Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 4}&2
\end{array}} \right]\], then \[{A^2} - 5A = \]
A. \[I\]
B. \[14I\]
C. \[0\]
D. None of these

Answer 1

Hint: In the given problem we are given a matrix A. On squaring A, we obtain the value of \[{A^2}\] and on multiplying matrix A with scalar 5 we obtain the value of 5A. By subtracting both the matrices we obtain the value of \[{A^2} - 5A\].

Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]

Complete step by step solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 4}&2
\end{array}} \right]\]
On squaring to get \[{A^2}\]we get,
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 4}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 4}&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {29}&{ - 25} \\
  { - 20}&{24}
\end{array}} \right]\]
On multiplying the scalar 5 with the matrix A we get,
\[5A = 5\left[ {\begin{array}{*{20}{c}}
  3&{ - 5} \\
  { - 4}&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {15}&{ - 25} \\
  { - 20}&{10}
\end{array}} \right]\]
\[{A^2} - 5A = \left[ {\begin{array}{*{20}{c}}
  {29}&{ - 25} \\
  { - 20}&{24}
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  {15}&{ - 25} \\
  { - 20}&{10}
\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}
  {29 - 15}&{ - 25 - ( - 25)} \\
  { - 20 - ( - 20)}&{24 - 10}
\end{array}} \right]\]\[ = \left[ {\begin{array}{*{20}{c}}
  {14}&0 \\
  0&{14}
\end{array}} \right]\]\[ = 14\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]\]\[ = 14I\]

Option B. is the correct answer.

Note:To solve the given problem, one must know to multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix. When multiplying a matrix with a scalar, the scalar gets multiplied by all the elements of the matrix.