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If \[A=\left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]\], find the determinant of the matrix \[{{A}^{2}}-2A\].

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Last updated date: 22nd Jul 2024
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Answer
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Hint: Here in this question, we need to find the determinant of the matrix \[{{A}^{2}}-2A\]. Before solving this, we need to look at the definition of matrix. After that, we will consider the given data and given expression, firstly we are going to take the determinant to the \[{{A}^{2}}-2A\], then evaluate the answer.

Complete step by step answer:
Matrix is defined as the rectangular arrangement of numbers (real or complex) which may be represented as
\[\left( \begin{matrix}
   {{a}_{11}} & \ldots & {{a}_{1n}} \\
   \vdots & \ddots & \vdots \\
   {{a}_{m1}} & \cdots & {{a}_{mn}} \\
\end{matrix} \right)\]
Matrix is enclosed by \[\left( {} \right)\] or \[\left[ {} \right]\].
Compact from the above matrix is represented by \[{{\left[ {{a}_{ij}} \right]}_{m\times n}}\]or \[A=\left[ {{a}_{ij}} \right]\].

Let us solve the given question,
Given data \[A=\left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]\],
Given expression, \[{{A}^{2}}-2A\]
To find determinant \[{{A}^{2}}-2A\]
Now,
\[\left| {{A}^{2}}-2A \right|=\left| A\left( A-2I \right) \right|\]
(Taking A common on Right-hand-side)
Writing the determinants separately, on the basis of \[\left( \left| AB \right|=\left| A \right|\left| B \right| \right)\], then we get
\[\Rightarrow \left| A \right|\left| A-2I \right|\]
We are going to substituting the matrix \[A=\left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]\] and identity matrix \[I=\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\] is an identity matrix same as \[2\times 2\] on above expression \[{{A}^{2}}-2A\],
\[\Rightarrow \left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]\times \left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]-2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]\]
We will multiply the first two matrices first and then we will multiply the resultant to remaining matrix, then we get
\[\Rightarrow \left[ \left( 1\times 1-2\times 3 \right) \right]\times \left| \left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]-2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] \right|\]
Above matrix is obtained by \[2\times 2\]matrix multiplication,
\[\Rightarrow \left( 1-6 \right)\times \left| \left[ \begin{matrix}
   1 & 3 \\
   2 & 1 \\
\end{matrix} \right]-2\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right] \right|\]
On solving,
\[\Rightarrow -5\times \left| \left[ \begin{matrix}
   1-2 & 3-0 \\
   2-0 & 1-2 \\
\end{matrix} \right] \right|\]
On further evaluation,
\[\Rightarrow -5\times \left| \left[ \begin{matrix}
   -1 & 3 \\
   2 & -1 \\
\end{matrix} \right] \right|\]
Finding the determinant of the above matrix,
\[\Rightarrow -5\times \left( \left( -1\times -1 \right)-2\times 3 \right)\]
Multiplying the above terms,
\[\Rightarrow -5\times \left( 1-6 \right)\]
\[\Rightarrow -5\times -5\]
Therefore, \[{{A}^{2}}-2A=25\].

Note: It is important to note that when we consider two matrices to be equal then in order to hold the equality every corresponding element in both the matrices should be equal. For matrix multiplication, the number of columns present in the first matrix should be equal to the number of rows present in the second matrix.