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Miscellaneous Exercise 2(A) · Q47

Q.If A=[x000y000z]A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} is a nonsingular matrix then find A−1A^{-1} by elementary row transformations. Hence, find the inverse of [20001000−1]\begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1: A=[x000y000z]A=\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix} is nonsingular, so ∣A∣=xyz≠0|A|=xyz\neq0, meaning x,y,zx,y,z are all individually nonzero.

Step 2: Start from AA−1=IAA^{-1}=I: [x000y000z]A−1=[100010001]\begin{bmatrix}x&0&0\\0&y&0\\0&0&z\end{bmatrix}A^{-1}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.

Step 3: Use R1→1xR1R_1\to\tfrac1xR_1, R2→1yR2R_2\to\tfrac1yR_2, R3→1zR3R_3\to\tfrac1zR_3 (each is a legal scalar-multiplication transformation, valid since x,y,z≠0x,y,z\neq0): the left side becomes I3I_3, and the right side becomes [1x0001y0001z]\begin{bmatrix}\tfrac1x&0&0\\0&\tfrac1y&0\\0&0&\tfrac1z\end{bmatrix}.

Step 4: So A−1=[1x0001y0001z]A^{-1}=\begin{bmatrix}\tfrac1x&0&0\\0&\tfrac1y&0\\0&0&\tfrac1z\end{bmatrix} -- confirming the general rule that the inverse of a diagonal matrix is diagonal, with each entry replaced by its reciprocal. …

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