**Exercises****2.1.** In each of the following cases, compute a basis matrix for the null space of the matrix \( A \) and express the points \( x_i \) as \( x_i = p_i + q_i \) where \( p_i \) is in the null space of \( A \) and \( q_i \) is in the range space of \( A^T \). **(i)** \[ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad x_1 = \begin{pmatrix} 1 \\ 3 \\ 1 \\ 2 \end{pmatrix}, \quad x_2 = \begin{pmatrix} 0 \\ -2 \\ -3 \\ 4 \end{pmatrix}.\]

(a) The given matrix is: A=11111-1-110101 Calculating row reduced echelon form of given matrix: Applying R2→R2-R1: A≈11110-2-200101 applying R2=R2-2: A≈111101100101 Applying R1→R1-R2 and R3→R3-R2 A≈1001011000-11 Applying R3=-1R3: A≈10010110001-1 Applying R2→R2-R3: A≈10010101001-1 Therefore there are three pivot elements and hence the rank of matrix is 3.Now for null space. AX=010010101001-1xywz=0000x+z=0⇒x=-zy+z=0⇒y=-zw+z=0⇒w=-z Hence every vector X in null space can be written as: X=xywz=-z-z-zz=z-1-1-11 Therefore null space is spanned by a single vector. Therefore basis for null space is B1=-1-1-11.(b) Now consider AT: AT=1101-111-10111 Therefore row reduced form of AT can be written as: AT=100010001000 Therefore there are three pivot element, and hence dimension of range space of AT is 3. Therefore the basis of range space of AT is: B2=1000,0100,0010.Now given: x1=1312, therefore it can be written as: 1312=a-1-1-11+b1000+c0100+d0010=-a+b-a+c-a+da Hence a=2, b=3, c=5, d=3 Therefore: 1312=2-1-1-11+31000+50100+30010=-2-2-22+3530=p1+q1 Hence, p1=-2-2-22,q1= 3530.Now given: x2=0-2-34, therefore it can be written as: 0-2-34=a-1-1-11+b1000+c0100+d0010=-a+b-a+c-a+da Hence a=4, b=4, c=2, d=1 Therefore: 0-2-34=4-1-1-11+41000+20100+10010=-4-4-44+4210=p1+q1 Hence, p2=-4-4-44, q2=4210.