<p><em>A</em> is a \(2 \times 2\) matrix such that \(A\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}-1\\2\end{bmatrix}\) and \(A^2\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}\). The sum of the elements of <em>A</em> is</p>
Step-by-Step Solution
Key Concept: Use the two given conditions to set up equations for matrix A's elements, then leverage the relationship between A·v and A²·v where v is the given column vector to find the matrix systematically.
<p><strong>Step 1:</strong> Let $A = \begin{bmatrix}a & b\\c & d\end{bmatrix}$. From $A\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}-1\\2\end{bmatrix}$:</p><p>$\begin{bmatrix}a-b\\c-d\end{bmatrix} = \begin{bmatrix}-1\\2\end{bmatrix}$ gives us: $a-b = -1$ and $c-d = 2$</p><p><strong>Step 2:</strong> From $A^2\begin{bmatrix}1\\-1\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$, we use $A^2\begin{bmatrix}1\\-1\end{bmatrix} = A\left(A\begin{bmatrix}1\\-1\end{bmatrix}\right) = A\begin{bmatrix}-1\\2\end{bmatrix}$:</p><p>$A\begin{bmatrix}-1\\2\end{bmatrix} = \begin{bmatrix}-a+2b\\-c+2d\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$</p><p>This gives us: $-a+2b = 1$ and $-c+2d = 0$</p><p><strong>Step 3:</strong> Solve the system:</p><p>From equations (1) and (3): $a-b=-1$ and $-a+2b=1$</p><p>Adding: $b = 0$, so $a = -1$</p><p>From equations (2) and (4): $c-d=2$ and $-c+2d=0$</p><p>Adding: $d = 2$, so $c = 4$</p><p><strong>Step 4:</strong> Therefore $A = \begin{bmatrix}-1 & 0\\4 & 2\end{bmatrix}$</p><p>Sum of elements = $-1 + 0 + 4 + 2 = 5$</p><p>∴ Answer: A</p>
Correct Answer: A