a = 0, 1, c = 0, and b is arbitrary

```
Find all real numbers a, b and c such that the following matrix is in reduced row echelon form.
⎛
⎜
⎝
a 1 0 1
0 0 1 b
0 0 c 0
⎞
⎟
⎠
```

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Answer from Sia

Posted 3 months ago

Solution by Steps

step 1

To determine the values of $a$, $b$, and $c$ such that the given matrix is in reduced row echelon form (RREF), we need to ensure that the matrix satisfies the conditions for RREF

step 2

The given matrix is:
$\begin{pmatrix}
a & 1 & 0 & 1 \\
0 & 0 & 1 & b \\
0 & 0 & c & 0 \\
\end{pmatrix}$

step 3

In RREF, each leading entry in a row must be 1, and all entries in the column containing a leading 1 must be zero

step 4

For the second row, the leading entry is already 1 in the third column. Therefore, $b$ can be any real number

step 5

For the third row, the leading entry must be 1. Therefore, $c$ must be 1

step 6

For the first row, the leading entry must be 1. Therefore, $a$ must be 0

step 7

The values of $a$, $b$, and $c$ that make the matrix in RREF are:
$a = 0, \quad b = \text{any real number}, \quad c = 1$

Answer

$a = 0$, $b$ can be any real number, $c = 1$

Key Concept

Reduced Row Echelon Form (RREF)

Explanation

In RREF, each leading entry in a row must be 1, and all entries in the column containing a leading 1 must be zero. This determines the values of $a$, $b$, and $c$.

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