Answer to Question #128937 in Linear Algebra for Sym

Question

Answer to Question #128937 in Linear Algebra for Sym

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

Expert's answer

A = "\\begin{bmatrix}\n 0 & 2 & 3 & -4 & 1 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 2 & 2 & -5 & 2 & 4 \\\\\n 2 & 0 & -6 & 9 & 7 \\\\\n\\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: "\\begin{bmatrix}\n 2 & 2 & -5 & 2 & 4 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 2 & 3 & -4 & 1 \\\\\n 2 & 0 & -6 & 9 & 7 \\\\\n\\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]; "\\begin{bmatrix}\n 2 & 2 & -5 & 2 & 4 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 2 & 3 & -4 & 1 \\\\\n 0 & -2 & -1 & 7 & 3 \\\\\n\\end{bmatrix}"

Divide row 1 by 2

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

Divide row 2 by 2

[R2 = R2/2]; "\\begin{bmatrix}\n 1 & 1 & -5\/2 & 1 & 2 \\\\\n 0 & 1 & 3\/2 & -2 & 1\/2 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n\\end{bmatrix}"

Subtract row 2 from row 1

[R1 = R1 - R2]; "\\begin{bmatrix}\n 1 & 0 & -4 & 3 & 3\/2 \\\\\n 0 & 1 & 3\/2 & -2 & 1\/2 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n\\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]; "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 19\/2 \\\\\n 0 & 1 & 3\/2 & -2 & 1\/2 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n\\end{bmatrix}"

Subtract row 3 from row 4

[R4 = R4 - R3]; "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 19\/2 \\\\\n 0 & 1 & 3\/2 & -2 & 1\/2 \\\\\n 0 & 0 & 2 & 3 & 4 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n\\end{bmatrix}"

Divide row 3 by 2

[R3 = R3/2]; "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 19\/2 \\\\\n 0 & 1 & 3\/2 & -2 & 1\/2 \\\\\n 0 & 0 & 1 & 3\/2 & 2 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n\\end{bmatrix}"

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

[R2 = R2 - (3/2)R3]; "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 19\/2 \\\\\n 0 & 1 & 0 & -17\/4 & -5\/2 \\\\\n 0 & 0 & 1 & 3\/2 & 2 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n\\end{bmatrix}"

SINCE ALL ROWS BELOW ARE ZERO THEN THE ANSWER IS,

rref(A)

= "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 19\/2 \\\\\n 0 & 1 & 0 & -17\/4 & -5\/2 \\\\\n 0 & 0 & 1 & 3\/2 & 2 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n\\end{bmatrix}\n\n\u200b"

or in decimal form

= "\\begin{bmatrix}\n 1 & 0 & 0 & 9 & 9.5 \\\\\n 0 & 1 & 0 & -4.25 & -2.5 \\\\\n 0 & 0 & 1 & 1.5 & 2 \\\\\n 0 & 0 & 0 & 0 & 0 \\\\\n\\end{bmatrix}"

The rank is = 3