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Exercise 2.2 · Q21

Q.Find the inverse of the following matrix. [122−1]\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1: A=[122−1]A=\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}, ∣A∣=1(−1)−2(2)=−5≠0|A|=1(-1)-2(2)=-5\neq0, so A−1A^{-1} exists.

Step 2: Start AA−1=IAA^{-1}=I: [122−1]A−1=[1001]\begin{bmatrix}1&2\\2&-1\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.

Step 3: R2→R2−2R1R_2\to R_2-2R_1: [120−5]A−1=[10−21]\begin{bmatrix}1&2\\0&-5\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\-2&1\end{bmatrix}.

Step 4: R2→−15R2R_2\to-\tfrac15R_2: [1201]A−1=[1025−15]\begin{bmatrix}1&2\\0&1\end{bmatrix}A^{-1}=\begin{bmatrix}1&0\\\tfrac25&-\tfrac15\end{bmatrix}. …

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