Consider the system of equations AX = 0 where
\[ A = \left[ \begin{array}{c c} a & b \\ c & d \end{array} \right] \]
is a \(2 \times 2\) matrix over the field \(F\) . Prove the following.
If every entry of \(A\) is 0, then every pair \((x_1, x_2)\) is a solution of \(AX = 0\) .
If \(ad - bc \neq 0\), the system \(AX = 0\) has only the trivial solution \(x_{1} = x_{2} = 0\) .
If \(ad - bc = 0\) and some entry of \(A\) is different from 0, then there is a solution \((x_1^0, x_2^0)\) such that \((x_1, x_2)\) is a solution if and only if there is a scalar \(y\) such that \(x_1 = yx_1^0, x_2 = yx_2^0\) .
Solution
Part (a).
1. If \(A = 0\), then \(AX = 0\) for every \(X = (x_1, x_2)\).
the zero matrix sends every vector to \(0\).
Part (b).
1. If \(ad - bc \neq 0\), then \(A\) is invertible.
\(\det A = ad - bc \neq 0\).
2. Hence \(AX = 0\) implies \(X = A^{-1} \cdot 0 = 0\).
multiply both sides by \(A^{-1}\).
3. Therefore the only solution is the trivial one.
2.
Part (c).
1. If \(ad - bc = 0\) and \(A \neq 0\), then \(\operatorname{rank}(A) = 1\).
\(A\) is nonzero but singular, so its rank is \(1\) (a nonzero \(2 \times 2\) matrix of determinant \(0\) has rank \(1\)).
2. Hence the solution space \(\{X : AX = 0\}\) is \(1\)-dimensional.
by rank–nullity, \(\dim \ker A = 2 - \operatorname{rank}(A) = 1\).
3. Let \((x_1^0, x_2^0)\) be any nonzero solution.
such a solution exists since \(\ker A \neq 0\).
4. Then \((x_1, x_2)\) is a solution iff \((x_1, x_2) = y(x_1^0, x_2^0)\) for some scalar \(y\).
the solution space is \(1\)-dimensional and spanned by \((x_1^0, x_2^0)\).
5. Q.E.D.
1 (a), 3 (b), and 4 (c).