Let \( A=\begin{bmatrix}a & 0\\b & 0\end{bmatrix} \) and \( A^2=\begin{bmatrix}4 & c\\2 & 0\end{bmatrix} \), \( a,b,c\in Z \). Then the minimum value of \( a+b+c \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → -3
\( \displaystyle A^2=\begin{bmatrix}a & 0\\b & 0\end{bmatrix}\begin{bmatrix}a & 0\\b & 0\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}a^2 & 0\\ab & 0\end{bmatrix} \)
Comparing with \( \begin{bmatrix}4 & c\\2 & 0\end{bmatrix} \):
\( \displaystyle a^2=4 \)
\( \displaystyle c=0 \)
\( \displaystyle ab=2 \)
With integers a and b:
\( \displaystyle a=2,\ b=1:\ a+b+c=3 \)
\( \displaystyle a=-2,\ b=-1:\ a+b+c=-3 \)
The minimum value is −3.
Why the other options are incorrect
Option (1) 0 — Incorrect. No integer pair gives 0; the only possible sums are 3 and −3.
Option (2) 3 — Incorrect. 3 is the maximum value, from a = 2, b = 1.
Option (4) -1 — Incorrect. −1 is not a possible value of a + b + c.