The question asks for the sum of the diagonal entries of a 3 × 3 matrix M, given three conditions relating the matrix M to specific column vectors.

Understanding Matrix Operations and Trace

The sum of the diagonal entries of a square matrix is called the trace of the matrix. For a 3 × 3 matrix M,

\( M = \begin{bmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{bmatrix} \)

The trace is given by \( \text{Trace}(M) = m_{11} + m_{22} + m_{33} \).

The given conditions are of the form \( M \mathbf{v} = \mathbf{w} \), where \( \mathbf{v} \) are column vectors and \( \mathbf{w} \) are the resulting vectors. These conditions essentially define how the matrix M acts on certain vectors.

Finding Matrix Columns Using Vector Equations

Let the standard basis vectors in 3D space be \( \mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), and \( \mathbf{e}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \). Multiplying a matrix M by \( \mathbf{e}_i \) yields the \( i \)-th column of M.

We are given:

• \( M\mathbf{v}_1 = M\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} = \mathbf{w}_1 \)
• \( M\mathbf{v}_2 = M\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} = \mathbf{w}_2 \)
• \( M\mathbf{v}_3 = M\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} = \mathbf{w}_3 \)

From the first equation, \( M\mathbf{e}_2 = \mathbf{w}_1 = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \). This directly tells us the second column of M:

\( \begin{bmatrix} m_{12} \\ m_{22} \\ m_{32} \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \)

So, we know \( m_{22} = 2 \).

Expressing Basis Vectors as Linear Combinations

To find \( m_{11} \) and \( m_{33} \), we need to express \( \mathbf{e}_1 \) and \( \mathbf{e}_3 \) as linear combinations of the given vectors \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \).

Finding \( \mathbf{e}_1 \):

Let's try combining \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \):

\( \mathbf{v}_1 + \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \mathbf{e}_1 \)

Using the linearity property of matrix multiplication, \( M\mathbf{e}_1 = M(\mathbf{v}_1 + \mathbf{v}_2) = M\mathbf{v}_1 + M\mathbf{v}_2 \):

\( M\mathbf{e}_1 = \mathbf{w}_1 + \mathbf{w}_2 = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} \)

Since \( M\mathbf{e}_1 \) gives the first column of M:

\( \begin{bmatrix} m_{11} \\ m_{21} \\ m_{31} \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} \)

Thus, \( m_{11} = 0 \).

Finding \( \mathbf{e}_3 \):

Let's try to express \( \mathbf{e}_3 \) using \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \). We need coefficients \( c_1, c_2, c_3 \) such that \( c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3 = \mathbf{e}_3 \).

\( c_1 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \)

This leads to the system of equations:

• $c_2 + c_3 = 0$
• $c_1 - c_2 + c_3 = 0$
• $c_3 = 1$

Solving this system:

• From $c_3 = 1$, we get $c_2 + 1 = 0 \implies c_2 = -1$.
• Substituting $c_1, c_2, c_3$ into the second equation: $c_1 - (-1) + 1 = 0 \implies c_1 + 2 = 0 \implies c_1 = -2$.

So, \( \mathbf{e}_3 = -2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 \).

Now, use linearity: \( M\mathbf{e}_3 = M(-2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3) = -2M\mathbf{v}_1 - M\mathbf{v}_2 + M\mathbf{v}_3 \)

\( M\mathbf{e}_3 = -2\mathbf{w}_1 - \mathbf{w}_2 + \mathbf{w}_3 \)

\( M\mathbf{e}_3 = -2 \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} \)

\( M\mathbf{e}_3 = \begin{bmatrix} 2 \\ -4 \\ -6 \end{bmatrix} - \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 - 1 + 0 \\ -4 - 1 + 0 \\ -6 + 1 + 12 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 7 \end{bmatrix} \)

Since \( M\mathbf{e}_3 \) gives the third column of M:

\( \begin{bmatrix} m_{13} \\ m_{23} \\ m_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 7 \end{bmatrix} \)

Thus, \( m_{33} = 7 \).

Calculating the Sum of Diagonal Entries

We have found the diagonal entries:

• \( m_{11} = 0 \)
• \( m_{22} = 2 \)
• \( m_{33} = 7 \)

The sum of the diagonal entries (the trace) is:

\( \text{Trace}(M) = m_{11} + m_{22} + m_{33} = 0 + 2 + 7 = 9 \)

Final Answer Check

The calculated sum of the diagonal entries is 9, which matches one of the given options.