Problem 262 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0\end{array}\right]\), find \(A^{2}\) and show that \(A^{2}=A^{-1}\). Is \(A\) a periodic matrix? If yes, find its period.

Step-by-Step Solution

Answer

Squared A matrix: \left[\begin{array}{ccc}0 & 0 & 1 \ 2 & -2 & 2 \ 1 & -1 & 1\end{array}\right]\. The \(A\) matrix does not have an inverse because its determinant equals zero. Hence, \(A^2\) does not equal \(A^{-1}\). Therefore \(A\) is not a periodic matrix and its period is undefined.

Step 1: Calculate A^2

To calculate \(A^2\), multiply matrix \(A\) by itself: \(A^2 = A*A = \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0\end{array}\right] * \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 1 \ 2 & -2 & 2 \ 1 & -1 & 1\end{array}\right]\)

Step 2: Calculate A^-1

To calculate \(A^{-1}\), use the formula \(A^{-1} = \frac{1}{|A|}*adj(A)\), where \(|A|\) is the determinant of \(A\) and \(adj(A)\) is the adjoint of \(A\). In this case, \(|A|= 1* (-1-(0*2)) - (-1*1) + 1*(0) = -1 + 1 = 0\). Because the \(|A|\) equals zero, \(A\) does not have an inverse. Therefore, \(A^2\) does not equal \(A^{-1}\).

Step 3: Determine if A is a periodic matrix

A matrix is periodic iff there exists a positive integer k such that A^k equals the identity matrix. In this case, since \(A^2\) does not result in an identity matrix and the matrix doesn't have an inverse, it is not a periodic matrix and its period is undefined.