Question - Determining the Scalar Multiple of a Matrix

To solve for \( \alpha \), we need to set up an equation based on the given matrix equality:

\( \alpha^2 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 1 & 2 \\ 0 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} + 3X \)

We can see that the matrix \( X \) is also the matrix we are multiplying by \( \alpha^2 \). So, let's denote it as \( X \) and find \( 3X \):

\( 3X = 3 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 6 & 9 \\ 0 & 3 & 6 \\ 0 & 0 & 3 \end{bmatrix} \)

Now, let's put \( 3X \) back into the matrix equality and compare the corresponding entries:

\( \alpha^2 \times \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 1 & 2 \\ 0 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} = \begin{bmatrix} 5 & -10 & 2 \\ 0 & -6 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 6 & 9 \\ 0 & 3 & 6 \\ 0 & 0 & 3 \end{bmatrix} \)

Thus, we obtain the following matrix equation:

\( \begin{bmatrix} \alpha^2 + 3 & 2\alpha^2 + 1 & 3\alpha^2 + 2 \\ 0 & \alpha^2 - 10 & 2\alpha^2 + 2 \\ 0 & 0 & \alpha^2 + 2 \end{bmatrix} = \begin{bmatrix} 5 + 3 & -10 + 6 & 2 + 9 \\ 0 & -6 + 3 & 2 + 6 \\ 0 & 0 & 2 + 3 \end{bmatrix} \)

\( \begin{bmatrix} \alpha^2 + 3 & 2\alpha^2 + 1 & 3\alpha^2 + 2 \\ 0 & \alpha^2 - 10 & 2\alpha^2 + 2 \\ 0 & 0 & \alpha^2 + 2 \end{bmatrix} = \begin{bmatrix} 8 & -4 & 11 \\ 0 & -3 & 8 \\ 0 & 0 & 5 \end{bmatrix} \)

By comparing the corresponding entries, we can now write a system of equations. For instance, by comparing the (1,1) entries, we have:

\( \alpha^2 + 3 = 8 \)

\( \alpha^2 = 5 \)

\( \alpha = \pm \sqrt{5} \)

Since \( \alpha \) has been determined from the first entry, we do not need to proceed with the rest, as it will be the same for all corresponding entries.

The principal square root is often taken as the solution for \( \alpha \), thus:

\( \alpha = \sqrt{5} \)