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Matrices and Determinants question

2019 · Shift 2 · Q25

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JEE Advanced
Mathematics
Matrices and Determinants
2019 · Shift 2 · Q25

Matrices and Determinants question

2019 · Shift 2 · Q25

JEE AdvancedMathematicsMatrices and DeterminantsMultiple correct+4 / −1

Let x

\in

R and let

P = \left[ {\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \\ \end{matrix} } \right]

,

Q = \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \\ \end{matrix} } \right]

and R = PQP

-

1, which of the following options is/are correct?

A
There exists a real, number x such that PQ = QP
B
For $x = 0$ , if $R \left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right] = 6\left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right]$ , then a + b =5
C
For x = 1, there exists a unit vector $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$ for which $R\left[ {\begin{matrix} \alpha \\ \beta \\ \gamma \\ \end{matrix} } \right] = \left[ {\begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} } \right]$
D
$\det R = \det \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \\ \end{matrix} } \right] + 8$ , for all x $\in$ R

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