Question

Find

−1 using column transformations :

i) A = [

5 3

3 −2

]​

Answer 1

Given :    $A=\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right]$

To find : A⁻¹ using column transformations

Solution:

AI  = A

$\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right] \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] = \left[\begin{array}{ccc}5&3\\3&-2\end{array}\right]$

AA⁻¹  = I

C₂ → 5C₂ - 3C₁

$\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right] \left[\begin{array}{ccc}1&-3\\0&5\end{array}\right] = \left[\begin{array}{ccc}5&0\\3&-19\end{array}\right]$

C₁ → 19C₁ +3C₂

$\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right] \left[\begin{array}{ccc}10&-3\\15&5\end{array}\right] = \left[\begin{array}{ccc}95&0\\0&-19\end{array}\right]$

C₁ →  C₁/95

$\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right] \left[\begin{array}{ccc}\frac{2}{19} &-3\\ \frac{3}{19} &5\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&-19\end{array}\right]$

C₂ →  C₂/-19

$\left[\begin{array}{ccc}5&3\\3&-2\end{array}\right] \left[\begin{array}{ccc}\frac{2}{19} & \frac{3}{19}\\ \frac{3}{19} & -\frac{5}{19}\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

AA⁻¹  = I

$A^{-1}=\left[\begin{array}{ccc}\frac{2}{19} & \frac{3}{19}\\ \frac{3}{19} & -\frac{5}{19}\end{array}\right]$

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