Reduced Row Echelon Form Of 4x4 Matrix A Reyes

Let's Compute the Reduced Row Echelon form of the following 4x4 matrix:

(1)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 4 & 3 \\ 2 & 4 & 1 & 0 \\ -1 & 3 & 2 & 1 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

It's best to have a pivot(pivots must be one) in the (1,1) position and we already have one.
So let make all entries under that pivot a zero starting with the 2nd row.
To do that we need to do the following row operation: -2R1+R2=R2
This gives us:

(2)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 4 & 3 \\ 0 & 2 & -7 & -6 \\ -1 & 3 & 2 & 1 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

Excellent! lets keep making all entries under the pivot zeros.
Next we need to do: R1+R3=R3
The result is:

(3)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 4 & 3 \\ 0 & 2 & -7 & -6 \\ 0 & 4 & 6 & 4 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

Alrighty lets keep it up. Let's get a pivot in the 2nd row now… It must be a non-zero number
and the number 2 in the (2,2) position will suffice. Let's make it a 1 by dividing row 2 by 2.
(R2/2)=R2 gives us:

(4)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 4 & 3 \\ 0 & 1 & -7/2 & -3 \\ 0 & 4 & 6 & 4 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

Okay looking good. I hope you get the point. We are using multiples of the pivot to make the entries above
and below it equal zero… That being said, I am only going to write row operations from here on.

-4R2+R3=R3
This will result in:

(5)
\begin{align} \left(\begin{array}{cc} 1 & 1 & 4 & 3 \\ 0 & 1 & -7/2 & -3 \\ 0 & 0 & 20 & 16 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

Now lets make the entry above the pivot(2,2) a zero.
-R2+R1=R1

(6)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 15/2 & 6 \\ 0 & 1 & -7/2 & -3 \\ 0 & 0 & 20 & 16 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

Now we can make a pivot in the (3,3) entry:
(R3/20)=R3

(7)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 15/2 & 6 \\ 0 & 1 & -7/2 & -3 \\ 0 & 0 & 1 & 4/5 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

let's continue the trend:
7R3/2+R2=R2 gives us:

(8)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 15/2 & 6 \\ 0 & 1 & 0 & -1/5 \\ 0 & 0 & 1 & 4/5 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

-15R3/2+R1=R1:

(9)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1/5 \\ 0 & 0 & 1 & 4/5 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

-4R4/5+R3=R3:

(10)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1/5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

R4/5+R2=R2:

(11)
\begin{align} \left(\begin{array}{cc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right) \end{align}

I know it was tedious… but now our mission has been accomplished!! YEEEHAW!

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