Given The Matrix:${ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \ 3 & 4 & 5 & 5 \ -1 & 2 & 1 & -1 \end{array}\right] }$Which Operation Will Make The Lower Left Element The Largest?A. { R_1 \leftrightarrow R_3 $} B . \[ B. \[ B . \[ R_2

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Introduction

In linear algebra, matrices are used to represent systems of equations and perform various operations. One common operation is to transform a matrix into a specific form, such as row echelon form or reduced row echelon form. However, in this article, we will focus on a specific problem: given a matrix, which operation will make the lower left element the largest?

The Given Matrix

The given matrix is:

$$\left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ 3 & 4 & 5 & 5 \\ -1 & 2 & 1 & -1 \end{array}\right]$$

Understanding the Problem

To maximize the lower left element, we need to perform a series of row operations on the matrix. The goal is to make the element in the third row and first column as large as possible.

Row Operations

There are three types of row operations:

1. Swap two rows: Swap the positions of two rows.
2. Multiply a row by a scalar: Multiply a row by a non-zero scalar.
3. Add a multiple of one row to another: Add a multiple of one row to another row.

Option A: Swap Rows

Option A is to swap rows 1 and 3. This operation is denoted as:

$$R_1 \leftrightarrow R_3$$

After swapping rows 1 and 3, the matrix becomes:

$$\left[\begin{array}{ccc|c} -1 & 2 & 1 & -1 \\ 3 & 4 & 5 & 5 \\ 2 & -1 & -4 & 7 \end{array}\right]$$

Option B: Multiply Row 2 by a Scalar

Option B is to multiply row 2 by a scalar. Let's multiply row 2 by -1:

$$R_2 \rightarrow -R_2$$

After multiplying row 2 by -1, the matrix becomes:

$$\left[\begin{array}{ccc|c} -1 & 2 & 1 & -1 \\ -3 & -4 & -5 & -5 \\ 2 & -1 & -4 & 7 \end{array}\right]$$

Option C: Add a Multiple of Row 1 to Row 3

Option C is to add a multiple of row 1 to row 3. Let's add 2 times row 1 to row 3:

$$R_3 \rightarrow R_3 + 2R_1$$

After adding 2 times row 1 to row 3, the matrix becomes:

$$\left[\begin{array}{ccc|c} -1 & 2 & 1 & -1 \\ -3 & -4 & -5 & -5 \\ 0 & 3 & -2 & 9 \end{array}\right]$$

Comparing the Results

Now, let's compare the results of each option:

• Option A: The lower left element is -1.
• Option B: The lower left element is -3.
• Option C: The lower left element is 0.

Conclusion

Based on the calculations, we can see that Option C, which involves adding a multiple of row 1 to row 3, results in the largest lower left element. Therefore, the correct answer is:

Option C: $R_3 \rightarrow R_3 + 2R_1$

This operation makes the lower left element the largest.

Final Matrix

The final matrix after performing the correct operation is:

$$\left[\begin{array}{ccc|c} -1 & 2 & 1 & -1 \\ -3 & -4 & -5 & -5 \\ 0 & 3 & -2 & 9 \end{array}\right]$$

Takeaways

In this article, we learned how to maximize the lower left element in a matrix by performing a series of row operations. We compared the results of three different options and found that Option C, which involves adding a multiple of row 1 to row 3, results in the largest lower left element. This operation can be useful in various applications, such as solving systems of linear equations and finding the inverse of a matrix.