If \( A=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} \), such that \( \mathrm{A}^2-4\mathrm{A}+\lambda\mathrm{I}=0 \), where I is identity matrix of order 2, then '\( \lambda \)' is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1
Here 0 on the right side is the zero matrix O of order 2.
\( \displaystyle \mathrm{A}^2=\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix} \)
\( \displaystyle \mathrm{A}^2-4\mathrm{A}=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}-\begin{bmatrix}8 & 12\\4 & 8\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix} \)
\( \displaystyle =-\mathrm{I} \)
So
\( \displaystyle \mathrm{A}^2-4\mathrm{A}+\lambda\mathrm{I}=(\lambda-1)\mathrm{I} \)
and this is the zero matrix only when λ = 1.
Shortcut: By the Cayley-Hamilton theorem, a 2 × 2 matrix satisfies \( \mathrm{A}^2-(\mathrm{tr\,A})\mathrm{A}+|\mathrm{A}|\,\mathrm{I}=\mathrm{O} \). Here tr A = 2 + 2 = 4 and
\( \displaystyle \lambda=|\mathrm{A}|=4-3=1 \)
NCERT (2026-27 edition):
Adjoint and Inverse of a Matrix
