Given the matrix A and the vector b = (1, 1, 1)T, where A = (see attached photo)a) Express the Gram-Schmidt orthogonalization of the columns of A inthe form of A = QR.b) Use your QR factorization from part (a) to find the least-squares solutionx to Ax = b. The image displays a matrix labeled \( A \), which is a \( 3 \times 2 \) matrix. The matrix is presented as follows:\[ A = \begin{bmatrix} 1 & 1 \\ 2 & 3 \\ 2 & 1 \end{bmatrix} \]Each element within the matrix is organized into rows and columns. There are three rows and two columns:- The first row contains the elements 1 and 1.- The second row contains the elements 2 and 3.- The third row contains the elements 2 and 1.This structure is commonly used in linear algebra for various applications such as solving systems of equations, transformations, and more.

Given: A=112321 (a) Let v1=122, v2=131 As v1,v2 are not orthogonal. So, using Gram Schmidt Process to make them orthogonal. Suppose v1=u1=122 and, u2=v2-v2,u1u1,u1u1 1 v2,u1=131,122=11+32+12=9u1,u1=122,122=11+22+22=9 Putting these values in (1), we get u2=131-99122 =01-1 Then, u1, u2 are orthogonal. Consider the orthonormal matrix Q=u1∥u1∥ u2∥u2∥ ∥u2∥=u2,u21/2=2 So, Q=1/302/31/22/3-1/2 Since, Q is orthonormal. So, A=QRQTQ=I Consider , QTA=QTQR=IR=R R=QTA=1/32/32/301/2-1/2112321 =3302 Hence, for above calculated Q, R, we have Now, given b=1,1,1T To find solution to Ax=b ⇒x=A-1b =QR-1b =R-1Q-1b AB-1=B-1A-1 =R-1QTb QQT=I⇒QT=Q-1 Now, R-1=1322-303=1/3-1/201/2 So, x=1/3-1/201/21/32/32/301/2-1/2111 =1/3-1/201/25/30 =5/90So, x=[5/90]