Let
\[ A = \left[ \begin{array}{c c} a & b \\ c & d \end{array} \right]. \]
Prove, using elementary row operations, that \(A\) is invertible if and only if \((ad - bc) \neq 0\) .
Solution
1. Criterion for invertibility via elementary row operations:
1.1. A square matrix \(A \in M_2(F)\) is invertible if and only if it is row-equivalent to the identity matrix \(I_2\) (i.e. its reduced row echelon form is \(I_2\)).
Hoffman–Kunze Theorem 1.13 on invertible matrices and row equivalence.
2. Case 1: \(a \neq 0\):
2.1. Multiply row 1 by \(\frac{1}{a}\):\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \xrightarrow{R_1 \leftarrow \frac{1}{a} R_1} \begin{pmatrix} 1 & b/a \\ c & d \end{pmatrix}. \]
elementary row operation (scaling by non-zero scalar).
2.2. Add \(-c\) times row 1 to row 2:\[ \begin{pmatrix} 1 & b/a \\ c & d \end{pmatrix} \xrightarrow{R_2 \leftarrow R_2 - c R_1} \begin{pmatrix} 1 & b/a \\ 0 & d - c(b/a) \end{pmatrix} = \begin{pmatrix} 1 & b/a \\ 0 & \frac{ad - bc}{a} \end{pmatrix}. \]
elementary row operation (row addition).
2.3. If \(ad - bc = 0\), the matrix has a row of zeros \(\begin{pmatrix} 1 & b/a \\ 0 & 0 \end{pmatrix}\), so its rank is 1 and it cannot be row-reduced to \(I_2\).Thus \(A\) is not invertible.
a matrix with a zero row in row echelon form cannot be row-equivalent to \(I_2\).
2.4. If \(ad - bc \neq 0\), the entry \(\delta = \frac{ad - bc}{a} \neq 0\).Scale row 2 by \(\frac{1}{\delta}\):\[ \begin{pmatrix} 1 & b/a \\ 0 & \delta \end{pmatrix} \xrightarrow{R_2 \leftarrow \frac{1}{\delta} R_2} \begin{pmatrix} 1 & b/a \\ 0 & 1 \end{pmatrix} \xrightarrow{R_1 \leftarrow R_1 - \frac{b}{a} R_2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2. \]Thus \(A\) is row-equivalent to \(I_2\), so \(A\) is invertible.
sequence of elementary row operations producing \(I_2\).
3. Case 2: \(a = 0\):
3.1. If \(a = 0\), then \(ad - bc = -bc\).
\(0 \cdot d - bc = -bc\).
3.2. If \(c = 0\), the first column of \(A = \begin{pmatrix} 0 & b \\ 0 & d \end{pmatrix}\) is zero, so \(A\) has non-trivial nullspace and rank \(\le 1\), meaning \(A\) is not invertible.Note that \(ad - bc = -0 \cdot b = 0\).
a matrix with a zero column cannot be row-reduced to \(I_2\).
3.3. If \(c \neq 0\), swap rows 1 and 2:\[ \begin{pmatrix} 0 & b \\ c & d \end{pmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{pmatrix} c & d \\ 0 & b \end{pmatrix} \xrightarrow{R_1 \leftarrow \frac{1}{c} R_1} \begin{pmatrix} 1 & d/c \\ 0 & b \end{pmatrix}. \]
elementary row operations.
3.4. If \(b = 0\), the second row is zero, so \(A\) is not invertible, and \(ad - bc = -0 \cdot c = 0\).
rank 1 matrix cannot be row-reduced to \(I_2\).
3.5. If \(b \neq 0\), then \(ad - bc = -bc \neq 0\).Scale \(R_2\) by \(1/b\) and clear the \((1,2)\) entry:\[ \begin{pmatrix} 1 & d/c \\ 0 & b \end{pmatrix} \xrightarrow{R_2 \leftarrow \frac{1}{b} R_2} \begin{pmatrix} 1 & d/c \\ 0 & 1 \end{pmatrix} \xrightarrow{R_1 \leftarrow R_1 - \frac{d}{c} R_2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2. \]Thus \(A\) is invertible.
sequence of elementary row operations producing \(I_2\).
4. Conclusion: In all cases, \(A\) is row-equivalent to \(I_2\) if and only if \(ad - bc \neq 0\).
2 and 3.
Q.E.D.