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Chapter 10: Problem 262

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.

Short Answer

Expert verified
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 by step solution

01

Calculate A^2

To calculate \(A^2\), multiply matrix \(A\) by itself: \(A^2 = A*A = \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0\end{array}\right] * \left[\begin{array}{ccc}1 & -1 & 1 \ 2 & -1 & 0 \ 1 & 0 & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 1 \ 2 & -2 & 2 \ 1 & -1 & 1\end{array}\right]\)
02

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 \(A^{-1}\).
03

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.

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Most popular questions from this chapter

PROVING IDENTITIES BY DETERMINANTS. $$ \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\ (x-1)^{2} & x^{2} & (x+1)^{2} \\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right|=-8 $$

For a fixed positive integer \(n\), if \(D=\mid \begin{array}{ccc}n ! & (n+1) ! & (n+2) ! \\ (n+1) ! & (n+2) ! & (n+3) ! \\ (n+2) ! & (n+3) ! & (n+4) !\end{array}\) then show that \(\left[\frac{D}{(n !)^{3}}-4\right]\) is divisible by \(n\).

EQUATIONS CONTAINING DETERMINANTS. $$ \left|\begin{array}{ccc} x+2 & 2 x+3 & 3 x+4 \\ 2 x+3 & 3 x+4 & 4 x+5 \\ 3 x+5 & 5 x+8 & 10 x+17 \end{array}\right|=0 $$

EVALUATING DETERMINANTS. $$ \left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right| $$

If \(a^{-1}+b^{-1}+c^{-1}=0\), prove that \(\left|\begin{array}{ccc}1+a & 1 & 1 \\\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\).
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