## Related questions with answers

Consider the system of equations AX = 0 where

$A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$

is a $2 \times 2$ matrix over the field F. Prove the following. (a) If every entry of A is 0, then every pair $\left(x_{1}, x_{2}\right)$ is a solution of AX = 0. (b) If $a d-b c \neq 0$, the system AX = 0 has only the trivial solution $x_{1}=x_{2}=0$. (c) If ad - bc = 0 and some entry of A is different from 0, then there is a solution $\left(x_{1}^{0}, x_{2}^{0}\right)$ such that $\left(x_{1}, x_{2}\right)$ is a solution if and only if there is a scalar y such that $x_{1}=y x_{1}^{0}, x_{2}=y x_{2}^{0}$.

Solution

Verified$\textbf{Given:}$

$A=\begin{bmatrix} a & b\\c & d \end{bmatrix}$

and the system $A\mathbf{x}=0$, where $\mathbf{x}=(x_1,x_2)$.

$\textbf{(a) To prove:}$ If every entry of $A$ is zero, then every pair $(x_1,x_2)$ is a solution of $A \mathbf{x}=0$.

$\textbf{Proof:}$ If every entry is zero, then

$A\mathbf{x}=0 \implies \begin{bmatrix} 0 & 0\\0 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix} \implies 0x_1+0x_2=0$

which holds true for all pairs $(x_1,x_2)$ and hence, every pair $(x_1,x_2)$ is a solution of $A\mathbf{x}=0$.

$\textbf{(b) To prove:}$ If $ad-bc \neq 0$, then the system $A\mathbf{x}=0$ has only trivial solution $x_1=x_2=0$.

$\textbf{Proof:}$ Since $ad-bc \neq 0$, therefore $|A| \neq 0$ and so $A^{-1}$ exists.

Now,

$A\mathbf{x}=0 \implies A^{-1}A\mathbf{x}=A^{-1}0=0 \implies \mathbf{x}=0$

So, solution is trivial, viz., $x_1=x_2=0$.

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