5 5 votes Let \(N\) be the matrix\[N=\begin{bmatrix}1&2\\3&4\end{bmatrix}.\]Which of the following matrix equations does \(N\) satisfy?\(N^{2}-5N-2I=0\) \(N^{2}+5N-2I=0\) \(N^{2}-5N+2I=0\) \(N^{2}+5N+2I=0\) Linear Algebra goclasses goclasses-cs-dpp goclasses-cs-dpp-day-214 goclasses-da-dpp goclasses-da-dpp-day-116 linear-algebra goclasses-linear-algebra-practice-questions matrix + – GO Classes 567 views answer comment Share Follow Print See 1 comment 1 1 comment reply Raj Kushwaha commented Mar 15 reply Follow flag option A 0 0 replyShare Please log in or register to add a comment.
2 2 votes Let $N=\begin{bmatrix}1&2\\3&4\end{bmatrix}$.First compute trace and determinant:$\text{trace}(N)=1+4=5$$\det(N)=1\cdot 4-2\cdot 3=4-6=-2$For a $2\times 2$ matrix, the characteristic polynomial is$p(\lambda)=\lambda^2-(\text{trace}(N))\lambda+\det(N)$So,$p(\lambda)=\lambda^2-5\lambda-2$By the Cayley-Hamilton theorem, $p(N)=0$, hence$N^2-5N-2I=0$Therefore, the correct option is $\boxed{\text{A}}$ GO Classes answered Mar 9 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
Let $N=\begin{bmatrix}1&2\\3&4\end{bmatrix}$.First compute trace and determinant:$\text{trace}(N)=1+4=5$$\det(N)=1\cdot 4-2\cdot 3=4-6=-2$For a $2\times 2$ matrix, the characteristic polynomial is$p(\lambda)=\lambda^2-(\text{trace}(N))\lambda+\det(N)$So,$p(\lambda)=\lambda^2-5\lambda-2$By the Cayley-Hamilton theorem, $p(N)=0$, hence$N^2-5N-2I=0$Therefore, the correct option is $\boxed{\text{A}}$
0 0 votes just find characterstic polynomial of N and put lambda = N that's it Raj Kushwaha answered Mar 15 Raj Kushwaha comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Answer: A Meticulous_March answered Mar 19 Meticulous_March comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Option (A). N^2 - 5N - 2I = 0 SV27 answered Apr 6 SV27 comment Share Follow 0 reply Please log in or register to add a comment.