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Question 108 of 121

Q.Find the inverse of the matrix, A=[12−2−1300−21]A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} using elementary row transformations.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 4mImportance★★★★★
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Augment AA with the identity matrix and apply row operations until the left block becomes II; the right block is then A−1A^{-1}.

Write A=[12−2−1300−21]A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} and augment with II:

[12−2100−1300100−21001]\left[\begin{array}{ccc|ccc} 1 & 2 & -2 & 1 & 0 & 0 \\ -1 & 3 & 0 & 0 & 1 & 0 \\ 0 & -2 & 1 & 0 & 0 & 1 \end{array}\right]

Step 1: R2→R2+R1R_2 \to R_2 + R_1:

[12−210005−21100−21001]\left[\begin{array}{ccc|ccc} 1 & 2 & -2 & 1 & 0 & 0 \\ 0 & 5 & -2 & 1 & 1 & 0 \\ 0 & -2 & 1 & 0 & 0 & 1 \end{array}\right]

Step 2: R3→5R3+2R2R_3 \to 5R_3 + 2R_2:

[12−210005−2110001225]\left[\begin{array}{ccc|ccc} 1 & 2 & -2 & 1 & 0 & 0 \\ 0 & 5 & -2 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 2 & 5 \end{array}\right]

Step 3: R2→R2+2R3R_2 \to R_2 + 2R_3 and R1→R1+2R3R_1 \to R_1 + 2R_3:

[12054100505510001225]\left[\begin{array}{ccc|ccc} 1 & 2 & 0 & 5 & 4 & 10 \\ 0 & 5 & 0 & 5 & 5 & 10 \\ 0 & 0 & 1 & 2 & 2 & 5 \end{array}\right]

Step 4: R2→15R2R_2 \to \dfrac{1}{5}R_2: …

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