b) x/[} 5&-1 0&2/3 ]=2[3 0]

To solve the equation $x/[\begin{matrix} 5&-1\\ 0&2/3\end{matrix} ]=2[3\quad 0]$, we need to multiply both sides of the equation by the inverse of the matrix on the left side.<br /><br />The inverse of a matrix $A$ is denoted as $A^{-1}$ and satisfies the equation $AA^{-1} = I$, where $I$ is the identity matrix.<br /><br />In this case, the matrix on the left side is $[\begin{matrix} 5&-1\\ 0&2/3\end{matrix} ]$. To find its inverse, we can use the formula for the inverse of a 2x2 matrix:<br /><br />$A^{-1} = \frac{1}{ad-bc}[\begin{matrix} d&-b\\ -c&a\end{matrix} ]$<br /><br />where $A = [\begin{matrix} a&b\\ c&d\end{matrix} ]$.<br /><br />Applying this formula, we get:<br /><br />$A^{-1} = \frac{1}{(5)(2/3)-(-1)(0)}[\begin{matrix} 2/3&1\\ 0&5\end{matrix} ] = \frac{1}{10/3}[\begin{matrix} 2/3&1\\ 0&5\end{matrix} ] = [\begin{matrix} 1/5&3/10\\ 0&3/2\end{matrix} ]$<br /><br />Now, we can multiply both sides of the equation by $A^{-1}$:<br /><br />$x = 2[3\quad 0] \cdot [\begin{matrix} 1/5&3/10\\ 0&3/2\end{matrix} ]$<br /><br />Performing the matrix multiplication, we get:<br /><br />$x = [\begin{matrix} 6/5&3/5\\ 0&3\end{matrix} ]$<br /><br />Therefore, the solution to the equation is $x = [\begin{matrix} 6/5&3/5\\ 0&3\end{matrix} ]$.