2 problems 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"
For any square matrix \(\mathbf{A}\) such that \(\mathbf{I-A}\) is non-singular (where \(\mathbf{I}\) is the unit matrix), the matrix \(\mathbf{B}\) is defined by \[ \mathbf{B}=(\mathbf{I+A})(\mathbf{I-A})^{-1}. \] Prove that \(\mathbf{B}^{\mathrm{T}}\mathbf{B}=\mathbf{I}\) if and only if \(\mathbf{A+A}^{\mathrm{T}}=\mathbf{O}\) (where \(\mathbf{O}\) is the zero matrix), explaining clearly each step of your proof. {[}You may quote standard results about matrices without proof.{]}
Solution: We use the following properties: \((\mathbf{AB})^T = \mathbf{B}^T\mathbf{A}^T\), \((\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}\), and \((\mathbf{A}^T)^{-1} = (\mathbf{A}^{-1})^T\) \begin{align*} &&\mathbf{I} &= \mathbf{B}^{\mathrm{T}}\mathbf{B} \\ \Leftrightarrow && {\mathbf{B}^{T}}^{-1} &= \mathbf{B} \\ \Leftrightarrow && (\mathbf{I+A})(\mathbf{I-A})^{-1} &= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{T}}^{-1} \\ &&&= {((\mathbf{I+A})(\mathbf{I-A})^{-1})^{-1}}^{T} \\ &&&= ((\mathbf{I-A})(\mathbf{I+A})^{-1})^{T} \\ &&&= {(\mathbf{I+A})^{-1}}^T(\mathbf{I-A})^T \\ &&&= {(\mathbf{I+A}^T)}^{-1}(\mathbf{I-A}^T) \\ \Leftrightarrow && (\mathbf{I+A}^T)(\mathbf{I+A}) &= (\mathbf{I-A}^T)(\mathbf{I-A}) \\ \Leftrightarrow && \mathbf{I}+\mathbf{A}+\mathbf{A}^T+\mathbf{A}^T\mathbf{A} &= \mathbf{I}-\mathbf{A}-\mathbf{A}^T+\mathbf{A}^T\mathbf{A} \\ \Leftrightarrow && 2( \mathbf{A}^T+\mathbf{A}) &= \mathbf{O} \\ \Leftrightarrow && \mathbf{A}+\mathbf{A}^T &= \mathbf{O} \end{align*}