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I was going through a homework problem which is essentially a math review in a kinematics frame. This group of problems start as follows,

Given $a_x$, constant acceleration, and initial conditions, find the position, $x(t)$.

Not too complicated, but there were a lot of them, so I solved two problems instead. By direct integration, I know that the solution is a parabola of the $a_xt^2/2+C_1t+C_2$

For the initial conditions $x(t_1)=x_1$ and $v(t_2)=v_2$, find $x(t)$ if acceleration is constant, $a_x$ For the initial conditions $x(t_1)=x_1$ and $x(t_2)=x_2$, find $x(t)$ if acceleration is constant, $a_x$

For the first I have a solution it's all fine and dandy, I wrote the system of equations as a matrix and inverting a 2x2 is easy. $$ \bigg(\begin{matrix}C_1 \\ C_2\end{matrix}\bigg) = \bigg(\begin{matrix}1 & 0 \\ -t_1 & 1 \end{matrix}\bigg) \bigg(\begin{matrix} v_2-a_xt_2 \\ x_1-a_xt_1^2/2 \end{matrix}\bigg) $$

But the other problem got a weird thing going on, I'll show my work because I might just have an error.

$$ \begin{matrix} x_1 = a_xt_1^2/2+c_1t_1+c_2 \\ x_2 = a_xt_2^2/2+c_1t_2+c_2 \end{matrix} $$ Which can be written as, $$ \bigg(\begin{matrix}t_1 & 1 \\ t_2 & 1 \end{matrix}\bigg) \bigg(\begin{matrix}C_1 \\ C_2\end{matrix}\bigg) = \bigg(\begin{matrix} x_1-a_xt_1^2/2 \\ x_2-a_xt_2^2/2 \end{matrix}\bigg) $$

We left divide and the inverse of a 2x2 happens to be one over the determinant of the matrix times the original matrix with the negative of the off diagonals $$ \bigg(\begin{matrix}C_1 \\ C_2\end{matrix}\bigg) = \frac{1}{t_1-t_2}\bigg(\begin{matrix}t_1 & -1 \\ -t_2 & 1 \end{matrix}\bigg) \bigg(\begin{matrix} x_1-a_xt_1^2/2 \\ x_2-a_xt_2^2/2 \end{matrix}\bigg) $$

I now have a dispute with my units. $C_1$ now has units as follows, $$ \frac{1}{time}~(time~distance+distance) $$

What happened to the units? Where in matrix inversion (cofactoring happens once so the units should be the same from that, transposing the matrix doesn't seem like it should change units either, and the determinant units seemed fine) is there something I'm missing that would affect the units but not other coefficients?

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That's not how you take the inverse of a 2x2 matrix.

$$\pmatrix{t_1 & 1 \\ t_2 & 1}^{-1} = \frac{1}{t_1-t_2}\pmatrix{1& -t_2 \\ -1 & t_1}^T = \frac{1}{t_1-t_2} \pmatrix{1 & -1 \\ -t_2 & t_1} $$

The inverse of a 2x2 matrix is one over the determinant times the transpose of the cofactor matrix. Most generally,

$$ \pmatrix{a & b \\c & d}^{-1} = \frac{1}{ad-bc} \pmatrix{d & -c \\ -b & a}^T = \frac{1}{ad-bc} \pmatrix{d & -b \\ -c & a}$$

So you flip the top left and bottom right, multiply the top right and bottom left by -1, and divide by the determinant.

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  • $\begingroup$ I knew it was something ridiculous on my part. Thanks! $\endgroup$ – Orion Yeung Sep 2 '17 at 23:37

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