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Problem 357

# Reduce the matrix $$\begin{array}{rrrr} & 11 & 4 & 1 & 3 \\ & 10 & -1 & 3 & -1 \\ & A=\mid 3 & 1 & 0 & 2 \\ & 1 & -2 & 5 & 1 \end{array} \mid$$ to triangular form.

Expert verified
The matrix A in upper triangular form is: $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & 7 & -15 & 5 \\ 0 & 0 & -\frac{20}{7} & -\frac{71}{7}\\ 0 & 0 & \frac{16}{7} & \frac{36}{7} \end{pmatrix}$$
See the step by step solution

## Step 1: Swap row 1 and row 4

To make calculations easier, we will start by swapping row 1 with row 4. This is called a row interchange. $$R_1 \leftrightarrow R_4$$ $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 10 & -1 & 3 & -1 \\ 3 & 1 & 0 & 2 \\ 11 & 4 & 1 & 3 \end{pmatrix}$$ Step 2: Eliminate elements below pivot element in the first column

## Step 2: Eliminate elements below the pivot element in the first column

Perform necessary row operations to eliminate elements below the pivot element in the first column: $$R_2 - 10R_1 \rightarrow R_{2}$$ $$R_3 - 3R_1 \rightarrow R_{3}$$ $$R_4 - 11R_1 \rightarrow R_{4}$$ $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & 19 & -47 & -11 \\ 0 & 7 & -15 & 5 \\ 0 & 26 & -54 & 2 \end{pmatrix}$$ Step 3: Swap row 2 and row 3

## Step 3: Swap row 2 and row 3

To make calculations easier, we will now swap row 2 with row 3. This is called a row interchange. $$R_2 \leftrightarrow R_3$$ $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & 7 & -15 & 5 \\ 0 & 19 & -47 & -11 \\ 0 & 26 & -54 & 2 \end{pmatrix}$$ Step 4: Eliminate elements below pivot element in the second column

## Step 4: Eliminate elements below the pivot element in the second column

Perform necessary row operations to eliminate elements below the pivot element in the second column: $$R_3 - \frac{19}{7} R_2 \rightarrow R_{3}$$ $$R_4 - \frac{26}{7} R_2 \rightarrow R_{4}$$ $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & 7 & -15 & 5 \\ 0 & 0 & \frac{16}{7} & \frac{36}{7} \\ 0 & 0 & -\frac{20}{7} & -\frac{71}{7} \end{pmatrix}$$ Step 5: Swap row 3 and row 4

## Step 5: Swap row 3 and row 4

To make calculations easier, we will now swap row 3 with row 4. This is called a row interchange. $$R_3 \leftrightarrow R_4$$ $$A=\begin{pmatrix} 1 & -2 & 5 & 1 \\ 0 & 7 & -15 & 5 \\ 0 & 0 & -\frac{20}{7} & -\frac{71}{7}\\ 0 & 0 & \frac{16}{7} & \frac{36}{7} \end{pmatrix}$$ Now, the given matrix A is in the upper triangular form.

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