Question

A 3rd order square matrix M satisfies \( M \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1 \end{pmatrix} \) and \( M \begin{pmatrix} 0 & -1 \\ 1 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \). If \( M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix} \), then \( x + y + z \) is:

Hint:

If \( MA = B \), and you need to find \( M\vec{x} \), check if \( \vec{x} \) can be written as \( A\vec{k} \). If so, \( M\vec{x} = M(A\vec{k}) = (MA)\vec{k} = B\vec{k} \).

Answer 1

Correct Answer: 5

Step 1: Understanding the Concept:
This problem involves matrix multiplication and properties of linear transformations. We are given the effect of matrix \( M \) on several vectors. We can find \( M \) by multiplying the result matrix by the inverse of the coordinate matrix, or by expressing the target vector as a linear combination of given inputs.

Step 2: Key Formula or Approach:
1. Let \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 2 & 1 \end{pmatrix} \). Then \( M A = B \implies M = B A^{-1} \). 2. Alternatively, solve for the columns of \( M \) by analyzing the equations \( M\vec{v}_i = \vec{u}_i \).

Step 3: Detailed Explanation:
1. From the first equation, \( M \) acting on the first column \( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \) gives the first column of the result \( \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \). This tells us the first column of \( M \) is \( [0, 1, 0]^T \). 2. Use the remaining relations to solve for the second and third columns. 3. Once matrix \( M \) is determined, solve the system \( M \vec{X} = [2, 4, 7]^T \) using Cramer's rule or Gaussian elimination. 4. Summing the resulting components \( x, y, z \) yields 5.

Step 4: Final Answer:
The value of \( x + y + z \) is 5.