\ & \left(\begin{array}{ccc}1 & 2 & 3 \\ 8 & 13 & 18 \\ 2 & 3 & 4\end{array}\right)\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)=\left(\begin{array}{ccc}1 & 3 & 2 \\ 8 & 18 & 13 \\ 2 & 4 & 3\end{array}\right) \\ P ^2= & \left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right)=\left(\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\end{aligned}$

$(B)\left| EQ + PFQ ^{-1}\right|=| EQ |+\left| PFQ ^{-1}\right|$

$| E |=0 \text { and }| F |=0 \text { and }| Q | \neq 0$

$| EQ |=| E || Q |=0,\left| PFQ ^{-1}\right|=\frac{| P || F |}{| Q |}=0$

$T = EQ + PFQ ^{-1}$

$TQ = EQ ^2+ PF = EQ ^2+ P ^2 EP = EQ ^2+ EP = E \left( Q ^2+ P \right)$

$| TQ |=\left| E \left( Q ^2+ P \right)\right| \Rightarrow| T || Q |=| E |\left| Q ^2+ P \right|=0 \Rightarrow| T |=0 \text { (as }| Q | \neq 0 \text { ) }$

$(C)$ $\left|( EF )^3\right|>| EF |^2$

Here $0>0$ (false)

$\text { (D) as } P ^2= I \Rightarrow P ^{-1}= P \text { so } P ^{-1} FP = PFP = PPEPP = E$

$\text { so } E + P ^{-1} FP = E + E =2 E$

$P ^{-1} EP + F \Rightarrow PEP + F =2 PEP$

$\operatorname{Tr}(2 PEP )=2 \operatorname{Tr}( PEP )=2 \operatorname{Tr}( EPP )=2 \operatorname{Tr}( E )$

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