Using elementary row transformations, find the inverse of the matrix $A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}]$

Solution

$A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}] | A | = 1 (- 25 + 28) - 2 (- 10 + 14 + 3 (- 8 + 10) = 3 - 2 (4) + 3 (2) = 9 - 8 = 1 \neq 0 A^{- 1} exists A . A^{- 1} = I$

$[\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}] A^{- 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] R_{2} \Rightarrow R_{2} - 2 R_{1}; R_{3} \Rightarrow R_{3} + 2 R_{1} [\begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] A^{- 1} = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}] R_{1} \Rightarrow R_{1} - 2 R_{2} [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}] A^{- 1} = [\begin{matrix} 5 & - 2 & 0 \\ - 2 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}] R_{1} \Rightarrow R_{1} - R_{3}; R_{2} \Rightarrow R_{2} - R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A^{- 1} = [\begin{matrix} 3 & - 2 & - 1 \\ - 4 & 1 & - 1 \\ 2 & 0 & 1 \end{matrix}] I . A^{- 1} = A^{- 1} = [\begin{matrix} 3 & - 2 & - 1 \\ - 4 & 1 & - 1 \\ 2 & 0 & 1 \end{matrix}]$

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Matrices

Using elementary row transformations, find the inverse of the matrix $A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}]$

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If a matrix A has 12 elements, what arc the possible orders it can have 7 What if it has 7 elements ?

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Matrices

Using elementary row transformations, find the inverse of the matrix A = 123257-2-4-5

Some More Questions From Matrices Chapter

If a matrix A has 12 elements, what arc the possible orders it can have 7 What if it has 7 elements ?

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Using elementary row transformations, find the inverse of the matrix $A = [\begin{matrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ - 2 & - 4 & - 5 \end{matrix}]$