Simplify the Matrix [[1,1,-1,1],[2,3,a,3],[1,a,3,2]]

Problem

[[1,1,−1,1],[2,3,a,3],[1,a,3,2]]

Solution

1. Set up the matrix for row reduction to reach row echelon form.

A=[[1,1,−1,1],[2,3,a,3],[1,a,3,2]]

2. Eliminate the first entries in the second and third rows using the first row.

(R_2)←(R_2)−2*(R_1)

(R_3)←(R_3)−(R_1)

[[1,1,−1,1],[0,1,a+2,1],[0,a−1,4,1]]

3. Eliminate the second entry in the third row using the second row.

(R_3)←(R_3)−(a−1)*(R_2)

[[1,1,−1,1],[0,1,a+2,1],[0,0,4−(a−1)*(a+2),1−(a−1)]]

4. Simplify the expressions in the third row.

4−(a2+a−2)=−a2−a+6

1−a+1=2−a

[[1,1,−1,1],[0,1,a+2,1],[0,0,−a2−a+6,2−a]]

5. Factor the quadratic expression in the third row.

−a2−a+6=−(a+3)*(a−2)

[[1,1,−1,1],[0,1,a+2,1],[0,0,−(a+3)*(a−2),2−a]]

6. Perform back-substitution to reach reduced row echelon form, assuming a≠2 and a≠−3

(R_1)←(R_1)−(R_2)

[[1,0,−a−3,0],[0,1,a+2,1],[0,0,−(a+3)*(a−2),2−a]]

7. Divide the third row by 2−a (assuming a≠2.

(R_3)←(R_3)/(2−a)

[[1,0,−a−3,0],[0,1,a+2,1],[0,0,a+3,1]]

8. Eliminate the third column entries in the first and second rows.

(R_1)←(R_1)+(R_3)

(R_2)←(R_2)−(a+2)/(a+3)*(R_3)

[[1,0,0,1],[0,1,0,1−(a+2)/(a+3)]]

9. Simplify the final entries.

1−(a+2)/(a+3)=(a+3−a−2)/(a+3)=1/(a+3)

[[1,0,0,1],[0,1,0,1/(a+3)],[0,0,a+3,1]]

Final Answer

[[1,1,−1,1],[2,3,a,3],[1,a,3,2]]⇒[[1,0,0,1],[0,1,0,1/(a+3)],[0,0,1,1/(a+3)]]

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