Assertion (A) : If $P=\begin{bmatrix}1 & 4 \\ 2 & 3\end{bmatrix}$, then $P^{2}-4P-5I=0$.
Reason (R) : Every square matrix satisfies its own characteristic equation.
In the light of the above statements, choose the most appropriate answer from the options given below :
Given the matrix $P = \begin{bmatrix}1 & 4 \\ 2 & 3\end{bmatrix}$, we need to verify the equation $P^{2}-4P-5I=0$.
Since $P^{2}-4P-5I$ results in the zero matrix, Assertion (A) is correct.
Reason (R) states that "Every square matrix satisfies its own characteristic equation." This statement is a fundamental theorem in linear algebra known as the Cayley-Hamilton Theorem. It holds true for all square matrices.
Therefore, Reason (R) is correct.
To understand the connection, let's find the characteristic equation for matrix $P$. The characteristic equation is given by $\det(P - \lambda I) = 0$.
For $P=\begin{bmatrix}1 & 4 \\ 2 & 3\end{bmatrix}$, $P - \lambda I = \begin{bmatrix}1-\lambda & 4 \\ 2 & 3-\lambda\end{bmatrix}$.
The determinant is: $\det(P - \lambda I) = (1-\lambda)(3-\lambda) - (4 \times 2)$
$= 3 - \lambda - 3\lambda + \lambda^2 - 8$
$= \lambda^2 - 4\lambda - 5$
The characteristic equation is $\lambda^2 - 4\lambda - 5 = 0$. The Cayley-Hamilton theorem (Reason R) asserts that if we substitute the matrix $P$ for $\lambda$, the equation $P^2 - 4P - 5I = 0$ must hold true. This is precisely what Assertion (A) states.
Thus, Reason (R) correctly explains Assertion (A).