Solveeit Logo
Login

Question

Mathematics Question on Matrices

Find the inverse of each of the matrices,if it exists.[45\34]\begin{bmatrix}4&5\\\3&4\end{bmatrix}

Answer

The correct answer is A−1=[4−5−34]A^{-1}=\begin{bmatrix}4&-5\\\\-3&4\end{bmatrix}
Let A=[45\34]A=\begin{bmatrix}4&5\\\3&4\end{bmatrix} We know that A=IAA = IA
so [45\34]=[10\01]\begin{bmatrix}4&5\\\3&4\end{bmatrix}=\begin{bmatrix}1&0\\\0&1\end{bmatrix}
  ⟹  [11\34]=[1−1\01]A(R1→R1−R2)\implies \begin{bmatrix}1&1\\\3&4\end{bmatrix}=\begin{bmatrix}1&-1\\\0&1\end{bmatrix}A (R_1\rightarrow R_1-R_2)
  ⟹  [11\01]=[1−1−34]A(R2→R2−3R1)\implies\begin{bmatrix}1&1\\\0&1\end{bmatrix}=\begin{bmatrix}1&-1\\\\-3&4\end{bmatrix} A (R_2\rightarrow R_2-3R_1)
  ⟹  [11\01]=[4−5\34]A(R1→R1−R2)\implies\begin{bmatrix}1&1\\\0&1\end{bmatrix}=\begin{bmatrix}4&-5\\\3&4\end{bmatrix} A (R_1\rightarrow R_1-R_2)
therefore A−1=[4−5−34]A^{-1}=\begin{bmatrix}4&-5\\\\-3&4\end{bmatrix}