Solveeit Logo
Login

Question

Mathematics Question on Determinants

If A=[3−34 2−34 0−11]A = \begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} , then A−1A^{-1} equal to:

A

AA

B

A2A^2

C

A3A^3

D

A4A^4

Answer

A3A^3

Explanation

Solution

Given: A=[3−34 2−34 0−11]A = \begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} ⇒ ∣A∣=∣3−34 2−34 0−11∣\Rightarrow \ |A| = \begin{vmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{vmatrix} ⇒ ∣A∣=3(−3+4)+3(2−0)+4(−2)\Rightarrow \ |A| =3 (- 3 + 4) + 3(2 - 0) + 4(- 2) =3+6−8=1= 3 + 6- 8 = 1 ⇒ ∣A∣=1≠0\Rightarrow \ |A| = 1 \neq 0 Thus A is non singular matrix and therefore it is invertible. Let cijc_{ij} be cofactor of aija_{ij} in A. Then c11=∣−34 −11∣=−3+4=1c_{11} = \begin{vmatrix}-3&4\\\ -1&1\end{vmatrix} = -3 + 4 = 1 c12=−∣24 01∣=−2,c_{12} = - \begin{vmatrix}2&4\\\ 0&1\end{vmatrix}= - 2, c13=∣2−3 0−1∣=−2c_{13} = \begin{vmatrix}2&-3\\\ 0&-1\end{vmatrix} = -2 c21=−∣−34 −11∣=−(−3+4)=−1 \text{c}_{21} = - \begin{vmatrix}-3&4\\\ -1&1\end{vmatrix}= - \left(-3+4\right) = - 1 c22=∣34 01∣=3 c_{22} = \begin{vmatrix}3&4\\\ 0&1\end{vmatrix} = 3 c23=−∣3−3 0−1∣=−(−3)=3 c_{23} = - \begin{vmatrix}3&-3\\\ 0&-1\end{vmatrix} = -\left(-3\right) = 3 c31=∣−34 −34∣=0 c_{31} = \begin{vmatrix}-3&4\\\ -3&4\end{vmatrix} = 0 c32=∣34 24∣=−(12−8)=−4,c_{32} = \begin{vmatrix}3&4\\\ 2&4\end{vmatrix} = -\left(12-8\right)= -4, c33=∣3−3 2−3∣=9+6=−3c_{33} = \begin{vmatrix}3&-3\\\ 2&-3\end{vmatrix} = 9 + 6 = - 3 ∴ adjA=[1−2−2 −133 0−4−3]T=[1−10 −23−4 −23−3]\therefore \ adj A = \begin{bmatrix}1&-2&-2\\\ -1&3&3\\\ 0&-4&-3\end{bmatrix} ^{T} = \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix} ⇒A−1=1∣A∣adjA \Rightarrow A^{-1} = \frac{1}{\left|A\right|} adj A =[1−10 −23−4 −23−3]....(1)= \begin{bmatrix}1&-1&0\\\ - 2&3&-4\\\ -2&3&-3\end{bmatrix} ....\left(1\right) Now, A2=[3−34 2−34 0−11][3−34 2−34 0−11]A^{2} = \begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} \begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} [∵A2=A.A]\left[\because A^{2} = A.A\right] =[9−6−9+9−412−12+4 6−6−6+9−48−12+4 −23−1−4+1]=[3−44 0−10 −22−3]= \begin{bmatrix}9-6&-9+9-4&12-12+4\\\ 6-6&-6+9-4&8-12+4\\\ -2&3-1&-4+1\end{bmatrix} = \begin{bmatrix}3&-4&4\\\ 0&-1&0\\\ -2&2&-3\end{bmatrix} and A3=A2.AA^{3} = A^{2} .A =[3−44 0−10 −22−3][3−34 2−34 0−11]= \begin{bmatrix}3&-4&4\\\ 0&-1&0\\\ -2&2&-3\end{bmatrix} \begin{bmatrix}3&-3&4\\\ 2&-3&4\\\ 0&-1&1\end{bmatrix} ⇒A3=[9−8−9+12−412−16+4 −23−4 −6+46−6+3−8+8−3]\Rightarrow A^{3} = \begin{bmatrix}9-8&-9+12-4&12-16+4\\\ -2&3&-4\\\ -6+4&6-6+3&-8+8-3\end{bmatrix} =[1−10 −23−4 −23−3]= \begin{bmatrix}1&-1&0\\\ -2&3&-4\\\ -2&3&-3\end{bmatrix} ....(2) from equation (1) and (2), we have A−1=A3A^{-1} = A^3