Question #128937
Given the matrix
0 2 3 -4 1
0 0 2 3 4
2 2 -5 2 4
2 0 -6 9 7
Reduce to row echelon form
Find the rank of the matrix
1
Expert's answer
2020-08-10T17:25:23-0400


A = [02341002342252420697]\begin{bmatrix} 0 & 2 & 3 & -4 & 1 \\ 0 & 0 & 2 & 3 & 4 \\ 2 & 2 & -5 & 2 & 4 \\ 2 & 0 & -6 & 9 & 7 \\ \end{bmatrix}


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: [22524002340234120697]\begin{bmatrix} 2 & 2 & -5 & 2 & 4 \\ 0 & 0 & 2 & 3 & 4 \\ 0 & 2 & 3 & -4 & 1 \\ 2 & 0 & -6 & 9 & 7 \\ \end{bmatrix}

make zeros in column 1 except the at entry row 1, column 1 (pivot entry)


Subtract row 1 from row 4

[R4= R4 - R1]; [22524002340234102173]\begin{bmatrix} 2 & 2 & -5 & 2 & 4 \\ 0 & 0 & 2 & 3 & 4 \\ 0 & 2 & 3 & -4 & 1 \\ 0 & -2 & -1 & 7 & 3 \\ \end{bmatrix}


Divide row 1 by 2

[R1 = R1/2] [115/212002340234102173]\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}


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; [115/212023410023402173]\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}


Make zeros in column 2 except the at entry row 2, column 2 (pivot entry).

Add row 2 to row 4

[R2 =R4 + R2]; [115/212023410023400234]\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}


Divide row 2 by 2

[R2 = R2/2]; [115/212013/221/20023400234]\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}


Subtract row 2 from row 1

[R1 = R1 - R2]; [10433/2013/221/20023400234]\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}


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]; [100919/2013/221/20023400234]\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}


Subtract row 3 from row 4

[R4 = R4 - R3]; [100919/2013/221/20023400000]\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}


Divide row 3 by 2

[R3 = R3/2]; [100919/2013/221/20013/2200000]\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}


Subtract row 3 multiplied by 3/2 to row 2.

[R2 = R2 - (3/2)R3]; [100919/201017/45/20013/2200000]\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}


SINCE ALL ROWS BELOW ARE ZERO THEN THE ANSWER IS,

rref(A)

= [100919/201017/45/20013/2200000]\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} ​

or in decimal form

= [10099.50104.252.50011.5200000]\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}




The rank is = 3



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