Respuesta :
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]
