If \( \mathrm{A}=\begin{bmatrix}a & b\\c & \displaystyle\frac{1+bc}{a}\end{bmatrix} \), then \( a\mathrm{A}^{-1} \) is equal to which of the following ?
[ I is the identity matrix of order 2 ]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (a^2+bc+1)\mathrm{I}-a\mathrm{A} \)
The determinant of A is
\( \displaystyle |\mathrm{A}|=a\cdot\frac{1+bc}{a}-bc \)
\( \displaystyle =1+bc-bc=1 \)
Since \( |\mathrm{A}|=1 \), the inverse is just the adjoint (swap the diagonal entries, change the sign of the other two):
\( \displaystyle \mathrm{A}^{-1}=\begin{bmatrix}\frac{1+bc}{a} & -b\\-c & a\end{bmatrix} \)
\( \displaystyle a\mathrm{A}^{-1}=\begin{bmatrix}1+bc & -ab\\-ac & a^2\end{bmatrix} \)
The options have the form \( k\mathrm{I}-a\mathrm{A} \), where \( a\mathrm{A}=\begin{bmatrix}a^2 & ab\\ac & 1+bc\end{bmatrix} \). With \( k=a^2+bc+1 \):
\( \displaystyle (a^2+bc+1)\mathrm{I}-a\mathrm{A}=\begin{bmatrix}a^2+bc+1-a^2 & -ab\\-ac & a^2+bc+1-(1+bc)\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1+bc & -ab\\-ac & a^2\end{bmatrix} \)
This is exactly \( a\mathrm{A}^{-1} \).
Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies
\( \displaystyle \mathrm{A}^2-(\mathrm{tr}\,\mathrm{A})\mathrm{A}+|\mathrm{A}|\mathrm{I}=\mathrm{O} \).
Multiplying by \( \mathrm{A}^{-1} \) and using \( |\mathrm{A}|=1 \):
\( \displaystyle \mathrm{A}^{-1}=(\mathrm{tr}\,\mathrm{A})\mathrm{I}-\mathrm{A} \)
\( \displaystyle =\left(a+\frac{1+bc}{a}\right)\mathrm{I}-\mathrm{A} \)
\( \displaystyle a\mathrm{A}^{-1}=(a^2+1+bc)\mathrm{I}-a\mathrm{A} \)