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In matrix equation $[A] \{X\}=\{R\}$,

$[A] = \begin{bmatrix} 4 & 8 & 4 \\ 8 & 16 & -4 \\ 4 & -4 & 15 \end{bmatrix} \{X\} = \begin{Bmatrix} 2 \\ 1 \\ 4 \end{Bmatrix} \text{ and} \{ R \}  = \begin{Bmatrix}  32 \\ 16 \\ 64 \end{Bmatrix}$

One of the eigen values of matrix $[A]$ is

  1. $4$
  2. $8$
  3. $15$
  4. $16$
in Linear Algebra 24.6k points
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Let $A$ be a square matrix of order $n$ and $ \lambda$ is one of its eigenvalues. Let $X$ be an eigenvector associated to eigenvalue $ \lambda $ then we must have equation
$AX = \lambda X$ --------- Equation $(1)$

In Question, It is given that   $AX=R$     ----------Equation $(2)$
Now, From Equation $(1)$ and $(2),$ $\lambda X = R$

So, $\lambda \begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} = \begin{Bmatrix} 32\\ 16\\64 \end{Bmatrix}$

$\Rightarrow$ $\lambda \begin{Bmatrix} 2\\ 1\\ 4 \end{Bmatrix} =16 \begin{Bmatrix} 2\\ 1\\4 \end{Bmatrix}$

$\Rightarrow$ $\lambda = 16$

So, Answer should be $(D)$
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