given
3 by 3 matrix X [10/3 -5/3 0][-5/3 11/3 -2][0 -2 4],
3 by 1 matrix Y [Q2 Q3 Q4],
3 by 1 matrix Z [300000 0 600000],

If 2*1e5[matrix X][matrix Y]=[matrix Z]
find the values of Q2, Q3 and Q4

Given the matrices:

\[
\mathbf{X} = \begin{bmatrix} \frac{10}{3} & -\frac{5}{3} & 0 \\ -\frac{5}{3} & \frac{11}{3} & -2 \\ 0 & -2 & 4 \end{bmatrix},
\quad
\mathbf{Y} = \begin{bmatrix} Q_2 \\ Q_3 \\ Q_4 \end{bmatrix},
\quad
\mathbf{Z} = \begin{bmatrix} 300000 \\ 0 \\ 600000 \end{bmatrix}
\]

The equation given is:

\[
2 \times 10^5 \times \mathbf{X} \times \mathbf{Y} = \mathbf{Z}
\]

First, simplify the equation by dividing both sides by \(2 \times 10^5\):

\[
\mathbf{X} \times \mathbf{Y} = \frac{1}{2 \times 10^5} \times \mathbf{Z} = \begin{bmatrix} \frac{300000}{2 \times 10^5} \\ \frac{0}{2 \times 10^5} \\ \frac{600000}{2 \times 10^5} \end{bmatrix} = \begin{bmatrix} \frac{3}{2} \\ 0 \\ 3 \end{bmatrix}
\]

Now, we have the equation:

\[
\mathbf{X} \times \mathbf{Y} = \begin{bmatrix} \frac{3}{2} \\ 0 \\ 3 \end{bmatrix}
\]

This leads to the following system of equations:

\[
\frac{10}{3} Q_2 - \frac{5}{3} Q_3 = \frac{3}{2} \quad \text{(first row)}
\]
\[
-\frac{5}{3} Q_2 + \frac{11}{3} Q_3 - 2 Q_4 = 0 \quad \text{(second row)}
\]
\[
-2 Q_3 + 4 Q_4 = 3 \quad \text{(third row)}
\]

### Solving for \(Q_2\), \(Q_3\), and \(Q_4\):

**Step 1: Solve the third equation for \(Q_4\):**

\[
4Q_4 = 2Q_3 + 3 \implies Q_4 = \frac{2Q_3 + 3}{4}
\]

**Step 2: Substitute \(Q_4\) into the second equation:**

\[
-\frac{5}{3}Q_2 + \frac{11}{3}Q_3 - 2\left(\frac{2Q_3 + 3}{4}\right) = 0
\]

Simplify the second equation:

\[
-\frac{5}{3}Q_2 + \frac{11}{3}Q_3 - \frac{Q_3 + 3}{2} = 0
\]

Multiply the equation by 6 to eliminate fractions:

\[
-10Q_2 + 22Q_3 - 3(Q_3 + 3) = 0
\]

\[
-10Q_2 + 19Q_3 - 9 = 0 \implies 10Q_2 = 19Q_3 - 9
\]

\[
Q_2 = \frac{19Q_3 - 9}{10}
\]

**Step 3: Substitute \(Q_2\) into the first equation:**

\[
\frac{10}{3}\left(\frac{19Q_3 - 9}{10}\right) - \frac{5}{3}Q_3 = \frac{3}{2}
\]

Simplify:

\[
\frac{19Q_3 - 9}{3} - \frac{5}{3}Q_3 = \frac{3}{2}
\]

Multiply the equation by 6 to eliminate fractions:

\[
2(19Q_3 - 9) - 10Q_3 = 9
\]

Simplify:

\[
38Q_3 - 18 - 10Q_3 = 9 \implies 28Q_3 = 27 \implies Q_3 = \frac{27}{28}
\]

**Step 4: Substitute \(Q_3\) back into the equations for \(Q_2\) and \(Q_4\):**

\[
Q_2 = \frac{19 \times \frac{27}{28} - 9}{10} = \frac{\frac{513}{28} - 9}{10} = \frac{\frac{513}{28} - \frac{252}{28}}{10} = \frac{261}{280} = \frac{261}{280} = \frac{261}{280}
\]

Now, substitute \(Q_3 = \frac{27}{28}\) into \(Q_4\):

\[
Q_4 = \frac{2 \times \frac{27}{28} + 3}{4} = \frac{\frac{54}{28} + \frac{84}{28}}{4} = \frac{\frac{138}{28}}{4} = \frac{69}{56}
\]

**Final answers**:
\[
Q_2 = \frac{261}{280}, \quad Q_3 = \frac{27}{28}, \quad Q_4 = \frac{69}{56}
\]

Would you like more details on any steps, or have any questions?

---

1. What if \(Q_3\) was an integer value?
2. How do the values of \(Q_2\), \(Q_3\), and \(Q_4\) change if \(\mathbf{Z}\) changes?
3. Can you explain how to solve a similar problem using matrix inversion?
4. What is the determinant of matrix \(\mathbf{X}\)?
5. How would a change in the scalar coefficient (e.g., \(2 \times 10^5\)) affect the solution?
6. What are the properties of the matrix \(\mathbf{X}\)?
7. How could you confirm the solution using a different method?
8. What would happen if matrix \(\mathbf{X}\) were singular?

**Tip:** Understanding matrix multiplication and systems of linear equations is crucial when dealing with problems involving matrices.

Learn how to solve a system of linear equations using matrix algebra. This problem involves matrices X, Y, and Z, where matrix multiplication and Gaussian elimination methods are applied to find values Q2, Q3, and Q4. Suitable for advanced mathematics students.

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