Example # 2: (Solving a Matrix in R3) Solve the linear system using matrices x+3y+4z19=0 x+2y+z=12 x+y+z8=O Solution: :1: as: *I :1: 1 0 0| # It is not always possible to reduce a matrix of the form * * *| * to one of the form 0 1 0| # using :1: * * l :1: 0 0 1 l # row reduction. If one of the equations is a linear combination of the other two, a will 1 0 *| # occur at some point. It may be possible to reduce the matrix to a form such as 0 1 *| # . Then the 0 0 0| 0 system is , and the corresponding planes intersect in a line. Parametric equations of this line constitute the solution of the system. Example # 1: Solve the system using matrices and interpret the solution geometrically. 2x+7y+22=3 6x+y4z=1 2x+9y+3z=4 Solution: Example # 1: Solve the system using matrices and interpret the solution geometrically. 2x+7y+22=3 6x+y4z=1 2x+9y+3z=4 Solution: Example # 2: Solve the following system of linear equations by reducing the corresponding augmented matrix to row-reduced echelon form. wx+yz=4 w+x+y+z=0 w3x+y3z=8 w+x+yz=0 Solution