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Exercise 2.2 · Q24

Q.Find the inverse of the following matrix. [20−1510013]\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1: A=[20−1510013]A=\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}. Expanding along row 1: ∣A∣=2(1⋅3−0⋅1)−0+(−1)(5⋅1−1⋅0)=2(3)−1(5)=6−5=1≠0|A|=2(1\cdot3-0\cdot1)-0+(-1)(5\cdot1-1\cdot0)=2(3)-1(5)=6-5=1\neq0.

Step 2: Using row transformations on [A∣I]\begin{bmatrix}A\mid I\end{bmatrix} -- clear column 1 below the (1,1) pivot with R2→R2−52R1R_2\to R_2-\tfrac52R_1, then continue clearing column 2 and column 3 in turn, finally scaling every row so the diagonal is 11 -- the right-hand block converges to [3−11−156−55−22]\begin{bmatrix}3&-1&1\\-15&6&-5\\5&-2&2\end{bmatrix}. …

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