Let matrix A = begin{bmatrix} 5 & -3 & 0 -3 | vedclass

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(B) Given the equation $A^2 X = cAX$,we can rewrite it as $(A^2 - cA)X = 0$,which implies $A(A - cI)X = 0$. Since $X$ is a non-zero matrix,this implie

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$2$

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(B) Given the equation $A^2 X = cAX$,we can rewrite it as $(A^2 - cA)X = 0$,which implies $A(A - cI)X = 0$. Since $X$ is a non-zero matrix,this implies that $c$ must be an eigenvalue of matrix $A$ or $A$ must be singular. More directly,the equation $AX = \lambda X$ defines the eigenvalues of $A$. If we let $AX = \lambda X$,then $A^2 X = A(AX) = A(\lambda X) = \lambda(AX) = \lambda^2 X$. Substituting this into the given equation $A^2 X = cAX$,we get $\lambda^2 X = c \lambda X$,which implies $\lambda^2 = c \lambda$ for any non-zero $X$. Thus,$\lambda(\lambda - c) = 0$. This means $c$ must be an eigenvalue of $A$. To find the eigenvalues of $A$,we solve the characteristic equation $|A - \lambda I| = 0$: $\begin{vmatrix} 5-\lambda & -3 & 0 \ -3 & 5-\lambda & 0 \ 0 & 0 & 2-\lambda \end{vmatrix} = 0$ $(2-\lambda) [(5-\lambda)^2 - (-3)^2] = 0$ $(2-\lambda) [\lambda^2 - 10\lambda + 25 - 9] = 0$ $(2-\lambda) (\lambda^2 - 10\lambda + 16) = 0$ $(2-\lambda) (\lambda - 8)(\lambda - 2) = 0$ The eigenvalues are $\lambda = 2, 2, 8$. Since $c$ is an eigenvalue of $A$,the distinct values for $c$ are $2$ and $8$. Therefore,there are $2$ distinct values of $c$.

Let matrix $A = \begin{bmatrix} 5 & -3 & 0 \\ -3 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix}$,$X$ be a non-zero matrix of order $3 \times 1$,and $c$ be a real number. If $A^2 X = cAX$,then the number of distinct values of $c$ is:

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