Problems

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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"

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Solution: Claim \(G\) is a group under matrix multiplication

• (Closure) Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are matrices of that form, then \(\begin{pmatrix} a_1 & b_1 \\ 0 & c_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ 0 & c_2 \end{pmatrix} = \begin{pmatrix} a_1a_2 & a_1b_2 + b_1c_2 \\ 0 & c_1c_2 \end{pmatrix}\), this is clearly of the required form since if \(a_1, a_2, c_1, c_2 \neq 0\) then \(a_1a_2, c_1c_2 \neq 0\)
• (Associative) By inheritance from matrix multiplication
• (Identity) Consider \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) also clearly of the required form.
• (Inverse) Consider \((ac)^{-1}\begin{pmatrix} c & -b \\ 0 & a \end{pmatrix}\), since \(ac \neq 0\) we can assume it has an inverse mod \(5\). therefore we have another matrix of the required form.

There are \(4\) possible values for \(a\) and \(c\) and \(5\) possible values for \(b\), so \(4 \times 4 \times 5 = 80\) elements, so the group is order \(80\). \(G\) is not commutative, consider \(\begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 2 \end{pmatrix}\) \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 2 \end{pmatrix}\) The elements of order \(1\) or \(2\) satisfy \(\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = \begin{pmatrix} a^{-1} & -ba^{-1}c^{-1} \\ 0 & c^{-1} \end{pmatrix}\) Therefore \(a^2 = 1, c^2 = 1 \Rightarrow a, c = 1, 4\) and \(b = -ba^{-1}c^{-1} \Rightarrow b = 0\) or , \(ac = -1\), so we have \((a,b,c) = (1,0,1), (4,0,4), (1, *, 4), (4, *, 1)\) So there are \(12\) elements of order \(1\) or \(2\). But this can't be a subgroup since \(12 \not \mid 80\)

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Solution: Since \(0 = \det \mathbf{0} = \det \mathbf{AB} = \det \mathbf{A} \det\mathbf{B}\) at least one of \(\det \mathbf{A}\) or \(\det \mathbf{B}\) is zero. If \(\det \mathbf{B} \neq 0\) then \(\mathbf{B}\) is invertible, and multiplying on the right by \(\mathbf{B}^{-1}\) gives us \(\mathbf{A} = \mathbf{0}\). If \(\mathbf{A} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix}\), then \(\det \mathbf{A} = \det \mathbf{B} = 0\), but \(\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \mathbf{0}\) Since \(\mathbf{M}\) commutes with itself and the identity matrix, this is equivalent to factorising the polynomial over the reals. Therefore $$\mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}-2\mathbf{I})(\mathbf{M}+\mathbf{I})(\mathbf{M}+3\mathbf{I}),$$ Since we now know at least one of \(\det (\mathbf{M}-2\mathbf{I})\), \(\det (\mathbf{M}+\mathbf{I})\), \(\det (\mathbf{M}+3\mathbf{I})\), we should look at cases: Since at least one of those must be non-zero, we must have the following cases: \((a,b) = (2,-1), (-1,2), (-1,-3), (-3,-1), (2,-3), (-3,2)\) In each of those cases, we will have: \(\begin{pmatrix} 0 & c \\ 0 & b+k \end{pmatrix}\begin{pmatrix} a+l & c \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\) and so all of those solutions are valid. So \(c\) can be anything as long as \((a,b)\) are in that set of solutions