\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Let \[ \vect {M} = \begin {pmatrix} a & b \\ c & d \end {pmatrix}, \vect {N} = \begin {pmatrix} e & f \\ g & h \end {pmatrix}, \] and hence we have \[ \tr \vect {M} = a + d, \tr \vect {N} = e + h. \]
Notice that \[ \vect {M} \vect {N} = \begin {pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end {pmatrix}, \vect {N} \vect {M} = \begin {pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end {pmatrix}, \] which means \[ \tr (\vect {M} \vect {N}) = ae + bg + cf + dh, \tr (\vect {N} \vect {M}) = ae + cf + bg + dh, \] and hence \(\tr (\vect {M} \vect {N}) = \tr (\vect {N} \vect {M})\) as desired.
We also have \[ \vect {M} + \vect {N} = \begin {pmatrix} a + e & b + f \\ c + g & d + h \end {pmatrix}, \] meaning \(\tr (\vect {M} + \vect {N}) = a + e + d + h = (a + d) + (e + h) = \tr \vect {M} + \tr \vect {N}\).
We have \(\det \vect {M} = ad - bc\), and hence \[ \DiffOp {t} \det \vect {M} = \dot {a} d + a \dot {d} - \dot {b} c - b \dot {c}. \]
Hence, \[ \LHS = \frac {1}{ad - bd} \left (\dot {a} d + a \dot {d} - \dot {b} c - b \dot {c}\right ). \]
On the other hand, \[ \DiffFrac {\vect {M}}{t} = \begin {pmatrix} \dot {a} & \dot {b} \\ \dot {c} & \dot {d} \end {pmatrix}, \vect {M}^{-1} = \frac {1}{ad - bc} \begin {pmatrix} d & -b \\ -c & a \end {pmatrix}, \] and hence \begin {align*} \vect {M}^{-1} \DiffFrac {\vect {M}}{t} & = \frac {1}{ad - bc} \begin {pmatrix} d & -b \\ -c & a \end {pmatrix} \begin {pmatrix} \dot {a} & \dot {b} \\ \dot {c} & \dot {d} \end {pmatrix} \\ & = \frac {1}{ad - bc} \begin {pmatrix} \dot {a} d - b \dot {c} & \dot {b} d - b \dot {d} \\ -\dot {a}c + a\dot {c} & -\dot {b}c + a\dot {d} \end {pmatrix}. \end {align*}
Hence, \begin {align*} \RHS & = \tr \left (\vect {M}^{-1} \DiffFrac {\vect {M}}{t}\right ) \\ & = \frac {1}{ad - bc} \left (\dot {a} d - b \dot {c} - \dot {b} c + a\dot {d}\right ) \\ & = \frac {1}{ad - bc} \left (\dot {a} d + a \dot {d} - b \dot {c} - \dot {b} c\right ) \\ & = \LHS , \end {align*}
as desired.
\(\det \vect {M} \neq 0\) since \(\vect {M}\) is non-singular, and hence left-multiplying by \(\vect {M}^{-1}\) on both sides gives us \[ \vect {M}^{-1} \DiffFrac {\vect {M}}{t} = \vect {N} - \vect {M}^{-1} \vect {N} \vect {M}. \]
Taking trace on both sides, we have \begin {align*} \frac {1}{\det \vect {M}} \DiffOp {t} \det \vect {M} & = \tr \left (\vect {M}^{-1} \DiffFrac {\vect {M}}{t}\right ) \\ & = \tr \left (\vect {N} - \vect {M}^{-1} \vect {N} \vect {M}\right ) \\ & = \tr \vect {N} - \tr \left (\vect {M}^{-1} \vect {N} \vect {M}\right ) \\ & = \tr \vect {N} - \tr \left (\left (\vect {M}^{-1} \vect {N}\right ) \vect {M}\right ) \\ & = \tr \vect {N} - \tr \left (\vect {M} \left (\vect {M}^{-1} \vect {N}\right )\right ) \\ & = \tr \vect {N} - \tr \left (\left (\vect {M} \vect {M}^{-1}\right ) \vect {N}\right ) \\ & = \tr \vect {N} - \tr \left (\vect {I} \vect {N}\right ) \\ & = \tr \vect {N} - \tr \vect {N} \\ & = 0. \end {align*}
Hence, \(\DiffOp {t} \det \vect {M} = 0\), which means \(\det \vect {M}\) is a constant independent of \(t\).
Directly taking trace on both sides, we have \begin {align*} \tr \DiffFrac {\vect {M}}{t} & = \tr (\vect {M} \vect {N} - \vect {N} \vect {M}) \\ & = \tr (\vect {M} \vect {N}) - \tr (\vect {N} \vect {M}) \\ & = 0, \end {align*}
and note \[ \tr \DiffFrac {\vect {M}}{t} = \DiffOp {t} \tr \vect {M}, \] and hence \[ \DiffOp {t} \tr \vect {M} = 0, \] meaning \(\tr \vect {M}\) is a constant independent of \(t\).
Notice that \[ \tr \left (\vect {M}^2\right ) = \tr \left [\begin {pmatrix} a & b \\ c & d \end {pmatrix}\begin {pmatrix} a & b \\ c & d \end {pmatrix}\right ] = a^2 + bc + bc + d^2 = a^2 + 2bc + d^2. \]
Since \(\tr \vect {M}\) and \(\det \vect {M}\) are both independent of \(t\), we must have \begin {align*} (\tr \vect {M})^2 - 2 \det \vect {M} & = (a + d)^2 - 2 (ad - bc) \\ & = a^2 + 2ad + d^2 - 2ad + 2bc \\ & = a^2 + 2bc + d^2 \\ & = \tr \left (\vect {M}^2\right ) \end {align*}
is independent of \(t\) as well.
Let \[ \vect {M} = \begin {pmatrix} A + x & b \\ c & D - x \end {pmatrix}, \] the diagonal ones being so since the trace is independent of \(t\). Here, \(x\) is a function of \(t\).
By differentiating, \[ \DiffFrac {\vect {M}}{t} = \begin {pmatrix} \dot {x} & \dot {b} \\ \dot {c} & -\dot {x} \end {pmatrix}, \] and the right-hand side satisfies \begin {align*} \vect {M} \vect {N} - \vect {N} \vect {M} & = \begin {pmatrix} A + x & b \\ c & D - x \end {pmatrix} \begin {pmatrix} t & t \\ & t \end {pmatrix} - \begin {pmatrix} t & t \\ & t \end {pmatrix} \begin {pmatrix} A + x & b \\ c & D - x \end {pmatrix} \\ & = \begin {pmatrix} t (A + x) & (A + x) t + bt \\ ct & ct + (D - x) t \end {pmatrix} - \begin {pmatrix} t (A + x) + ct & bt + t(D - x) \\ ct & t(D - x) \end {pmatrix} \\ & =\begin {pmatrix} -ct & (A - D + 2x)t \\ 0 & ct. \end {pmatrix} \end {align*}
Comparing the components, we see that \(\dot {c} = 0\), meaning that \(c\) is a constant: \(c = C\).
Hence, \(\dot {x} = -Ct\), which solves to \(x = -\frac {Ct^2}{2}\), since \(x = 0\) when \(t = 0\).
This means \[ \dot {b} = (A - D + 2x) t = (A - D - Ct^2)t, \] and hence \[ b = \frac {(A - D) t^2}{2} - \frac {Ct^4}{4} + B \] since \(b = B\) when \(t = 0\).
Hence, \[ \vect {M} = \begin {pmatrix} A - Ct^2 / 2 & (A - D) t^2 / 2 - Ct^4 / 4 \\ C & D + Ct^2 / 2 \end {pmatrix} \] is the solution given the conditions.
By rearranging, we have \[ \vect {N} = \vect {M}^{-1} \DiffFrac {\vect {M}}{t}. \]
Hence, let \[ \vect {M} = \begin {pmatrix} 1 + e^t & \\ & 1 - e^t \end {pmatrix}, \] we have \[ \tr \vect {M} = 2 \] which is non-zero and independent of \(t\).
Hence, \[ \vect {M}^{-1} = \frac {1}{1 - e^{2t}} \begin {pmatrix} 1 - e^t & \\ & 1 + e^t \end {pmatrix}, \DiffFrac {\vect {M}}{t} = \begin {pmatrix} e^t & \\ & -e^t \end {pmatrix}, \] so \begin {align*} \vect {N} & = \frac {1}{1 - e^{2t}} \begin {pmatrix} 1 - e^t & \\ & 1 + e^t \end {pmatrix} \begin {pmatrix} e^t & \\ & -e^t \end {pmatrix} \\ & = \frac {1}{1 - e^{2t}} \begin {pmatrix} e^t (1 - e^t) & \\ & -e^t (1 + e^t) \end {pmatrix}, \end {align*}
which gives \[ \tr \vect {N} = \frac {e^{2t}}{e^{2t} - 1} \] which is clearly non-zero.