Question

If $A = \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$ , How do you find $F(A)$ where $f(x) = {x^2} - 2x + 3$ ?

Answer 1

Hint: Firstly substitute the matrix $A$ in $f(x) = {x^2} - 2x + 3$.For ${A^2}$do matrix multiplication which is $A \times A$ and then coming to the next term,$2\;A$ multiply the constant to the entire matrix. Multiply the last constant $3$to an identity matrix to evaluate all the $3$terms to get a solution which is also a matrix.

Complete step-by-step answer:
The given matrix is $A = \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$
To find $f(A)$, put $A$ in place of $x$in the equation,$f(x) = {x^2} - 2x + 3$
$ \Rightarrow f(A) = {A^2} - 2A + 3$
Whenever a constant is given in an equation where we must solve the matrices, we multiply that constant with the Identity matrix $I$.
$ \Rightarrow f(A) = {A^2} - 2A + 3I$
Now, solving each term starting with ${A^2}$,
$ \Rightarrow A \times A$
Writing it in the form of matrices.
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$You can only do matrix multiplication operation only if the “number of columns of the first matrix equals the number of rows of the second matrix”.
Since the matrices are of $2 \times 2;2 \times 2$ dimensions,
Matrix multiplication is possible.
Notations,
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right) \Leftrightarrow \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$For multiplication of matrices, we first multiply the elements of the rows of the first matrix with corresponding columns in the second matrix.
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right) \times \left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {( - 1 \times - 1) + (2 \times 3)}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)$
Now, we continue this for all the elements.
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {( - 1 \times - 1) + (2 \times 3)}&{( - 1 \times 2) + (2 \times 1)} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {( - 1 \times - 1) + (2 \times 3)}&{( - 1 \times 2) + (2 \times 1)} \\
  {(3 \times - 1) + (1 \times 3)}&{(3 \times 2) + (1 \times 1)}
\end{array}} \right)$
Now, evaluate the matrix
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {(1) + (6)}&{( - 2) + (2)} \\
  {( - 3) + (3)}&{(6) + (1)}
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  7&0 \\
  0&7
\end{array}} \right)$
The value of ${A^2}$ is $\left( {\begin{array}{*{20}{c}}
  7&0 \\
  0&7
\end{array}} \right)$
Now, finding the second term which is $2\;A$
$ \Rightarrow 2\left( {\begin{array}{*{20}{c}}
  { - 1}&2 \\
  3&1
\end{array}} \right)$Multiply all the values in the matrix with $2$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - 1 \times 2}&{2 \times 2} \\
  {3 \times 2}&{1 \times 2}
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  { - 2}&4 \\
  6&2
\end{array}} \right)$
Now the third term,$3\;I$
$ \Rightarrow 3\left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  3&0 \\
  0&3
\end{array}} \right)$
Now, put all terms together and evaluate.
$f(A) = {A^2} - 2A + 3I$
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  7&0 \\
  0&7
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  { - 2}&4 \\
  6&2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  3&0 \\
  0&3
\end{array}} \right)$
We can now directly perform all the operations to get a single matrix.
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {(7 - ( - 2) + 3)}&{(0 - 4 + 0)} \\
  {(0 - 6 + 0)}&{(7 - 2 + 3)}
\end{array}} \right)$$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {12}&{ - 4} \\
  { - 6}&8
\end{array}} \right)$
$\therefore f(A) = {A^2} - 2A + 3 = \left( {\begin{array}{*{20}{c}}
  {12}&{ - 4} \\
  { - 6}&8
\end{array}} \right)$

Note:
Always check before multiplying any two matrices if the number of columns of the first matrix is equal to the number of rows of the second matrix. If it is not equal matrix multiplication is not possible. Whenever there is a constant in $f(x)$ function, it should be multiplied with an identity matrix to solve it with other terms.