A = [ 0 2 3 − 4 1 0 0 2 3 4 2 2 − 5 2 4 2 0 − 6 9 7 ] \begin{bmatrix}
0 & 2 & 3 & -4 & 1 \\
0 & 0 & 2 & 3 & 4 \\
2 & 2 & -5 & 2 & 4 \\
2 & 0 & -6 & 9 & 7 \\
\end{bmatrix} ⎣ ⎡ 0 0 2 2 2 0 2 0 3 2 − 5 − 6 − 4 3 2 9 1 4 4 7 ⎦ ⎤
Since element at row 1 and column 1 (pivot element) equals 0, we need to swap rows
Then find the first non zero element in the column 1 under the pivot entry
The first non zero element is row 3
Swap rows 1 and 3: [ 2 2 − 5 2 4 0 0 2 3 4 0 2 3 − 4 1 2 0 − 6 9 7 ] \begin{bmatrix}
2 & 2 & -5 & 2 & 4 \\
0 & 0 & 2 & 3 & 4 \\
0 & 2 & 3 & -4 & 1 \\
2 & 0 & -6 & 9 & 7 \\
\end{bmatrix} ⎣ ⎡ 2 0 0 2 2 0 2 0 − 5 2 3 − 6 2 3 − 4 9 4 4 1 7 ⎦ ⎤
make zeros in column 1 except the at entry row 1, column 1 (pivot entry)
Subtract row 1 from row 4
[R4= R4 - R1]; [ 2 2 − 5 2 4 0 0 2 3 4 0 2 3 − 4 1 0 − 2 − 1 7 3 ] \begin{bmatrix}
2 & 2 & -5 & 2 & 4 \\
0 & 0 & 2 & 3 & 4 \\
0 & 2 & 3 & -4 & 1 \\
0 & -2 & -1 & 7 & 3 \\
\end{bmatrix} ⎣ ⎡ 2 0 0 0 2 0 2 − 2 − 5 2 3 − 1 2 3 − 4 7 4 4 1 3 ⎦ ⎤
Divide row 1 by 2
[R1 = R1/2] [ 1 1 − 5 / 2 1 2 0 0 2 3 4 0 2 3 − 4 1 0 − 2 − 1 7 3 ] \begin{bmatrix}
1 & 1 & -5/2 & 1 & 2 \\
0 & 0 & 2 & 3 & 4 \\
0 & 2 & 3 & -4 & 1 \\
0 & -2 & -1 & 7 & 3 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 0 2 − 2 − 5/2 2 3 − 1 1 3 − 4 7 2 4 1 3 ⎦ ⎤
Since element at row 2 and column 2 (pivot element) equals 0, we need to swap rows
Then find the first non zero element in the column 2 under the pivot entry
The first non zero element is row 3
swap rows 2 with 3; [ 1 1 − 5 / 2 1 2 0 2 3 − 4 1 0 0 2 3 4 0 − 2 − 1 7 3 ] \begin{bmatrix}
1 & 1 & -5/2 & 1 & 2 \\
0 & 2 & 3 & -4 & 1 \\
0 & 0 & 2 & 3 & 4 \\
0 & -2 & -1 & 7 & 3 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 2 0 − 2 − 5/2 3 2 − 1 1 − 4 3 7 2 1 4 3 ⎦ ⎤
Make zeros in column 2 except the at entry row 2, column 2 (pivot entry).
Add row 2 to row 4
[R2 =R4 + R2]; [ 1 1 − 5 / 2 1 2 0 2 3 − 4 1 0 0 2 3 4 0 0 2 3 4 ] \begin{bmatrix}
1 & 1 & -5/2 & 1 & 2 \\
0 & 2 & 3 & -4 & 1 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 2 & 3 & 4 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 2 0 0 − 5/2 3 2 2 1 − 4 3 3 2 1 4 4 ⎦ ⎤
Divide row 2 by 2
[R2 = R2/2]; [ 1 1 − 5 / 2 1 2 0 1 3 / 2 − 2 1 / 2 0 0 2 3 4 0 0 2 3 4 ] \begin{bmatrix}
1 & 1 & -5/2 & 1 & 2 \\
0 & 1 & 3/2 & -2 & 1/2 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 2 & 3 & 4 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 1 1 0 0 − 5/2 3/2 2 2 1 − 2 3 3 2 1/2 4 4 ⎦ ⎤
Subtract row 2 from row 1
[R1 = R1 - R2]; [ 1 0 − 4 3 3 / 2 0 1 3 / 2 − 2 1 / 2 0 0 2 3 4 0 0 2 3 4 ] \begin{bmatrix}
1 & 0 & -4 & 3 & 3/2 \\
0 & 1 & 3/2 & -2 & 1/2 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 2 & 3 & 4 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 − 4 3/2 2 2 3 − 2 3 3 3/2 1/2 4 4 ⎦ ⎤
Make zeros in column 3 except the at entry row 3, column 3 (pivot entry).
Add row 3 multiplied by 2 to row 1.
[R3 = R1 + (2)R3]; [ 1 0 0 9 19 / 2 0 1 3 / 2 − 2 1 / 2 0 0 2 3 4 0 0 2 3 4 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 19/2 \\
0 & 1 & 3/2 & -2 & 1/2 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 2 & 3 & 4 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 0 3/2 2 2 9 − 2 3 3 19/2 1/2 4 4 ⎦ ⎤
Subtract row 3 from row 4
[R4 = R4 - R3]; [ 1 0 0 9 19 / 2 0 1 3 / 2 − 2 1 / 2 0 0 2 3 4 0 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 19/2 \\
0 & 1 & 3/2 & -2 & 1/2 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 0 3/2 2 0 9 − 2 3 0 19/2 1/2 4 0 ⎦ ⎤
Divide row 3 by 2
[R3 = R3/2]; [ 1 0 0 9 19 / 2 0 1 3 / 2 − 2 1 / 2 0 0 1 3 / 2 2 0 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 19/2 \\
0 & 1 & 3/2 & -2 & 1/2 \\
0 & 0 & 1 & 3/2 & 2 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 0 3/2 1 0 9 − 2 3/2 0 19/2 1/2 2 0 ⎦ ⎤
Subtract row 3 multiplied by 3/2 to row 2.
[R2 = R2 - (3/2)R3]; [ 1 0 0 9 19 / 2 0 1 0 − 17 / 4 − 5 / 2 0 0 1 3 / 2 2 0 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 19/2 \\
0 & 1 & 0 & -17/4 & -5/2 \\
0 & 0 & 1 & 3/2 & 2 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 0 0 1 0 9 − 17/4 3/2 0 19/2 − 5/2 2 0 ⎦ ⎤
SINCE ALL ROWS BELOW ARE ZERO THEN THE ANSWER IS,
rref(A)
= [ 1 0 0 9 19 / 2 0 1 0 − 17 / 4 − 5 / 2 0 0 1 3 / 2 2 0 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 19/2 \\
0 & 1 & 0 & -17/4 & -5/2 \\
0 & 0 & 1 & 3/2 & 2 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
⎣ ⎡ 1 0 0 0 0 1 0 0 0 0 1 0 9 − 17/4 3/2 0 19/2 − 5/2 2 0 ⎦ ⎤
or in decimal form
= [ 1 0 0 9 9.5 0 1 0 − 4.25 − 2.5 0 0 1 1.5 2 0 0 0 0 0 ] \begin{bmatrix}
1 & 0 & 0 & 9 & 9.5 \\
0 & 1 & 0 & -4.25 & -2.5 \\
0 & 0 & 1 & 1.5 & 2 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix} ⎣ ⎡ 1 0 0 0 0 1 0 0 0 0 1 0 9 − 4.25 1.5 0 9.5 − 2.5 2 0 ⎦ ⎤
The rank is = 3
Comments