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Area of Triangle with Coordinates of Vertices Given

Let P 1=|[ 1 ...

Question

Let
$P_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣,$ $P_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 0 & 0 & 1 0 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣,$ $P_{3} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣,$ $P_{4} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣,$ $P_{5} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 1 & 0 & 0 0 & 1 & 0 \end{matrix} ∣ ∣ ∣ ∣,$ $P_{6} = ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣,$ and $X = 6 \sum k = 1 P_{k} ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 3 1 & 0 & 2 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ P_{k}^{T},$ where $P_{k}^{T}$ denotes the transpose of the matrix $P_{k} .$ Then which of the following options is/are correct?

If X ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = α ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦, then α = 30

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X

is a symmetric matrix

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The sum of diagonal entries of

X

18

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X - 30 I

is an invertible matrix

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Solution

The correct options are
A $If X ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = α ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦, then α = 30$
B $X$ is a symmetric matrix
C The sum of diagonal entries of $X$ is $18$
$Let A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 3 1 & 0 & 2 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦$
$X = 6 \sum k = 1 (P_{k} A P_{k}^{T}) = (P_{1} A P_{1}^{T} + P_{2} A P_{2}^{T} + \dots + P_{6} A P_{6}^{T})$
$X^{T} = 6 \sum k = 1 (P_{k} A P_{k}^{T})^{T} = X [∵ A = A^{T}]$
$\Rightarrow X$ is a symmetric matrix.

$Let Q = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦$

$P_{1}^{T} Q = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = Q,$

Similarly, $P_{k}^{T} Q = Q$
$∴ X Q = 6 \sum k = 1 P_{k} A P_{k}^{T} Q$
$= 6 \sum k = 1 P_{k} A Q$
$= (6 \sum k = 1 P_{k}) A Q$
$6 \sum k = 1 P_{k} = ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦ and A Q = ⎡ ⎢ ⎣ \begin{matrix} 636 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow X Q = ⎡ ⎢ ⎣ \begin{matrix} 2 & 2 & 2 2 & 2 & 2 2 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 636 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 303030 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow X Q = 30 ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = 30 Q = α Q$
$∴ α = 30$
$Also, X ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = 30 ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow (X - 30 I) ⎡ ⎢ ⎣ \begin{matrix} 111 \end{matrix} ⎤ ⎥ ⎦ = 0$
$∴ (X - 30 I) Q = 0$
$∴$ The system has a non-trivial solution.
So, $| X - 30 I | = 0 \Rightarrow X - 30 I$ is a non-invertible matrix.

$Trace (X) = Trace (6 \sum k = 1 P_{k} A P_{k}^{T}) = 6 Trace (A) = 6 \times 3 = 18 [∵ Trace (A) = 2 + 0 + 1 = 3]$

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X = 6 \sum k = 1 P_{k} ⎡ ⎢ ⎣ \begin{matrix} 2 & 1 & 3 1 & 0 & 2 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦ P_{k}^{T},

P_{k}^{T}

P_{k} .

P

2 \times 2

[\begin{matrix} 1 & 0 \end{matrix}] P = - \frac{1}{\sqrt{2}} [\begin{matrix} 1 & 1 \end{matrix}]

[\begin{matrix} 0 & 1 \end{matrix}] P = \frac{1}{\sqrt{2}} [\begin{matrix} - 1 & 1 \end{matrix}]

0

I

2

A = a_{i j}

3,

a_{i j} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x; & if i = j, x \in R 1; & if | i - j | = 1 0; & otherwise \end{matrix},

2 \times 2

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