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Question

Mathematics Question on Determinants

If A=[23 5−2]A = \begin{bmatrix}2&3\\\ 5&-2\end{bmatrix} be such that A−1=kA,A^{-1} = kA, then kk is equal to

A

1919

B

119\frac{1}{19}

C

−19-19

D

−119- \frac {1}{19}

Answer

119\frac{1}{19}

Explanation

Solution

Given, A=[23 5−2]A=\left[\begin{matrix}2&3\\\ 5&-2\end{matrix}\right]
∴A−1=1−4−15[−2−3 −52]\therefore A^{-1}=\frac{1}{-4-15}\left[\begin{matrix}-2&-3\\\ -5&2\end{matrix}\right]
=−119[−2−3 −52]=119[23 5−2]=\frac{-1}{19}\left[\begin{matrix}-2&-3\\\ -5&2\end{matrix}\right]=\frac{1}{19}\left[\begin{matrix}2&3\\\ 5&-2\end{matrix}\right]
[multiplying −1-1 each element of a matrix]
=119A=\frac{1}{19}A
Given, A−1=kAA^{-1}=kA
∴k=119\therefore k=\frac{1}{19}