Problem

Calculate the values of the parameter $t$ such that the condition $∣A^3∣=8$ is satisfied, given that $A$ is the following matrix:

A = \begin{pmatrix}
      t - 1   &        t + 1    &     3 \\
t^2 - t       & t^2 + 2t        & t      \\
    - t + 1   &     - t  - 1    &    -2
\end{pmatrix}

Solution

To solve this problem, we need to find the values of $( t )$ for which the determinant of $( A^3 )$ is equal to $8$. This involves the following steps:

1. Calculate the determinant of $( A )$:

   We need to calculate $(Det(A))$ because $(\left| A^3 \right| = \left( Det(A) \right)^3)$.

2. Set up the equation:

We will set $(\left( Det(A) \right)^3 = 8)$ and solve for $( t )$.

3. Solve for $( t )$:

Simplify the equation and solve for the parameter $( t )$.

Step 1: Calculate the determinant of $( A )$

A = \begin{pmatrix}
      t - 1 &        t + 1 &    3 \\
t^2 - t     & t^2 + 2t     & t     \\
    - t + 1 &      - t - 1 &   - 2
\end{pmatrix}

The determinant of a $(3 \times 3)$ matrix $(A)$ can be calculated using the formula:

\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)

For the given matrix $( A )$:

\color{Gold}{\det(A)} = 
\color{tomato}{  (t - 1) 
          \left(  \color{red}{    (t^2 + 2t)(-2)    - (t)(-1 - t)               } \right) }
\color{lime}{  - (t + 1)
          \left(  \color{green}{  (t)(1 - t)        - (t^2 - t)(-2)             } \right) }
\color{cyan}{   + 3 
          \left(  \color{blue}{   (t^2 - t)(-1 - t) - (t^2 + 2t)(1 - t)         } \right) }

Now we compute each term individually.

\color{red}
(t^2 + 2t)(-2) - t(-1 - t) = -2t^2 - 4t + t + t^2 = -t^2 - 3t

\color{tomato}
(t - 1) \color{red}{(-t^2 - 3t)} = -t^3 - 3t^2 + t^2 + 3t = -t^3 - 2t^2 + 3t = -t·(t+3)·(t-1)

Calculate the second term:

\color{green}
(t^2 - t)(-2) - t(1 - t) = -2t^2 + 2t - t + t^2 = -t^2 + t = - t·(t-1)

So the second term is:

\color{lime}
-(t + 1)(-t^2 + t) = t^3 - t^2 - t^2 + t = t^3 - 2t^2 + t = t·(t+1)·(t-1)

Calculate the third term:

\color{blue}
(t^2 - t)(-1 - t) - (t^2 + 2t)(1 - t) = -t^2 - t^3 + t + t^2 - t^2 - 2t + t^3 + 2t^2 = t^2 - t = t(t-1)

So the third term is:

\color{cyan}
3(t^2 - t) = 3t^2 - 3t = 3·t·(t-1)

Adding these terms together gives:

\begin{eqnarray}

\color{gold}{ \det(A) }  \\ 
 &     \color{gold}{ = + }
      \color{tomato}{   (-t)·(t+3)·(t-1) }
      \color{gold}{    -                } 
      \color{lime}{      t·(t+1)·(t-1) }
       \color{gold}{    +              }
       \color{cyan}{     3t(t-1)       } \\
& \color{gold}{ 
  = t·(t-1)·( -(t+3) - (t+1) +3 )    } \\ & \color{gold}{ 
  = t·(t-1)·( -t - t - 3 -1 + 3 )    } \\ & \color{gold}{ 
  = t·(t-1)·( -2t  -1  )             } \\ & \color{gold}{ 
  = -t·(t-1)·(2t + 1)                
  }
\end{eqnarray}

From our premises we have that:

∣A^3∣=8

giben that

∣A^3∣= |A|^3

then

\color{gold}{
\begin{eqnarray}
-t·(t-1)·(2t + 1) = 2        \\
-t·(t(2t + 1)-1(2t + 1)) = 2 \\
-t·(2t^2 + t - 2t - 1) = 2 \\
-(2t^3 - t^2 - t) = 2 \\
-2t^3 + t^2 + t - 2 = 0 \\
\end{eqnarray}

\color{silver}{
\begin{eqnarray}
-2t^3  &   +  t^2   &  +   t   &   - 2  && : (t+1) \\
-2t^3  &   + -2t^2  &          &        &&  -2t^2   \\
       &   + 3t^2   &  +   t   &   - 2  &&            \\
       &   + 3t^2   &  +  3t   &        &&  3t          \\
       &            &  -  2t   &    -2  &&                \\
       &            &  -  2t   &    -2  &&  -2              \\
                              
\end{eqnarray}

(t+1)(-2t^2 + 3t - 2) = 0

\color{silver}{
\begin{eqnarray}
-2t^2  &   +  3t    &   - 2  && : (-t+2) \\
-2t^2  &   +  4t    &        &&   2t        \\
       &   -   t    &   - 2  &&               \\
       &   +   t    &   - 2  &&               \\


\end{eqnarray}