Define \(A=\begin{bmatrix}1 & 1\\3 & 0\end{bmatrix}\). Find a vertical vector V such that \((A^8+A^6+A^4+A^2+I)V=\begin{bmatrix}0\\11\end{bmatrix}\) (Where I is the identity matrix of order \(2\times2\))
\(\begin{bmatrix}0\\1/10\end{bmatrix}\)
The characteristic equation of \(A=\begin{bmatrix}1&1\\3&0\end{bmatrix}\) is \(\lambda^2-\lambda-3=0\), so by the Cayley-Hamilton theorem \(A^2=A+3I\).
Writing every power as \(A^n=a_nA+b_nI\) with \(a_{n+1}=a_n+b_n,\ b_{n+1}=3a_n\) and \(a_1=1,b_1=0\), computing successively gives (A²,A⁴,A⁶,A⁸) with coefficients \((a,b)\) equal to \((1,3),(7,12),(40,57),(217,291)\) respectively.
Summing \(A^8+A^6+A^4+A^2+I\): coefficient of A is \(217+40+7+1=265\); coefficient of I is \(291+57+12+3+1=364\). So \(S=265A+364I=\begin{bmatrix}629&265\\795&364\end{bmatrix}\).
Solving \(S\,V=\begin{bmatrix}0\\11\end{bmatrix}\) exactly by Cramer's rule (with \(\det S=18281\)) gives \(v_1=-\dfrac{2915}{18281}\), \(v_2=\dfrac{6919}{18281}\), which does not correspond exactly to any of the printed options (all of which require \(v_1=0\)).
Since \(S\) is not lower-triangular (its (1,2) entry, 265, is non-zero), no vector with a zero first entry can satisfy this system exactly, indicating an inconsistency between the given matrix/answer options in the source paper. Based on the closest-magnitude option to the exact second component, option (A) is selected, but this answer could not be fully verified against the printed choices.
Similar Questions
If \(A=\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}\) then value of \(\begin{bmatrix}1 & 0\\2 & 1\end{bmatrix}^{2007}\)
Important Questions from Operations on Matrices
The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
If \({\rm{E}}\left( {\rm{\theta }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta \;}}}&{\cos {\rm{\theta }}} \end{array}} \right]\) then E(α) E(β) is equal to
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
Find a matrix X such that 2A + B + X = 0 , where
\(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)
