Question
If \[A = \left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]\] is a $$2 \times 2$$ matrix and $$f\left( x \right) = {x^2} - x + 2$$ is a polynomial, then what is $$f\left( A \right) \,? $$
A.
\[\left[ {\begin{array}{*{20}{c}}
1&7\\
1&7
\end{array}} \right]\]
B.
\[\left[ {\begin{array}{*{20}{c}}
2&6\\
0&8
\end{array}} \right]\]
C.
\[\left[ {\begin{array}{*{20}{c}}
2&6\\
0&6
\end{array}} \right]\]
D.
\[\left[ {\begin{array}{*{20}{c}}
2&6\\
0&7
\end{array}} \right]\]
Answer :
\[\left[ {\begin{array}{*{20}{c}}
2&6\\
0&8
\end{array}} \right]\]
Solution :
Given that, \[A = \left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]\]
\[{A^2} = \left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&{2 + 6}\\
0&9
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&8\\
0&9
\end{array}} \right]\]
Since, $$f\left( x \right) = {x^2} - x + 2$$
Putting $$A$$ in place of $$x$$
$$f\left( A \right) = {A^2} - A + 2I$$
\[\begin{array}{l}
= \left[ {\begin{array}{*{20}{c}}
1&8\\
0&9
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
2&0\\
0&2
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
{1 - 1 + 2}&{8 - 2 + 0}\\
{0 - 0 + 0}&{9 - 3 + 2}
\end{array}} \right]\\
= \left[ {\begin{array}{*{20}{c}}
2&6\\
0&8
\end{array}} \right]
\end{array}\]