taiwan-gsat 2021 QB

taiwan-gsat · Other · ast__math-b 8 marks Matrices Matrix Power Computation and Application

Let matrix $A = \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 1 & 0 \\ 0 & 6 \end{array} \right] \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 } , B = \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 6 & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 }$, where $\left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 }$ is the inverse matrix of $\left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right]$. If $A + B = \left[ \begin{array} { l l } a & b \\ c & d \end{array} \right]$, then $a + b + c + d =$ (10)(11).

Let matrix $A = \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 1 & 0 \\ 0 & 6 \end{array} \right] \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 } , B = \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { l l } 6 & 0 \\ 0 & 1 \end{array} \right] \left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 }$, where $\left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right] ^ { - 1 }$ is the inverse matrix of $\left[ \begin{array} { c c } 1 & - 2 \\ 0 & 1 \end{array} \right]$. If $A + B = \left[ \begin{array} { l l } a & b \\ c & d \end{array} \right]$, then $a + b + c + d =$ (10)(11).

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