Let \(A = \begin{bmatrix}1 & 2 & 0\\ 0 & 1 & 2\\ 2 & 0 & 1\end{bmatrix}\) and \(B = \begin{bmatrix}1 & -2 & 4\\ 4 & 1 & -2\\ -2 & 4 & 1\end{bmatrix}\), so that \(AB = 9I\). Then the solution of the system of equations
\(x + 2z = 7\)
\(2x + y = 5\)
\(2y + z = 0\)
is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x = 3,\ y = -1,\ z = 2\)
Since \(AB = 9I\),
\( \displaystyle A^{-1} = \frac{1}{9}B \)
The coefficient matrix of the system is not A but its transpose: its rows are the columns of A.
\( \displaystyle A^{T}X = C,\quad A^{T} = \begin{bmatrix}1 & 0 & 2\\ 2 & 1 & 0\\ 0 & 2 & 1\end{bmatrix},\quad C = \begin{bmatrix}7\\ 5\\ 0\end{bmatrix} \)
Using \((A^{T})^{-1} = (A^{-1})^{T}\):
\( \displaystyle (A^{T})^{-1} = \frac{1}{9}B^{T} = \frac{1}{9}\begin{bmatrix}1 & 4 & -2\\ -2 & 1 & 4\\ 4 & -2 & 1\end{bmatrix} \)
Then
\( \displaystyle X = \frac{1}{9}\begin{bmatrix}1 & 4 & -2\\ -2 & 1 & 4\\ 4 & -2 & 1\end{bmatrix}\begin{bmatrix}7\\ 5\\ 0\end{bmatrix} \)
\( \displaystyle X = \frac{1}{9}\begin{bmatrix}7 + 20 + 0\\ -14 + 5 + 0\\ 28 - 10 + 0\end{bmatrix} \)
\( \displaystyle X = \frac{1}{9}\begin{bmatrix}27\\ -9\\ 18\end{bmatrix} = \begin{bmatrix}3\\ -1\\ 2\end{bmatrix} \)
Check: \(3 + 2(2) = 7\), \(2(3) - 1 = 5\), \(2(-1) + 2 = 0\).