Solveeit Logo
Login

Question

Mathematics Question on Determinants

Compute (AB)−1\left(AB\right)^{-1}, if A=[112 02−3 3−24]A = \left[\begin{matrix}1&1&2\\\ 0&2&-3\\\ 3&-2&4\end{matrix}\right] and B−1=[120 03−1 102]B^{-1}=\left[\begin{matrix}1&2&0\\\ 0&3&-1\\\ 1&0&2\end{matrix}\right]

A

119[16121 2111−7 10−23]\frac{1}{19}\left[\begin{matrix}16&12&1\\\ 21&11&-7\\\ 10&-2&3\end{matrix}\right]

B

119[161210 2111−2 1−73]\frac{1}{19}\left[\begin{matrix}16&12&10\\\ 21&11&-2\\\ 1&-7&3\end{matrix}\right]

C

119[16121 −21−117 10−23]\frac{1}{19}\left[\begin{matrix}16&12&1\\\ -21&-11&7\\\ 10&-2&3\end{matrix}\right]

D

119[16−211 21117 10−23]\frac{1}{19}\left[\begin{matrix}16&-21&1\\\ 21&11&7\\\ 10&-2&3\end{matrix}\right]

Answer

119[16121 2111−7 10−23]\frac{1}{19}\left[\begin{matrix}16&12&1\\\ 21&11&-7\\\ 10&-2&3\end{matrix}\right]

Explanation

Solution

∣A∣=∣112 02−3 3−24∣=1(8−6)−0+3(−3−4)=19≠0\left|A\right|=\left|\begin{matrix}1&1&2\\\ 0&2&-3\\\ 3&-2&4\end{matrix}\right|=1\left(8-6\right)-0+3\left(-3-4\right)=19\ne0(Expanding along C1)C_{1}) ∴A−1\therefore\quad A^{-1} exists. Now, adj(A) =[2−9−6 −8−25 −732]T=[2−8−7 −9−23 −652]=\left[\begin{matrix}2&-9&-6\\\ -8&-2&5\\\ -7&3&2\end{matrix}\right]^{T}=\left[\begin{matrix}2&-8&-7\\\ -9&-2&3\\\ -6&5&2\end{matrix}\right] ∴A−1=1∣A∣=adj (A)=−119[2−8−7 −9−23 −652]\therefore\quad A^{-1}=\frac{1}{\left|A\right|}=adj\,\left(A\right)=\frac{-1}{19} \left[\begin{matrix}2&-8&-7\\\ -9&-2&3\\\ -6&5&2\end{matrix}\right] So, (AB)−1=B−1A−1=[120 03−1 102]⋅−119[2−8−7 −9−23 −652]\left(AB\right)^{-1}=B^{-1}A^{-1}=\left[\begin{matrix}1&2&0\\\ 0&3&-1\\\ 1&0&2\end{matrix}\right]\cdot\frac{-1}{19}\left[\begin{matrix}2&-8&-7\\\ -9&-2&3\\\ -6&5&2\end{matrix}\right] =−119[2−18−0−8−4+0−7+6+0 0−27+60−6−50+9−2 2−0−12−8−0+10−7+10+4]=\frac{-1}{19} \left[\begin{matrix}2-18-0&-8-4+0&-7+6+0\\\ 0-27+6&0-6-5&0+9-2\\\ 2-0-12&-8-0+10&-7+10+4\end{matrix}\right] =−119[−16−12−1 −21−117 −102−3]=119[16121 2111−7 10−23]=\frac{-1}{19}\left[\begin{matrix}-16&-12&-1\\\ -21&-11&7\\\ -10&2&-3\end{matrix}\right]=\frac{1}{19} \left[\begin{matrix}16&12&1\\\ 21&11&-7\\\ 10&-2&3\end{matrix}\right]