A(t) is a 2×2 matrix of differentiable functions and X(t) is a 2×1 column matrix of differentiable functions, prove the product rule

Step1Let The matrices A(t) and X(t) beA(t)=[A11(t)A12(t)A21(t)A22(t)]X(t)=[x1(t)x2(t)]The product rule of differentiation states thatddt[A(t)X(t)]=A′(t)X(t)+A(t)X′(t)Step2A(t)X(t)=[A11(t)A12(t)A21(t)A22(t)][x1(t)x2(t)]=[A11(t)x1(t)A12(t)x2(t)A21(t)x1(t)A22(t)x2(t)]Now,ddt[A(t)X(t)]=[ddtA11(t)x1(t)A12(t)x2(t)ddtA21(t)x1(t)A22(t)x2(t)]=[ddt[A11(t)x1(t)A12(t)x2(t)]ddt[A21(t)x1(t)A22(t)x2(t)]]=[A11′(t)x1(t)+A11(t)x1′(t)+A12′(t)x2(t)+A12(t)x2′(t)A21′(t)x1(t)+A21(t)x1′(t)+A22′(t)x2(t)+A22(t)x2′(t)]Step 3Now, A′(t)=[A11′(t)A12′(t)A21(t)A22(t)]

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Aufopferaq

Answered question

2021-11-27

A(t) is a

2 \times 2

matrix of differentiable functions and X(t) is a

2 \times 1

column matrix of differentiable functions, prove the product rule

Answer & Explanation

Ched1950

Beginner2021-11-28Added 21 answers

Step1
Let The matrices A(t) and X(t) be
$A (\begin{matrix} t \end{matrix}) = [\begin{matrix} A_{11} (t) & A_{12} (t) \\ A_{21} (t) & A_{22} (t) \end{matrix}]$
$X (\begin{matrix} t \end{matrix}) = [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}]$
The product rule of differentiation states that
$\frac{d}{d t} [A (t) X (t)] = A^{'} (t) X (t) + A (t) X^{'} (t)$
Step2
$A (\begin{matrix} t \end{matrix}) X (\begin{matrix} t \end{matrix}) = [\begin{matrix} A_{11} (t) & A_{12} (t) \\ A_{21} (t) & A_{22} (t) \end{matrix}] [\begin{matrix} x_{1} (t) \\ x_{2} (t) \end{matrix}]$
$= [\begin{matrix} A_{11} (t) x_{1} (t) & A_{12} (t) x_{2} (t) \\ A_{21} (t) x_{1} (t) & A_{22} (t) x_{2} (t) \end{matrix}]$
Now,
$\frac{d}{d t} [A (t) X (t)]$
$= [\begin{matrix} \frac{d}{d t} A_{11} (t) x_{1} (t) & A_{12} (t) x_{2} (t) \\ \frac{d}{d t} A_{21} (t) x_{1} (t) & A_{22} (t) x_{2} (t) \end{matrix}]$
$= [\begin{matrix} \frac{d}{d t} [A_{11} (t) x_{1} (t) & A_{12} (t) x_{2} (t)] \\ \frac{d}{d t} [A_{21} (t) x_{1} (t) & A_{22} (t) x_{2} (t)] \end{matrix}]$
$= [\begin{matrix} A_{11}^{'} (t) x_{1} (t) + A_{11} (t) x_{1}^{'} (t) + & A_{12}^{'} (t) x_{2} (t) + A_{12} (t) x_{2}^{'} (t) \\ A_{21}^{'} (t) x_{1} (t) + A_{21} (t) x_{1}^{'} (t) + & A_{22}^{'} (t) x_{2} (t) + A_{22} (t) x_{2}^{'} (t) \end{matrix}]$
Step 3
Now, $A^{'} (\begin{matrix} t \end{matrix}) = [\begin{matrix} A_{11}^{'} (t) & A_{12}^{'} (t) \\ A_{21} (t) & A_{22} (t) \end{matrix}]$

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