Answer to Question #190367 in Linear Algebra for Regomoditswe Dibob

Question

(6.1) Find det(C) if (1) C =  λ λ + 1 λ λ − 1  (6.2) Use the cofactor expansion to determine 2 0 0 0 3 1 2 0 2 −5 0 4 1 3 0 3 (6.3) Consider the matrix A =  1 4 2 3  (a) Compute A −1 (b) Find det(A **−1 ) (c) Deduce a relation (if it exists) between det(A) and det(A **−1 

Answer 1

6.1. 

$C=\begin{bmatrix} \lambda& \lambda+1\\ \lambda& \lambda-1 \end{bmatrix}$

$Det(C)=\begin{vmatrix} \lambda& \lambda+1\\ \lambda& \lambda-1 \end{vmatrix}$

    $=\lambda(\lambda-1)-\lambda(\lambda+1) =-2\lambda$

6.2

        

The given determinant is $\begin{vmatrix} 2&0&0&0\\ 3&1&2&0\\ 2&-5&0&4\\ 1&3&0&3 \end{vmatrix}$

Use the cofactor expansion corresponding to the first row.

$=2 \begin{vmatrix} 1&2&0\\ -5&0&4\\ 3&0&3 \end{vmatrix} -0+0-0$

$=2 \begin{vmatrix} 1&2&0\\ -5&0&4\\ 3&0&3 \end{vmatrix} =2[1(0-0)-2(-15-12)+0] =2[54]=108$

6.3

$(a) \text{The given matrix is- } A=\begin{bmatrix} 1&4\\ 2&3\end{bmatrix}$

$det A=\begin{vmatrix} 1&4\\ 2&3\end{vmatrix}=3-8=-5$

$det A=-5 \neq 0.$

$\Rightarrow A^{-1}$ exist.

Now, $adj.(A)=\begin{bmatrix} 3&-2\\ 2&3\end{bmatrix}^T \Rightarrow adj(A)=\begin{bmatrix} 3&-4\\ -2&1\end{bmatrix}$

So, $A^{-1} =\dfrac{adj A}{det (A)}=\dfrac{1}{-5}\begin{bmatrix} 3&-4\\ -2&1\end{bmatrix}=\begin{bmatrix} \dfrac{-3}{5} & \dfrac{4}{5}\\\\ \dfrac{2}{5} & \dfrac{-1}{5}\end{bmatrix}$

(b) 

$det(A^{-1})=\begin{vmatrix} \dfrac{-3}{5} & \dfrac{4}{5}\\\\ \dfrac{2}{5} & \dfrac{-1}{5}\end{vmatrix}$

    $=\dfrac{3}{25}-\dfrac{8}{25}=\dfrac{3-8}{25}=\dfrac{-5}{25}=\dfrac{-1}{5}$

(c) We have, $det(A)=-5 \text{ and }det(A^{-1})=\dfrac{-1}{5}$

  $det(A).det(A^{-1})=(-5).(-\dfrac{1}{5})=1$

  Hence $det(A).det(A^{-1})=1$

This is the required relation.