Question

Let M be a $3 \times 3$ matrix such that $M \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}$ and $M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$. If $M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}$, then $x+y+z$ equals :

• A. 4
• B. 5
• C. 7
• D. 11

Answer 1

The Correct Option is C

To solve this problem, we need to determine the matrix \(M\) and understand its effect on given vectors. The problem provides us with the transformations of standard basis vectors, which gives us the columns of the matrix \(M\).

1. Identify the Basis Vectors and Transformations:
   1. \(M \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\)
   2. \(M \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\)
   3. \(M \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\)
2. Construct Matrix \(M\):

Using the given transformations, we can determine each column of \(M\):

\( M = \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix} \)

The vectors provide us:

• First column: \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\)
• Second column: \(\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}\)
• Third column: \(\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}\)

1. Apply Matrix \(M\) to Solve for \(x, y, z\):

We are given \(M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 7 \\ 11 \end{pmatrix}\). Let's multiply \(M\) with \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\):

• \(1 \cdot x + 0 \cdot y - 1 \cdot z = 1\)
• \(1 \cdot x + 1 \cdot y + 1 \cdot z = 7\)
• \(1 \cdot x + 2 \cdot y + 1 \cdot z = 11\)
• \(x - z = 1\)
• \(x + y + z = 7\)
• \(x + 2y + z = 11\)

1. Solving the System of Equations:
   • From equation \(x - z = 1\), we get \(x = z + 1\).
   • Substitute \(x = z + 1\) in the second equation: \((z + 1) + y + z = 7 \rightarrow y + 2z = 6.\)
   • Substitute \(x = z + 1\) in the third equation: \((z + 1) + 2y + z = 11 \rightarrow 2y + 2z = 10 \rightarrow y + z = 5.\)
   • Subtract equation \(y + z = 5\) from \(y + 2z = 6\):
     \(z = 1\) and consequently, \(y = 4\).
   • Using \(x = z + 1\), therefore \(x = 2\).
2. Conclusion:

The values of \(x, y, z\) are 2, 4, 1.

Hence, \(x + y + z = 2 + 4 + 1 = 7\).