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Matrices & Determinants: Matrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatrix Pmatr

JEE Maths question with a full step-by-step solution.

Question
AA is a 2×22\times 2 matrix such that A(1−1)=(−12)A\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix} and A2(1−1)=(10)A^2\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}. The sum of the elements of AA is:
A−1-1
B00
C22
D55correct
Solution
Step 1: Obtain equations from the first condition Let A=(abcd)A = \begin{pmatrix}a & b\\ c & d\end{pmatrix}. Then:
(abcd)(1−1)=(a−bc−d)=(−12)\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}a-b\\c-d\end{pmatrix} = \begin{pmatrix}-1\\2\end{pmatrix}
a−b=−1andc−d=2⋯(1)a - b = -1 \quad \text{and} \quad c - d = 2 \qquad \cdots (1)
 Step 2: Use the second condition
A2(1−1)=A ⁣(A(1−1))=A(−12)=(10)A^2\begin{pmatrix}1\\-1\end{pmatrix} = A\!\left(A\begin{pmatrix}1\\-1\end{pmatrix}\right) = A\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}
(abcd)(−12)=(−a+2b−c+2d)=(10)\begin{pmatrix}a & b\\ c & d\end{pmatrix}\begin{pmatrix}-1\\2\end{pmatrix} = \begin{pmatrix}-a+2b\\-c+2d\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}
−a+2b=1and−c+2d=0⋯(2)-a + 2b = 1 \quad \text{and} \quad -c + 2d = 0 \qquad \cdots (2)
 Step 3: Solve the system From (1)(1) and (2)(2): adding a−b=−1a-b=-1 and −a+2b=1-a+2b=1 gives b=0b = 0, hence a=−1a = -1. Adding c−d=2c-d=2 and −c+2d=0-c+2d=0 gives d=2d = 2, hence c=4c = 4.
a+b+c+d=−1+0+4+2=5a + b + c + d = -1 + 0 + 4 + 2 = 5
 Answer: (4)
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