In matrix equation [A] {X} = {R}

One of the eigen values of Matrix [A] is

The correct answer is 16.

To find one of the eigenvalues of the matrix [A], we can analyze the given matrix equation:

$\left[A\right]\left\{X\right\}=\left\{R\right\}$

From the provided image, the matrix $\left[A\right]$, the column vector $\left\{X\right\}$, and the resulting vector $\left\{R\right\}$ are defined as:

$\left[A\right]=\left[\begin{array}{ccc}4& 8& 4\\ 8& 16& -4\\ 4& -4& 15\end{array}\right]$

$\left\{X\right\}=\left[\begin{array}{c}2\\ 1\\ 4\end{array}\right]$

$\left\{R\right\}=\left[\begin{array}{c}32\\ 16\\ 64\end{array}\right]$

Let us observe the relationship between the result vector $\left\{R\right\}$ and the input vector $\left\{X\right\}$. We can factor out a common scalar factor of 16 from the vector $\left\{R\right\}$:

$\left\{R\right\}=\left[\begin{array}{c}32\\ 16\\ 64\end{array}\right]=16\left[\begin{array}{c}2\\ 1\\ 4\end{array}\right]=16\left\{X\right\}$

Substituting this relationship into the original matrix equation gives:

$\left[A\right]\left\{X\right\}=16\left\{X\right\}$

By definition, a non-zero vector $v$ is an eigenvector of a matrix $\left[A\right]$ corresponding to an eigenvalue $\lambda$ if:

$\left[A\right]v=\lambda v$

Comparing our equation $\left[A\right]\left\{X\right\}=16\left\{X\right\}$ to the eigenvalue definition, we see that $\left\{X\right\}$ acts as the eigenvector and the scalar value 16 is the corresponding eigenvalue:

$\lambda =16$

Therefore, one of the eigenvalues of Matrix [A] is 16.