1 problem found
The matrices \(\mathbf{A},\mathbf{B}\) and \(\mathbf{M}\) are given by \[ \mathbf{A}=\begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix}, \] where \(a,b,\ldots,r\) are real numbers. Given that \(\mathbf{M=AB},\) show that \(a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2\) and \(r=-3\) gives the unique solution for \(\mathbf{A}\) and \(\mathbf{B}.\) Evaluate \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1},\) Hence, or otherwise, solve the simultaneous equations \begin{alignat*}{1} x+3y+2z & =7\\ 4x+13y+5z & =18\\ 3x+8y+7z & =25. \end{alignat*}
Solution: \begin{align*} && \begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix} &= \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix}\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix} \\ &&&= \begin{pmatrix} a & ap & aq \\ b & pb + c & qb + cr\\ d & pd + e & qd + er +f \end{pmatrix} \\ \Rightarrow && a,b,d,p,q&=1,4,3,3,2\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 12 + c & 8+ cr\\ 3 & 9 + e & 6 + er +f \end{pmatrix} \\ \Rightarrow && c, e&=1,-1\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 13 & 8+ r\\ 3 & 8 & 6 -r +f \end{pmatrix} \\ \Rightarrow && r, f &= -3, -2 \end{align*} \begin{align*} \mathbf{A}^{-1} &= \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0\\ 3 & -1 & -2 \end{pmatrix}^{-1} \\ &=\frac{1}{-2} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \\ \\ \mathbf{B}^{-1} &= \begin{pmatrix} 1 & 3 & 2 \\ 0 & 1 & -3\\ 0 & 0 & 1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix} \\ \end{align*} We want to solve \(\mathbf{M}\mathbf{v} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}\), ie \begin{align*} \mathbf{v} &= \mathbf{M}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \mathbf{B}^{-1} \mathbf{A}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -14 \\ 20 \\ -6 \end{pmatrix} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 7 \\ -10 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} 4\\ -1 \\ 3 \end{pmatrix} \end{align*} This algorithm is called the "LU-decomposition"