If \( A=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} \) such that \( A^{2}+\beta A+\gamma I=0 \) then \( \beta+\gamma \) is equal to:
(where \( I \) is an identity matrix of order 2)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −7
Here 0 is the 2 × 2 zero matrix O.
\( \displaystyle A^{2}=\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}7 & 10\\15 & 22\end{bmatrix} \)
\( \displaystyle A^{2}+\beta A+\gamma I=\begin{bmatrix}7+\beta+\gamma & 10+2\beta\\15+3\beta & 22+4\beta+\gamma\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix} \)
From the (1, 2) entry:
\( \displaystyle 10+2\beta=0\ \Rightarrow\ \beta=-5 \)
From the (1, 1) entry:
\( \displaystyle 7-5+\gamma=0\ \Rightarrow\ \gamma=-2 \)
The other two entries check: 15 − 15 = 0 and 22 − 20 − 2 = 0.
\( \displaystyle \beta+\gamma=-5-2=-7 \)
Shortcut: By the Cayley-Hamilton theorem, \( A^{2}-(\mathrm{tr}\,A)A+|A|I=O \). Here tr A = 5 and |A| = 4 − 6 = −2, so
\( \displaystyle A^{2}-5A-2I=O \)
\( \displaystyle \beta=-5,\ \gamma=-2,\ \beta+\gamma=-7 \)