If \[A = \left[ {\begin{array}{*{20}{c}}\alpha &2\\2&\alpha \end{array}} \right]\], and \[\left| {{A^3}} \right| = 125\]. Then find the value of \[\alpha \].
A. \[ \pm 1\]
B. \[ \pm 2\]
C. \[ \pm 3\]
D. \[ \pm 4\]

Hint In the given question, a \[2 \times 2\] matrix and the determinant of a matrix are given. First we will calculate the determinant of \[A = \left[ {\begin{array}{*{20}{c}}\alpha &2\\2&\alpha \end{array}} \right]\]. By applying the formula \[{\left| A \right|^n} = \left| {{A^n}} \right|\], we will calculate the value \[\alpha \].

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

Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}\alpha &2\\2&\alpha \end{array}} \right]\] and the determinant of the matrix \[{A^3}\] is \[125\].
Let’s calculate the determinant of the matrix \[A\].
Apply the formula of the determinant of a \[2 \times 2\] matrix.
\[\left| A \right| = {\alpha ^2} - 4\] \[.....\left( 1 \right)\]
It is given that \[\left| {{A^3}} \right| = 125\].
Now apply the property of determinant \[\left| {{A^n}} \right| = {\left| A \right|^n}\].
\[\left| {{A^3}} \right| = {\left| A \right|^3}\]
\[ \Rightarrow \]\[125 = {\left| A \right|^3}\]
Take the cube root of the above equation.
\[5 = \left| A \right|\] \[.....\left( 2 \right)\]
Now equate the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
\[{\alpha ^2} - 4 = 5\]
\[ \Rightarrow \]\[{\alpha ^2} = 9\]
Take the square root of the above equation.
\[\alpha = \pm 3\]
Hence the correct option is C.

Note: Students often make a common mistake. They calculate \[{A^3}\] and determine it. It is a long procedure. The less complicit procedure is solve the question by using the formula \[{\left| A \right|^n} = \left| {{A^n}} \right|\].