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

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(B) Let $A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix}$.From $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} =

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infinitely many solutions

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(B) Let $A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix}$.From $A \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} c_1 \ c_2 \ c_3 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix}$,we get $c_1 = 1, c_2 = 1, c_3 = 2$.From $A \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} a_1 + c_1 \ a_2 + c_2 \ a_3 + c_3 \end{bmatrix} = \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}$,we get $a_1 = -2, a_2 = -1, a_3 = -1$.From $A \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} a_1 + b_1 \ a_2 + b_2 \ a_3 + b_3 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix}$,we get $b_1 = 3, b_2 = 2, b_3 = 1$.Thus,$A = \begin{bmatrix} -2 & 3 & 1 \ -1 & 2 & 1 \ -1 & 1 & 2 \end{bmatrix}$.Then $A - 2I = \begin{bmatrix} -4 & 3 & 1 \ -1 & 0 & 1 \ -1 & 1 & 0 \end{bmatrix}$.The determinant $|A - 2I| = -4(0 - 1) - 3(0 - (-1)) + 1(-1 - 0) = 4 - 3 - 1 = 0$.The system is $\begin{bmatrix} -4 & 3 & 1 \ -1 & 0 & 1 \ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 4 \ 1 \ 1 \end{bmatrix}$.This gives the equations:$1) -4x_1 + 3x_2 + x_3 = 4$$2) -x_1 + x_3 = 1 \Rightarrow x_3 = 1 + x_1$$3) -x_1 + x_2 = 1 \Rightarrow x_2 = 1 + x_1$Substituting $(2)$ and $(3)$ into $(1)$: $-4x_1 + 3(1 + x_1) + (1 + x_1) = -4x_1 + 3 + 3x_1 + 1 + x_1 = 4$. This simplifies to $4 = 4$,which is always true.Since the system is consistent and the determinant is $0$,there are infinitely many solutions.

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