1. We are given the matrix $$\begin{bmatrix} 1 & 0 & a+b \\ a & a & 0 \\ 0 & 0 & b \end{bmatrix}$$ and we need to find values of $a$ and $b$ such that this matrix is in row-reduced echelon form (RREF). 2. Recall the conditions for a matrix to be in RREF: - The leading entry (pivot) in each nonzero row is 1. - Each leading 1 is the only nonzero entry in its column. - The leading 1 in a lower row appears to the right of the leading 1 in the row above. - Any rows of all zeros are at the bottom of the matrix. 3. Analyze the matrix: - The first row has leading 1 at position (1,1). - The second row has first element $a$, second element $a$, and zero in third. - The third row has zeros except possibly $b$ at (3,3). 4. For RREF, the pivot in the second row must be 1 and must appear to the right of the first pivot (1,1). The first row pivot is at column 1, so the second pivot cannot be at column 1. - If $a \neq 0$, the second row starts with $a$, so the leading entry is in column 1, which violates the rule that pivots move to the right down the rows. - Therefore, $a=0$ so that the leading entry in the second row shifts right to the second column. 5. Now, with $a=0$, the matrix is: $$\begin{bmatrix} 1 & 0 & a+b \\ 0 & 0 & 0 \\ 0 & 0 & b \end{bmatrix}$$ - The second row is all zeroes. - The third row has $b$ in position (3,3). 6. Check the third row pivot: - The pivot is at column 3, consistent with the RREF. - For it to be a pivot 1, $b$ must be 1. 7. Check that the pivot columns have zeros elsewhere: - Column 1 pivot: only first row has 1, others zero. - Column 3 pivot: third row has 1, first and second rows must have zero in this column for RREF. - First row has $a+b$ in third column. 8. For the pivot column 3, only the pivot entry must be 1 and all other entries zero. So $a+b = 0$. 9. Since $a = 0$, $a+b=0$ implies $b=0$. 10. But we also need $b=1$ from step 6 for the pivot to be 1. 11. Contradiction unless the third row is zero (no pivot). 12. Let's consider if $b=0$, then third row is zero; then only two pivots: positions (1,1) and (second row) should be in column 2. 13. But second row has all zeros in this case. So that leaves only one pivot. 14. Test the options given for ($a$, $b$): - (0,1): $a=0$, $b=1$, check $a+b=1$ (not zero), conflict with step 8. - (1,0): $a=1$, $b=0$; violates step 4, no. - (-1,-1): $a=-1$, $b=-1$; first row pivot okay, second row pivot at column 1 not allowed, no. - (1,-1): $a=1$, $b=-1$; same as above no. - (0,0): $a=0$, $b=0$, third row zero, $a+b=0$, good. 15. So the only possible values are $a=0$, $b=0$. **Final answer:** $$a=0, b=0$$