Find the rank and nullity of the matrix \(A = \begin{bmatrix} 1 & 2 & 1 & 1 & 1 \\ 1 & 3 & 2 & 0 & 4 \\ 1 & 2 & 1 & 1 & 1 \end{bmatrix}\).
\(A = \begin{bmatrix} 1 & 2 & 1 & 1 & 1 \\ 1 & 3 & 2 & 0 & 4 \\ 1 & 2 & 1 & 1 & 1 \end{bmatrix}\)
First, the matrix needs to be transformed into the reduced row-echelon form.
To start with, our pivot on the first row of the matrix is the number in the first column as it is a one.
\(A = \begin{bmatrix} \boxed{1} & 2 & 1 & 1 & 1 \\ 1 & 3 & 2 & 0 & 4 \\ 1 & 2 & 1 & 1 & 1 \end{bmatrix}\)
This means the first row will be receiving no modifications during this pivot’s iteration. The second and third row will have the first row subtracted from them, as all three rows bear the same number and the pivot number is a one.
\(A = \begin{bmatrix} \boxed{1} & 2 & 1 & 1 & 1 \\ 1-1 & 3-2 & 2-1 & 0-1 & 4-1 \\ 1-1 & 2-2 & 1-1 & 1-1 & 1-1 \end{bmatrix}\)
\(A = \begin{bmatrix} \boxed{1} & 2 & 1 & 1 & 1 \\ 0 & 1 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
From here, the next row and column is observed to see if it has a 1. It does.
\(A = \begin{bmatrix} 1 & 2 & 1 & 1 & 1 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
As the last row is all zeros, it is untouched; the first row will be modified instead in order to make the value in the column of the pivot number equal to 0 when the pivot number is subtracted from it. To do this, the first row is halved.
\(A = \begin{bmatrix} 0.5 & 1 & 0.5 & 0.5 & 0.5 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
The pivot row is then subtracted from the first row.
\(A = \begin{bmatrix} 0.5-0 & 1-1 & 0.5-1 & 0.5+1 & 0.5-3 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
\(A = \begin{bmatrix} 0.5 & 0 & -0.5 & 1.5 & -2.5 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
The first row, to make the original pivot number return to a one, is doubled.
\(A = \begin{bmatrix} 0.5*2 & 0*2 & -0.5*2 & 1.5*2 & -2.5*2 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
\(A = \begin{bmatrix} 1 & 0 & -1 & 3 & -5 \\ 0 & 1 & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
As pivoting cannot be performed on a zero and the last row is full of zeros, this is the reduced row-echelon form of the original matrix, A.
\(A = \begin{bmatrix} \boxed{1} & 0 & -1 & 3 & -5 \\ 0 & \boxed{1} & 1 & -1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)
From this, it can be seen that matrix A has a rank of 2 due to there being two rows with leading ones. To calculate the nullity, the rank of the matrix \(r\) is subtracted from the total number of columns \(n\), written as \(n-r\). There are a total of five columns in this matrix.
\(r(A) = 2\)
\(n = 5\)
\(n(A) = n-r\)
\(n(A) = 5-2\)
\(n(A) = 3\)
Therefore, matrix A has a nullity of 3.
The matrix \(A = \begin{bmatrix} 1 & 2 & 1 & 1 & 1 \\ 1 & 3 & 2 & 0 & 4 \\ 1 & 2 & 1 & 1 & 1 \end{bmatrix}\) has a rank of 2 and a nullity of 3.