If $A = \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  3&2
\end{array}} \right]$ and $I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ , what is the value of ${A^2} - 6A$ ?
A. $3I$
B. $5I$
C. $ - 5I$
D. None of these

Hint: In the above question, we are provided with a matrix $A$ and we are asked to evaluate the value of ${A^2} - 6A$ . Perform Matrix Algebra and evaluate the value of ${A^2} - 6A$, then select the correct option in regard to the evaluated value.

Complete step by step Solution:
Given matrix:
$A = \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  3&2
\end{array}} \right]$
Now, we are asked to calculate the value of ${A^2} - 6A$.
Substituting the values,
${A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  3&2
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  3&2
\end{array}} \right] - 6 \times \left[ {\begin{array}{*{20}{c}}
  4&1 \\
  3&2
\end{array}} \right]$
Performing Matrix Algebra,
${A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  {16 + 3}&{4 + 2} \\
  {12 + 6}&{3 + 4}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {24}&6 \\
  {18}&{12}
\end{array}} \right]$
Simplifying further,
${A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  {19}&6 \\
  {18}&7
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {24}&6 \\
  {18}&{12}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {19 - 24}&{6 - 6} \\
  {18 - 18}&{7 - 12}
\end{array}} \right]$
This gives:
${A^2} - 6A = \left[ {\begin{array}{*{20}{c}}
  { - 5}&0 \\
  0&{ - 5}
\end{array}} \right]$
Taking $ - 5$ common from both the rows,
${A^2} - 6A = - 5\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
As $\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$ is an identity matrix, therefore, ${A^2} - 6A = - 5I$ .

Hence, the correct option is (C).

Note: When a scalar is multiplied with a matrix, the resultant matrix is the original matrix with each of its terms multiplied by the scalar. Consider a matrix $A$ such that $A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]$ and a scalar $k$ , then \[kA = \left[ {\begin{array}{*{20}{c}}
  {k{a_{11}}}&{k{a_{12}}}&{k{a_{13}}} \\
  {k{a_{21}}}&{k{a_{22}}}&{k{a_{23}}} \\
  {k{a_{31}}}&{k{a_{32}}}&{k{a_{33}}}
\end{array}} \right]\].