Answered

Fill in the missing entries by performing the indicated row operations to obtain the row-reduced matrix.

$\begin{array}{c}

{\left[\begin{array}{rrr|r}

0 & 1 & 2 & -3 \\

1 & 2 & -1 & 8 \\

1 & -2 & 2 & -2

\end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2}\left[\begin{array}{lll|l}

& & & \\

1 & -2 & 2 & -2

\end{array}\right]} \\

\\

\xrightarrow{R_3-R_1}\left[\begin{array}{llr|r}

1 & 2 & -1 & 8 \\

0 & 1 & 2 & -3

\end{array}\right] \\

\begin{array}{l}

R_1+\left(\frac{1}{2}\right) R_3 \\

R_3+4 R_2

\end{array}\left[\begin{array}{lll|l}

0 & 1 & 2 & -3

\end{array}\right] \\

\xrightarrow{(1 / 11) R_3}\left[\begin{array}{lll|l}

1 & 0 & \frac{1}{2} & 3 \\

0 & 1 & 2 & -3

\end{array}\right] \\

\xrightarrow[R_2-2 R_3]{R_1-\left(\frac{1}{2}\right) R_3}\left[\begin{array}{lll|l}

1 & 0 & 0 & 4 \\

0 & 1 & 0 & 1 \\

0 & 0 & 1 & -2

\end{array}\right]

\end{array}$

Verified Answer :

  • The initial augmented matrix \(\begin{bmatrix} 0 & 1 & 2 & -3 \ 1 & 2 & -1 & 8 \ 1 & -2 & 2 & -2 \end{bmatrix}\) is transformed through a series of elementary row operations.
  • Swap the first and second rows, then eliminate the leading coefficients to simplify the matrix columns.
  • Normalize the third row and use it to eliminate non-zero entries in the third column of the first and second rows.
  • The final reduced row-echelon form is \(\begin{bmatrix} 1 & 0 & 0 & 4 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & -2 \end{bmatrix}\), giving the final solution \(\boxed{x=4, y=1, z=-2}\).

Explanation

  1. Analysis of the Augmented Matrix

The problem requires us to perform row operations to transform an augmented matrix into its reduced row-echelon form. The initial augmented matrix is given as:
\[\left[\begin{array}{rrr|r} 0 & 1 & 2 & -3 \ 1 & 2 & -1 & 8 \ 1 & -2 & 2 & -2 \end{array}\right]\]
We need to verify and complete the sequence of operations step by step to reach the final reduced row-echelon form. This process involves elementary row operations such as swapping rows, adding/subtracting multiples of one row from another, and scaling a row by a constant.

  1. First Row Operations

The first step is to swap the first and second rows (\(R_1 \leftrightarrow R_2\)) to obtain a leading coefficient of 1 in the first column of the first row:
\[\left[\begin{array}{rrr|r} 0 & 1 & 2 & -3 \ 1 & 2 & -1 & 8 \ 1 & -2 & 2 & -2 \end{array}\right] \xrightarrow{R_1 \leftrightarrow R_2} \left[\begin{array}{rrr|r} 1 & 2 & -1 & 8 \ 0 & 1 & 2 & -3 \ 1 & -2 & 2 & -2 \end{array}\right]\]
Now we want to eliminate the leading coefficient in the first column of the third row. We perform \(R_3 \leftarrow R_3 - R_1\):

  • Element (3,1): \(1 - 1 = 0\)
  • Element (3,2): \(-2 - 2 = -4\)
  • Element (3,3): \(2 - (-1) = 3\)
  • Element (3,4): \(-2 - 8 = -10\)

Resulting in:
\[\left[\begin{array}{rrr|r} 1 & 2 & -1 & 8 \ 0 & 1 & 2 & -3 \ 0 & -4 & 3 & -10 \end{array}\right]\]

  1. Second and Third Row Operations

Next, we eliminate the entries in the second column above and below the pivot in the second row (the pivot is the 1 in the second row, second column).
We perform \(R_1 \leftarrow R_1 - 2R_2\) and \(R_3 \leftarrow R_3 + 4R_2\):

  • For \(R_1\): \(1 - 2(0) = 1\), \(2 - 2(1) = 0\), \(-1 - 2(2) = -5\), \(8 - 2(-3) = 14\).
  • For \(R_3\): \(0 + 4(0) = 0\), \(-4 + 4(1) = 0\), \(3 + 4(2) = 11\), \(-10 + 4(-3) = -22\).

The matrix becomes:
\[\left[\begin{array}{rrr|r} 1 & 0 & -5 & 14 \ 0 & 1 & 2 & -3 \ 0 & 0 & 11 & -22 \end{array}\right]\]
Then, we normalize the third row by performing \(R_3 \leftarrow \frac{1}{11}R_3\):
\[\left[\begin{array}{rrr|r} 1 & 0 & -5 & 14 \ 0 & 1 & 2 & -3 \ 0 & 0 & 1 & -2 \end{array}\right]\]

  1. Final Row Operations and Conclusion

Finally, we eliminate the entries in the third column above the pivot in the third row (the 1 in the third row, third column).
We perform \(R_1 \leftarrow R_1 + 5R_3\) and \(R_2 \leftarrow R_2 - 2R_3\):

  • For \(R_1\): \(1 + 5(0) = 1\), \(0 + 5(0) = 0\), \(-5 + 5(1) = 0\), \(14 + 5(-2) = 4\).
  • For \(R_2\): \(0 - 2(0) = 0\), \(1 - 2(0) = 1\), \(2 - 2(1) = 0\), \(-3 - 2(-2) = 1\).

This gives the final reduced row-echelon form:
\[\left[\begin{array}{rrr|r} 1 & 0 & 0 & 4 \ 0 & 1 & 0 & 1 \ 0 & 0 & 1 & -2 \end{array}\right]\]
This implies the solution to the system is \(x=4, y=1, z=-2\).

Examples

Solving systems of linear equations is a fundamental skill used in various fields, such as engineering and physics, to determine unknown values in interconnected systems. For example, if you are designing a electrical circuit, you might use Kirchhoff's laws to create a system of linear equations based on voltage and current, and by using row-reduction, you can find the current through each resistor, ensuring your circuit design is safe and functional.

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