solve this in matrix gaussian reduction: x+y+z=24, 2y+2z-x=0,y+9=x

To solve the system of equations using **Gaussian elimination**, let's first represent the system as a matrix. The given system is:

1. \( x + y + z = 24 \)
2. \( 2y + 2z - x = 0 \)
3. \( y + 9 = x \)

### Step 1: Rewrite the system in standard form

We rewrite the equations to ensure all variables are on one side and constants on the other.

1. \( x + y + z = 24 \)  → \( x + y + z = 24 \)
2. \( -x + 2y + 2z = 0 \)  → \( -x + 2y + 2z = 0 \)
3. \( x - y = 9 \)  → \( x - y + 0z = 9 \)

Now, let's represent this system in **augmented matrix** form:

\[
\begin{pmatrix}
1 & 1 & 1 & | & 24 \\
-1 & 2 & 2 & | & 0 \\
1 & -1 & 0 & | & 9
\end{pmatrix}
\]

### Step 2: Apply Gaussian elimination

#### First, use row operations to get a leading 1 in the first row:
The matrix is already in a good form, so we will focus on eliminating the first element in row 2 and row 3.

- Add row 1 to row 2: \( R_2 = R_2 + R_1 \)
- Subtract row 1 from row 3: \( R_3 = R_3 - R_1 \)

This results in the following matrix:

\[
\begin{pmatrix}
1 & 1 & 1 & | & 24 \\
0 & 3 & 3 & | & 24 \\
0 & -2 & -1 & | & -15
\end{pmatrix}
\]

#### Next, eliminate the leading element in the third row:
- Multiply row 2 by \(\frac{1}{3}\) to make the leading coefficient in the second row a 1:
  \[
  R_2 = \frac{1}{3}R_2
  \]

This gives:

\[
\begin{pmatrix}
1 & 1 & 1 & | & 24 \\
0 & 1 & 1 & | & 8 \\
0 & -2 & -1 & | & -15
\end{pmatrix}
\]

- Now, add 2 times row 2 to row 3: \( R_3 = R_3 + 2R_2 \)

This results in the matrix:

\[
\begin{pmatrix}
1 & 1 & 1 & | & 24 \\
0 & 1 & 1 & | & 8 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]

### Step 3: Perform back-substitution

From the matrix, we can now read off the equations:

1. \( z = 1 \)
2. \( y + z = 8 \) → \( y + 1 = 8 \) → \( y = 7 \)
3. \( x + y + z = 24 \) → \( x + 7 + 1 = 24 \) → \( x = 16 \)

### Solution:

\[
x = 16, \quad y = 7, \quad z = 1
\]

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

Here are some related questions:
1. What is Gaussian elimination, and why is it useful?
2. How does back-substitution work in solving systems of equations?
3. Can Gaussian elimination be applied to non-square matrices?
4. What is the difference between row echelon form and reduced row echelon form?
5. How does Gaussian elimination relate to matrix inverses?

**Tip:** Always check your solution by substituting values back into the original system to verify correctness.

solve this in matrix gaussian reduction: x+y+z=24, 2y+2z-x=0, y+9=x

Learn how to solve the system of linear equations x + y + z = 24, 2y + 2z - x = 0, and y + 9 = x using Gaussian elimination. This step-by-step solution demonstrates key linear algebra concepts, including matrix row operations and back-substitution, suitable for students in Grades 10-12.

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