(1 point) How many free variables does each augmented matrix have

(1 point) How many free variables does each augmented matrix have? 1 0 073 0 1-5 0 10 A. None B. One C. Two D. Three 2.0 15 A. None B. One C. Two D. Three 1 9 -9-10 0 A. None B. One C. Two D. Three WO 4. 0 1 0-2 A. None B. One C. Two D. Three

The Correct Answer and Explanation is :

To determine how many free variables an augmented matrix has, we first need to understand how to analyze an augmented matrix and the concepts of free variables and pivot variables.

A free variable corresponds to a column in the augmented matrix that does not contain a pivot, which is a leading 1 in a row (after row reduction, or Gaussian elimination). Pivot variables correspond to the columns with pivots. If there are columns without pivots, those variables are considered free, and they can take on any value. The number of free variables corresponds to the number of such columns.

Now, let’s analyze the given augmented matrices:

Matrix 1:

[
\begin{pmatrix}
1 & 0 & 0 & 7 \
3 & 0 & 1 & -5 \
0 & 1 & 0 & 10
\end{pmatrix}
]

To perform row reduction:

1. The first row is fine: ( (1, 0, 0, 7) ).
2. Subtract 3 times the first row from the second row to make the first entry of the second row 0:
   [
   R2 = R2 – 3R1 \implies (3 – 3, 0 – 0, 1 – 0, -5 – 21) = (0, 0, 1, -26)
   ]
3. The third row is already ( (0, 1, 0, 10) ), which is in proper form.

In this case, the pivot variables are in columns 1, 2, and 3 (there are no free variables). Answer: A. None.

Matrix 2:

[
\begin{pmatrix}
1 & 0 & 1 & 5 \
1 & 9 & -9 & -10
\end{pmatrix}
]

Performing row reduction:

1. Subtract the first row from the second to make the first entry of the second row 0:
   [
   R2 = R2 – R1 \implies (1 – 1, 9 – 0, -9 – 1, -10 – 5) = (0, 9, -10, -15)
   ]
2. Next, divide the second row by 9 to make the second pivot 1:
   [
   R2 = \frac{1}{9} R2 \implies (0, 1, -\frac{10}{9}, -\frac{5}{3})
   ]

Here, column 1 and column 2 contain pivots, but there’s a free variable in column 3. Answer: B. One.

Matrix 3:

[
\begin{pmatrix}
0 & 1 & 0 & -2
\end{pmatrix}
]
This is a 1×4 matrix, and we see only one non-zero entry in column 2, which is a pivot. Thus, the variable corresponding to column 2 is the only pivot. The remaining columns have free variables. Answer: B. One.

In summary, after performing Gaussian elimination, the number of free variables for each matrix is:

1. None
2. One
3. One

Each free variable corresponds to the columns that don’t contain pivots in the row-reduced form of the matrix.