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3 votes
Find the? inverse, if it? exists, for the given matrix.

[4 3]

[3 6]

1 Answer

5 votes

Answer:

Therefore, the inverse of given matrix is


=\begin{pmatrix}(2)/(5)&-(1)/(5)\\ -(1)/(5)&(4)/(15)\end{pmatrix}

Explanation:

The inverse of a square matrix
A is
A^(-1) such that


A A^(-1)=I where I is the identity matrix.

Consider,
A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]


\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}


\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\\e 0


\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}


=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}


\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc


4\cdot \:6-3\cdot \:3=15


=(1)/(15)\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}


=\begin{pmatrix}(2)/(5)&-(1)/(5)\\ -(1)/(5)&(4)/(15)\end{pmatrix}

Therefore, the inverse of given matrix is


=\begin{pmatrix}(2)/(5)&-(1)/(5)\\ -(1)/(5)&(4)/(15)\end{pmatrix}

User Lqc
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