Let P be an \[3 \times 3\]matrix where alpha belongs to R. Suppose \[Q = \left[ {{q_{ij}}} \right]\] is a matrix \[PQ = k{I_3}\] for some non-zero \[k \in R\]. If \[{q_{23}} = - k/8\] and \[\left| Q \right| = {k^2}/2\], then find the value of \[{\alpha ^2} + {k^2}\].
\[P = \left[ {\begin{array}{*{20}{c}}
  3&{ - 1}&{ - 2} \\
  2&0&\alpha \\
  3&{ - 5}&0
\end{array}} \right]\]

Hint: In this question, we have to determine the value of \[{\alpha ^2} + {k^2}\]. First we will find the value of alpha. For this, we have to use the relations such as \[PQ = k{I_3}\] and \[{q_{23}} = - k/8\].
After that, we can find the value of \[k\] using the relation \[\left| Q \right| = {k^2}/2\].

Formula used: We will use the following identity of iota for solving this example.
\[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj{\text{ A}}} \right)\] where, \[A\] is a matrix.
Complete step-by-step answer:
We know that \[PQ = kI\]
That gives us \[Q = k{P^{ - 1}}I\]
So, \[Q = \dfrac{k}{{\left| P \right|}}\left( {adj.{\text{ P}}} \right)I\]…… (1)
Let us find the value of \[\left| P \right|\]
\[
  \left| P \right| = \left| {\begin{array}{*{20}{c}}
  3&{ - 1}&{ - 2} \\
  2&0&\alpha \\
  3&{ - 5}&0
\end{array}} \right| \\
  \left| P \right| = 3\left( {\left( {0 \times 0} \right) - \left( {\left( { - 5} \right) \times \alpha } \right)} \right) - \left( { - 1} \right)\left( {\left( {2 \times 0} \right) - \left( {3 \times \alpha } \right)} \right) - 2\left( {\left( {2 \times \left( { - 5} \right)} \right) - \left( {3 \times 0} \right)} \right) \\
  \left| P \right| = 3\left( {0 + 5\alpha } \right) + \left( {0 - \left( {3\alpha } \right)} \right) - 2\left( { - 10 - 0} \right) \\
  \left| P \right| = 15\alpha - 3\alpha + 20 \\
  \left| P \right| = 12\alpha + 20 \\
 \]
Now, we will find the adjoint of a matrix P.
For this, first we will find the cofactor matrix of P that contains all the co-factors of matrix P.
For this, we will use \[{C_{ij}} = {\left( { - 1} \right)^{i + j}}{M_{ij}}\], where \[{M_{ij}}\]is the minor.
\[{C_{11}} = {\left( { - 1} \right)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  0&\alpha \\
  { - 5}&0
\end{array}} \right| = \left( 1 \right) \times \left[ {0 - \left( { - 5\alpha } \right)} \right] = 5\alpha \]
\[{C_{12}} = {\left( { - 1} \right)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  2&\alpha \\
  3&0
\end{array}} \right| = \left( { - 1} \right) \times \left[ {\left( {2 \times 0} \right) - \left( {3\alpha } \right)} \right] = 3\alpha \]
\[{C_{13}} = {\left( { - 1} \right)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  2&0 \\
  3&{ - 5}
\end{array}} \right| = \left( 1 \right) \times \left[ {\left( {2 \times \left( { - 5} \right)} \right) - \left( {3 \times 0} \right)} \right] = - 10\]
\[{C_{21}} = {\left( { - 1} \right)^{2 + 1}}\left| {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  { - 5}&0
\end{array}} \right| = \left( { - 1} \right) \times \left[ {\left( { - 1 \times 0} \right) - \left( {\left( { - 2} \right) \times \left( { - 5} \right)} \right)} \right] = 10\]
\[{C_{22}} = {\left( { - 1} \right)^{2 + 2}}\left| {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  3&0
\end{array}} \right| = \left( 1 \right) \times \left[ {\left( {3 \times 0} \right) - \left( {\left( { - 2} \right) \times \left( 3 \right)} \right)} \right] = 6\]
\[{C_{23}} = {\left( { - 1} \right)^{2 + 3}}\left| {\begin{array}{*{20}{c}}
  3&{ - 1} \\
  3&{ - 5}
\end{array}} \right| = \left( { - 1} \right) \times \left[ {\left( {3 \times \left( { - 5} \right)} \right) - \left( {\left( { - 1} \right) \times \left( 3 \right)} \right)} \right] = 12\]
\[{C_{31}} = {\left( { - 1} \right)^{3 + 1}}\left| {\begin{array}{*{20}{c}}
  { - 1}&{ - 2} \\
  0&\alpha
\end{array}} \right| = \left( 1 \right) \times \left[ {\left( { - 1 \times \left( \alpha \right)} \right) - \left( {\left( { - 2} \right) \times \left( 0 \right)} \right)} \right] = - \alpha \]
\[{C_{32}} = {\left( { - 1} \right)^{3 + 2}}\left| {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  2&\alpha
\end{array}} \right| = \left( { - 1} \right) \times \left[ {\left( {3 \times \left( \alpha \right)} \right) - \left( {\left( { - 2} \right) \times \left( 2 \right)} \right)} \right] = - 3\alpha - 4\]
\[{C_{33}} = {\left( { - 1} \right)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
  3&{ - 1} \\
  2&0
\end{array}} \right| = \left( 1 \right) \times \left[ {\left( {3 \times \left( 0 \right)} \right) - \left( {\left( { - 1} \right) \times \left( 2 \right)} \right)} \right] = 2\]
So, the desired cofactor matrix is given by \[\left[ {\begin{array}{*{20}{c}}
  {5\alpha }&{3\alpha }&{ - 10} \\
  {10}&6&{12} \\
  { - \alpha }&{ - 3\alpha - 4}&2
\end{array}} \right]\]
Now, we will obtain an adjoint matrix by interchanging the row and column of the cofactor matrix.
Thus, we get
\[Adj{\text{ P}} = \left[ {\begin{array}{*{20}{c}}
  {5\alpha }&{10}&{ - \alpha } \\
  {3\alpha }&6&{ - 3\alpha - 4} \\
  { - 10}&{12}&2
\end{array}} \right]\]
From equation (1), we get
\[Q = \dfrac{k}{{12\alpha + 20}}\left[ {\begin{array}{*{20}{c}}
  {5\alpha }&{10}&{ - \alpha } \\
  {3\alpha }&6&{ - 3\alpha - 4} \\
  { - 10}&{12}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]\]
This gives
\[Q = \dfrac{k}{{12\alpha + 20}}\left[ {\begin{array}{*{20}{c}}
  {5\alpha }&{10}&{ - \alpha } \\
  {3\alpha }&6&{ - 3\alpha - 4} \\
  { - 10}&{12}&2
\end{array}} \right]\]
But we know that \[{q_{23}} = \dfrac{{ - k}}{8}\]
\[{q_{23}} = \dfrac{{\left( { - 3\alpha - 4} \right)k}}{{\left( {12\alpha + 20} \right)}}\]
Hence, we get
\[\dfrac{{\left( { - 3\alpha - 4} \right)k}}{{\left( {12\alpha + 20} \right)}} = \dfrac{{ - k}}{8}\]
By simplifying, we get
\[\dfrac{{\left( {3\alpha + 4} \right)}}{{\left( {12\alpha + 20} \right)}} = \dfrac{1}{8}\]
\[8\left( {3\alpha + 4} \right) = 1\left( {12\alpha + 20} \right)\]
\[24\alpha + 32 = 12\alpha + 20\]
\[24\alpha - 12\alpha = 20 - 32\]
By simplifying further, we get
\[12\alpha = - 12\]
This gives
\[\alpha = - 1\]
Also, we know that \[\left| Q \right| = \dfrac{{{k^2}}}{2}\]
But \[\left| Q \right| = \dfrac{{{k^3}\left| I \right|}}{{\left| P \right|}}\]
Thus, we get
\[\dfrac{{{k^2}}}{2} = \dfrac{{{k^3}\left( 1 \right)}}{{\left( {12\alpha + 20} \right)}}\]
By simplifying, we get
\[\dfrac{1}{2} = \dfrac{k}{{4\left( {3\alpha + 5} \right)}}\]
\[1 = \dfrac{k}{{2\left( {3\alpha + 5} \right)}}\]
\[2\left( {3\alpha + 5} \right) = k\]
Now, put \[\alpha = - 1\] in the above equation.
So, we get
\[2\left( {3\left( { - 1} \right) + 5} \right) = k\]
\[2\left( { - 3 + 5} \right) = k\]
\[2\left( 2 \right) = k\]
Thus, we get \[k = 4\]
Let us find the value of \[{\alpha ^2} + {k^2}\].
\[{\alpha ^2} + {k^2} = {\left( { - 1} \right)^2} + {\left( 4 \right)^2}\]
\[{\alpha ^2} + {k^2} = 1 + 16\]
\[{\alpha ^2} + {k^2} = 17\]

Therefore, the value of \[{\alpha ^2} + {k^2}\]is \[17\].

Note: Many students make mistakes in finding the inverse of a matrix. Particularly, they may get confused with signs while calculating minors.