MatrixSolutions

The general idea of a matrix inverse is that if A is a matrix, its inverse, Failed to parse (syntax error): {\displaystyle \mathbf^{-1}} is the matrix which, when multiplied by A yields the identity matrix, I. I, in turn, is the matrix which, when multiplied by any matrix B just gives bavk B.

For the problem posed in the quiz:

find $\left[{\begin{array}{cc}1&4\\1&2\end{array}}\right]^{-1}$

I think that the easiest solution is to solve as a set of simultaneous equations. If $\mathbf {X} =\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right]$ is the inverse,

$\left[{\begin{array}{cc}1&4\\1&2\end{array}}\right]*\mathbf {X} =\left[{\begin{array}{cc}1&0\\0&1\end{array}}\right]$

Thus, the rules of matrix multiplication give us: $a+4c=1$
$b+4d=0-$
$-a+2c=0$
$-b+2d=1$

which is very easy to solve by substitution: $b=-4d$
$a=2c$
$2c+4c=1$
$c=1/6$
etc,...