If a matrix \( A=\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix} \) is such that \( \mathrm{A}^2+\alpha\mathrm{A}+\beta\mathrm{I}=0 \), where I is identity matrix of order 2, then which of following are TRUE?
(A) \( \alpha=-4,\ \beta=1 \)
(B) \( \alpha=4,\ \beta=-1 \)
(C) \( A^{-1}=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix} \)
(D) \( A^{-1}=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
Here the 0 on the right is the zero matrix O of order 2.
\( \displaystyle \mathrm{A}^2=\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix}\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}11 & 8\\4 & 3\end{bmatrix} \)
\( \displaystyle \mathrm{A}^2+\alpha\mathrm{A}+\beta\mathrm{I}=\begin{bmatrix}11+3\alpha+\beta & 8+2\alpha\\4+\alpha & 3+\alpha+\beta\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix} \)
\( \displaystyle 4+\alpha=0\ \Rightarrow\ \alpha=-4 \)
\( \displaystyle 3+\alpha+\beta=0\ \Rightarrow\ \beta=1 \)
(The other two entries check: 11 − 12 + 1 = 0 and 8 − 8 = 0.) So \( \mathrm{A}^2-4\mathrm{A}+\mathrm{I}=\mathrm{O} \). Multiplying by \( \mathrm{A}^{-1} \):
\( \displaystyle \mathrm{A}-4\mathrm{I}+\mathrm{A}^{-1}=\mathrm{O}\ \Rightarrow\ \mathrm{A}^{-1}=4\mathrm{I}-\mathrm{A} \)
\( \displaystyle \mathrm{A}^{-1}=\begin{bmatrix}4 & 0\\0 & 4\end{bmatrix}-\begin{bmatrix}3 & 2\\1 & 1\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix} \)
So α = −4, β = 1 and \( \mathrm{A}^{-1}=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix} \).
Checking the statements
(A) \( \alpha=-4,\ \beta=1 \) — Correct. The (2, 1) entry gives 4 + α = 0, so α = −4, and the (2, 2) entry then gives β = 1.
(B) \( \alpha=4,\ \beta=-1 \) — Incorrect. Both signs are wrong: α = −4 and β = 1.
(C) \( A^{-1}=\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix} \) — Correct. This is \( 4\mathrm{I}-\mathrm{A} \).
(D) \( A^{-1}=\begin{bmatrix}1 & 2\\1 & 3\end{bmatrix} \) — Incorrect. Its product with A is \( \begin{bmatrix}5 & 12\\2 & 5\end{bmatrix}\ne\mathrm{I} \).
Shortcut: By the Cayley-Hamilton theorem, every 2 × 2 matrix satisfies \( \mathrm{A}^2-(\mathrm{tr}\,\mathrm{A})\mathrm{A}+|\mathrm{A}|\mathrm{I}=\mathrm{O} \). Here tr A = 4 and |A| = 3 − 2 = 1, so α = −4 and β = 1 at once, and \( \mathrm{A}^{-1}=4\mathrm{I}-\mathrm{A} \).