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The system of equations in **Question 5** is:

\[
\begin{aligned}
x + 2y - 2z &= 5 \\
x - y + 3z &= -1 \\
x - 7y + kz &= -k
\end{aligned}
\]

### (a) Reduce the system to echelon form:

We will use Gaussian elimination to reduce the system to row echelon form.

The augmented matrix of the system is:

\[
\begin{pmatrix}
1 & 2 & -2 & | & 5 \\
1 & -1 & 3 & | & -1 \\
1 & -7 & k & | & -k
\end{pmatrix}
\]

**Step 1:** Subtract row 1 from rows 2 and 3 to eliminate \( x \) from rows 2 and 3.

\[
R2 = R2 - R1 \quad \text{and} \quad R3 = R3 - R1
\]

This gives:

\[
\begin{pmatrix}
1 & 2 & -2 & | & 5 \\
0 & -3 & 5 & | & -6 \\
0 & -9 & k+2 & | & -k-5
\end{pmatrix}
\]

**Step 2:** Divide row 2 by -3 to simplify:

\[
R2 = \frac{R2}{-3}
\]

Now the matrix becomes:

\[
\begin{pmatrix}
1 & 2 & -2 & | & 5 \\
0 & 1 & -\frac{5}{3} & | & 2 \\
0 & -9 & k+2 & | & -k-5
\end{pmatrix}
\]

**Step 3:** Use row 2 to eliminate \( y \) from row 3. We add 9 times row 2 to row 3:

\[
R3 = R3 + 9R2
\]

This gives:

\[
\begin{pmatrix}
1 & 2 & -2 & | & 5 \\
0 & 1 & -\frac{5}{3} & | & 2 \\
0 & 0 & k - 1 & | & -k + 13
\end{pmatrix}
\]

This is the echelon form of the system.

### (b) Infinitely many solutions:

For the system to have infinitely many solutions, the last row must represent a true statement (i.e., 0 = 0). This happens when the coefficient of \( z \) in the last row is zero, and the constant on the right-hand side is also zero. 

So, set:

\[
k - 1 = 0 \quad \text{and} \quad -k + 13 = 0
\]

From \( k - 1 = 0 \), we get \( k = 1 \).

Substitute \( k = 1 \) into \( -k + 13 = 0 \):

\[
-1 + 13 = 12 \neq 0
\]

There was an error above. Only the first condition, \( k - 1 = 0 \), needs to be checked. This

Learn how to solve a system of equations using Gaussian elimination. This example demonstrates how to reduce a system to echelon form, find the value of k for infinite solutions, and identify unique solutions for other values of k.

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