📘 Question
If
\[
A =
\begin{bmatrix}
1 & 2 & x \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix},
\quad
B =
\begin{bmatrix}
1 & -2 & y \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
and \(AB = I_3\), find \(x + y\).
✏️ Step-by-Step Solution
Step 1: Multiply \(A \cdot B\)
\[
AB =
\begin{bmatrix}
1 & 2 & x \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & -2 & y \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
\[
=
\begin{bmatrix}
1 & 0 & y + x \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
Step 2: Compare with identity matrix
\[
AB =
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\]
So,
\[
x + y = 0
\]
✅ Final Answer
\[
\boxed{0}
\]
💡 Key Concept
If \(AB = I\), then \(A\) and \(B\) are inverses. Compare product with identity matrix to find unknowns.