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Let's break down the given problem:

You are provided with the following matrix equation:

\[
\begin{bmatrix}
2 & 1 \\
-3 & 4
\end{bmatrix}
\mathbf{X} = 
\begin{bmatrix}
7 \\
6
\end{bmatrix}
\]

### (i) The order of the matrix \( \mathbf{X} \):

- The matrix on the left-hand side is a 2x2 matrix (2 rows and 2 columns).
- The matrix on the right-hand side is a 2x1 matrix (2 rows and 1 column).
  
Since the left-hand side is a multiplication of a 2x2 matrix and \( \mathbf{X} \), for the multiplication to be valid, \( \mathbf{X} \) must be a 2x1 matrix. Therefore, the order of the matrix \( \mathbf{X} \) is **2x1**.

### (ii) The matrix \( \mathbf{X} \):

To find \( \mathbf{X} \), we need to solve the system of equations that arises from the matrix multiplication:

\[
\begin{bmatrix}
2 & 1 \\
-3 & 4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
=
\begin{bmatrix}
7 \\
6
\end{bmatrix}
\]

This gives us two equations:
1. \( 2x_1 + x_2 = 7 \)
2. \( -3x_1 + 4x_2 = 6 \)

Let's solve this system.

From the first equation, solve for \( x_2 \):

\[
x_2 = 7 - 2x_1
\]

Substitute this into the second equation:

\[
-3x_1 + 4(7 - 2x_1) = 6
\]

Simplifying:

\[
-3x_1 + 28 - 8x_1 = 6
\]

\[
-11x_1 = -22
\]

\[
x_1 = 2
\]

Now, substitute \( x_1 = 2 \) back into the equation \( x_2 = 7 - 2x_1 \):

\[
x_2 = 7 - 2(2) = 3
\]

Thus, the matrix \( \mathbf{X} \) is:

\[
\mathbf{X} =
\begin{bmatrix}
2 \\
3
\end{bmatrix}
\]

---

Would you like further details or have any questions? Here are some additional questions to expand your understanding:

1. How do you check the validity of matrix multiplication?
2. Can the order of a matrix affect its ability to be multiplied by another matrix?
3. How would you solve a system of equations using matrix inverses?
4. What would change if the matrix on the right-hand side was different?
5. How does the determinant of the left-hand matrix influence the solvability of the system?

**Tip:** When solving systems of equations using matrices, always check the determinant of the coefficient matrix. If it’s zero, the system may have no unique solution.

Given a 2x2 matrix multiplied by X to give a 2x1 matrix, find the order of matrix X and solve for X.

Learn how to solve a system of equations using matrix multiplication. Find the order of matrix X and solve the system for matrix X in this detailed solution involving 2x2 and 2x1 matrices. Suitable for Grades 9-12.

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