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Question

Mathematics Question on Matrices

Find the inverse of each of the matrices, if it exists.[25\13]\begin{bmatrix}2&5\\\1&3\end{bmatrix}

Answer

The correct answer is A−1=[3−5−12]A^{-1}=\begin{bmatrix}3&-5\\\\-1&2\end{bmatrix}
Let A=[25\13]\begin{bmatrix}2&5\\\1&3\end{bmatrix} We know that A=IAA = IA
so [[25\13]=[10\01]A[\begin{bmatrix}2&5\\\1&3\end{bmatrix}=\begin{bmatrix}1&0\\\0&1\end{bmatrix}A
  ⟹  [152 13]=[120 01]A(R1→12R1)\implies\begin{bmatrix}1& \frac{5}{2}\\\ 1&3\end{bmatrix}=\begin{bmatrix}\frac{1}{2}& 0\\\ 0&1\end{bmatrix}A (R_1\rightarrow\frac{1}{2R_1})
  ⟹  [152 012]]=[120 −121]A(R2→R2−R1)\implies \begin{bmatrix}1& \frac{5}{2}\\\ 0& \frac{1}{2}\end{bmatrix}]=\begin{bmatrix}\frac{1}{2}& 0\\\ \frac{-1}{2}& 1\end{bmatrix}A (R_2\rightarrow R_2-R_1)
  ⟹  [10\012]=[3−5 −121]A(R1→R2−5R2)\implies\begin{bmatrix}1&0\\\0&\frac{1}{2}\end{bmatrix}=\begin{bmatrix}3&-5\\\ \frac{-1}{2}& 1\end{bmatrix}A (R_1\rightarrow R_2-5R_2)
  ⟹  [10\01]=[3−5−12]A(R2→2R2)\implies\begin{bmatrix}1&0\\\0&1\end{bmatrix}=\begin{bmatrix}3&-5\\\\-1&2\end{bmatrix}A (R_2\rightarrow2R_2)
therefore A−1=[3−5−12]A^{-1}=\begin{bmatrix}3&-5\\\\-1&2\end{bmatrix}