Question
Consider the 2 by 2 square matrix$$A=\left[\begin{array}{ll}a & b \\c & d\end{array}\right]$$If $D=a d-b c \neq 0$, show that $A$ is nonsingular and that$$A^{-1}=\frac{1}{D}\left[\begin{array}{rr}d & -b \\-c & a\end{array}\right]$$
Step 1
By the problem statement, we are given that \( D \neq 0 \). A matrix is nonsingular (invertible) if and only if its determinant is non-zero. Therefore, since \( D \neq 0 \), matrix \( A \) is nonsingular. Show more…
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