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sdas [7]
3 years ago
7

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

Mathematics
1 answer:
True [87]3 years ago
4 0

Answer:

Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

Step-by-step 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}\ne 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

=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

Therefore, the inverse of given matrix is

=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}

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