Question

Let \( A=\begin{bmatrix}1 & 2 & 3\\ 0 & 2 & 3\\ 0 & 0 & 1\end{bmatrix} \) and let \( P \) be a \( 3 \times 3 \) matrix. If \( AP=\begin{bmatrix}1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1\end{bmatrix} \) and \( \begin{bmatrix}1 & 0 & 2\end{bmatrix} P\begin{bmatrix}1\\ 0\\ 2\end{bmatrix}=[k]_{1 \times 1} \), then \( k \) is equal to ________.

Hint:

\( P=A^{-1}B \), so \( vPv^T=(vA^{-1})Bv^T \); find \( w=vA^{-1} \) by solving \( wA=v \).

Answer 1

Correct Answer: 6

Let \( B=AP \) so that \( P=A^{-1}B \) (\( \det A=2\neq0 \)). With \( v=[1\;0\;2] \), \( k=v A^{-1} B v^{T} \).
Let \( w=vA^{-1}=[w_1\;w_2\;w_3] \), i.e. \( wA=v \):
Column 1: \( w_1=1 \). Column 2: \( 2w_1+2w_2=0\Rightarrow w_2=-1 \). Column 3: \( 3w_1+3w_2+w_3=2\Rightarrow w_3=2 \).
So \( w=[1\;-1\;2] \).
\( wB=1\cdot[1\;0\;1]-1\cdot[1\;1\;0]+2\cdot[0\;1\;1]=[0\;1\;3] \).
\( k=[0\;1\;3]\cdot[1\;0\;2]^{T}=0+0+6=6 \).
Answer: 6