Find the? inverse, if it? exists, for the given matrix.[4 3][3 6]

Answer:Therefore, the inverse of given matrix is[tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]Step-by-step explanation:The inverse of a square matrix [tex]A[/tex] is [tex]A^{-1}[/tex] such that[tex]A A^{-1}=I[/tex] where I is the identity matrix.Consider, [tex]A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right][/tex][tex]\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}[/tex][tex]\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0[/tex][tex]\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}[/tex][tex]=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}[/tex][tex]\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc[/tex][tex]4\cdot \:6-3\cdot \:3=15[/tex][tex]=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}[/tex][tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]Therefore, the inverse of given matrix is[tex]=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}[/tex]