The image provides a matrix $ A $ and instructs on finding bases for the row space, column space, and null space of $ A $.The matrix $ A $ is given as:\[A = \begin{bmatrix} 1 & 1 & -8 \\ 0 & 2 & 1 \\ 1 & -1 & -9 \end{bmatrix}\]### Tasks:1. **row(A):** - Description: Find a basis for the row space of matrix $ A $. - Visual: Three blanks or placeholders, indicating room for basis vectors. - Arrows: Indicate that vectors are derived from the rows of the matrix.2. **col(A):** - Description: Find a basis for the column space of matrix $ A $. - Visual: Four blanks or placeholders, indicating vectors that fill this space. - Arrows: Indicate that vectors are derived from the columns of the matrix.3. **null(A):** - Description: Find a basis for the null space of matrix $ A $. - Visual: Four blanks or placeholders, indicating vectors that form the null space. - Arrows: Show the vector directions needed for the null space.### Explanation of Graphs/Diagrams:- Each space (row, column, and null) has a diagrammatic representation showing where basis vectors should be placed.- Green and gray arrows are used to indicate movement or the direction of derivation of these vectors from the matrix.- The desire is to fill these slots with appropriate vectors derived from computational processes (like row reduction or solving homogeneous equations) for each specific space.This visualization aids in understanding linear algebra concepts, particularly finding bases for different subspaces associated with a matrix.

Row Space: Non zero rows in the row reduced form is the basis of row space. Column Space: Since the pivot column form the basis of column space. Null Space: The solution AX=0 form the null space.Let us consider the given matrix A=11-80211-1-9. Apply row operation to determine pivot column of the matrix, 11-80211-1-9R3→R3-R111-80210-2-1R3→R3+R211-8021000R2→12R211-8010.5000R1→R1-R210-8.5010.5000 Non zero rows in the row reduced form is the basis of row space 108.5,010.5. Since the pivot column are 2 column 1 and 2. The column space has basis, 101,12-1 The solution AX=0 form the null space. 108.5010.5000abc=000⇒a+8.5c=0b+0.5c=0⇒a=-8.5cb=-0.5cabc=-8.5-0.51 Thus, the null space has basis -8.5-0.51. Answer: a) 108.5,010.5 b) 101,12-1 c) -8.5-0.51