Question

If \[A = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&1
\end{array}} \right]\], then the determinant of \[{A^2} - 2A\] is
A. \[5\]
B. \[25\]
C. \[ - 5\]
D. \[ - 25\]

Answer 1

Hint: The term\[{A^2} - 2A\] can be simplified by taking matrix A common and applying determinant. By evaluating the determinant of the two terms separately and multiplying we get the required solution. The determinant is obtained by cross multiplying the elements starting from the top left and then subtracting the products.

Formula used:
The determinant of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]\] is \[\left| A \right| = ad - bc\]

Complete step by step Solution:
We are given that,
\[{A^2} - 2A\]
\[\left| {{A^2} - 2A} \right| = \left| {A(A - 2I)} \right|\]
\[ = \left| A \right|\left| {A - 2I} \right|\] (Since \[\left| {AB} \right| = \left| A \right|\left| B \right|\])
\[ = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&1
\end{array}} \right] \times \left( {\left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&1
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]} \right)\]
\[ = \left( {1 - 6} \right) \times \left| {\left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&1
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  2&0 \\
  0&2
\end{array}} \right]} \right|\]
\[ = - 5 \times \left| {\left[ {\begin{array}{*{20}{c}}
  {1 - 2}&{3 - 0} \\
  {2 - 0}&{1 - 2}
\end{array}} \right]} \right|\]
\[ = - 5 \times \left[ {\begin{array}{*{20}{c}}
  { - 1}&3 \\
  2&{ - 1}
\end{array}} \right]\]
\[ = - 5 \times ( - 1 \times 1 \times - 6)\]
\[ = - 5 \times (1 - 6)\]
\[ = - 5 \times - 5\]
\[ = 25\]
\[\therefore \left| {{A^2} - 2A} \right| = 25\]

Therefore, the correct option is (B).

Note:When A is taken out from the term 2A, we get 2I. One must not forget to mention Identity matrix I when simplifying matrix terms. One must also note that the determinant is defined only if the matrix is a square matrix.