Let A be a 3 times 3 real matrix such that Ab | vedclass

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(A) The given equations show that the vectors $v_1 = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}$,$v_2 = \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}$,and

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unique solution

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(A) The given equations show that the vectors $v_1 = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}$,$v_2 = \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}$,and $v_3 = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}$ are eigenvectors of $A$ with corresponding eigenvalues $\lambda_1 = 2$,$\lambda_2 = 4$,and $\lambda_3 = 2$.Since these three vectors are linearly independent,they form a basis for $\mathbb{R}^3$.The matrix $A$ can be diagonalized as $A = PDP^{-1}$,where $D = 	ext{diag}(2, 4, 2)$ and $P$ is the matrix with columns $v_1, v_2, v_3$.The system is $(A-3I)X = B$. The matrix $(A-3I)$ has eigenvalues $\lambda_i - 3$,which are $2-3 = -1$,$4-3 = 1$,and $2-3 = -1$.Since none of the eigenvalues of $(A-3I)$ are $0$,the determinant $|A-3I| = (-1)(1)(-1) = 1 
eq 0$.Because the determinant is non-zero,the matrix $(A-3I)$ is invertible.Therefore,the system $(A-3I)X = B$ has a unique solution.

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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