Question
Find a \(2 \times 2\) matrix \(A\) such that \[ A \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = 6I_2 \]
Solution
Step 1: Assume \(A\)
\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \]Step 2: Multiply
\[ A \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} a+b & -2a+4b \\ c+d & -2c+4d \end{bmatrix} \]Step 3: Compare with \(6I_2\)
\[ \begin{bmatrix} a+b & -2a+4b \\ c+d & -2c+4d \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \]Step 4: Solve
From first row: \[ a+b=6,\quad -2a+4b=0 \Rightarrow b=2,\ a=4 \] From second row: \[ c+d=0,\quad -2c+4d=6 \Rightarrow d=1,\ c=-1 \]Final Answer
\[
A =
\begin{bmatrix}
4 & 2 \\
-1 & 1
\end{bmatrix}
\]