Question

[tex] If A= \left[\begin{array}{ccc}2&3\\5&-2\end{array}\right], Prove that A⁻¹=\frac{1}{19}A. [\tex]

Answer 1

Step-by-step explanation:

We know that

${A}^{ - 1} = \frac{ {A}^{T} }{ |A| }$

Given

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

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

$|A| =2 \times ( - 2) - 5 \times 3 \\ \\ |A| = - 4 - 15 = - 19$

${A}^{ - 1} = \frac{ {A}^{T} }{ |A| } \\ \\ \\ = \frac{ - 1}{19} {A}^{T}$

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

Hope it helps you