Answer :
Let's find the determinant of the given 3x3 matrix and solve for the values of [tex]\( x \)[/tex] that make the determinant zero. The matrix given is:
[tex]\[ \left|\begin{array}{ccc}5 x^2 + 18 x - 8 & x & x^2 \\ 0 & 1 & -1 \\ 0 & -1 & x + 1\end{array}\right| \][/tex]
Step 1: Calculate the determinant of the matrix.
To find the determinant, we can use the method of cofactor expansion along the first row. The determinant [tex]\( \Delta \)[/tex] is:
[tex]\[ \Delta = \left(5x^2 + 18x - 8\right) \cdot \left|\begin{array}{cc} 1 & -1 \\ -1 & x + 1\end{array}\right| - x \cdot \left|\begin{array}{cc} 0 & -1 \\ 0 & x + 1\end{array}\right| + x^2 \cdot \left|\begin{array}{cc} 0 & 1 \\ 0 & -1\end{array}\right| \][/tex]
Step 2: Calculate the 2x2 determinants (minors).
[tex]\[ \left|\begin{array}{cc} 1 & -1 \\ -1 & x + 1 \end{array}\right| = (1) \cdot (x + 1) - (-1) \cdot (-1) = (x + 1) - 1 = x \][/tex]
[tex]\[ \left|\begin{array}{cc} 0 & -1 \\ 0 & x + 1 \end{array}\right| = 0 \cdot (x + 1) - (-1) \cdot 0 = 0 \][/tex]
[tex]\[ \left|\begin{array}{cc} 0 & 1 \\ 0 & -1 \end{array}\right| = 0 \cdot (-1) - (1) \cdot 0 = 0 \][/tex]
Step 3: Substitute these minors back into the original determinant equation.
[tex]\[ \Delta = (5x^2 + 18x - 8) \cdot x - x \cdot 0 + x^2 \cdot 0 = (5x^2 + 18x - 8) \cdot x \][/tex]
[tex]\[ \Delta = 5x^3 + 18x^2 - 8x \][/tex]
Step 4: Set the determinant equal to zero and solve for [tex]\( x \)[/tex].
[tex]\[ 5x^3 + 18x^2 - 8x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(5x^2 + 18x - 8) = 0 \][/tex]
This gives us one solution:
[tex]\[ x = 0 \][/tex]
To find the other solutions, solve the quadratic equation [tex]\( 5x^2 + 18x - 8 = 0 \)[/tex].
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 18 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 5 \cdot (-8)}}{2 \cdot 5} \][/tex]
[tex]\[ x = \frac{-18 \pm \sqrt{324 + 160}}{10} \][/tex]
[tex]\[ x = \frac{-18 \pm \sqrt{484}}{10} \][/tex]
[tex]\[ x = \frac{-18 \pm 22}{10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-18 + 22}{10} = \frac{4}{10} = \frac{2}{5} \][/tex]
[tex]\[ x = \frac{-18 - 22}{10} = \frac{-40}{10} = -4 \][/tex]
Step 5: Compile all the solutions.
The values of [tex]\( x \)[/tex] that make the determinant zero are:
[tex]\[ x = -4, \quad x = 0, \quad x = \frac{2}{5} \][/tex]
So, the determinant of the matrix is [tex]\( 5x^3 + 18x^2 - 8x \)[/tex], and the solutions to the equation [tex]\( 5x^3 + 18x^2 - 8x = 0 \)[/tex] are [tex]\( x = -4, \ x = 0,\)[/tex] and [tex]\( x = \frac{2}{5} \)[/tex].
[tex]\[ \left|\begin{array}{ccc}5 x^2 + 18 x - 8 & x & x^2 \\ 0 & 1 & -1 \\ 0 & -1 & x + 1\end{array}\right| \][/tex]
Step 1: Calculate the determinant of the matrix.
To find the determinant, we can use the method of cofactor expansion along the first row. The determinant [tex]\( \Delta \)[/tex] is:
[tex]\[ \Delta = \left(5x^2 + 18x - 8\right) \cdot \left|\begin{array}{cc} 1 & -1 \\ -1 & x + 1\end{array}\right| - x \cdot \left|\begin{array}{cc} 0 & -1 \\ 0 & x + 1\end{array}\right| + x^2 \cdot \left|\begin{array}{cc} 0 & 1 \\ 0 & -1\end{array}\right| \][/tex]
Step 2: Calculate the 2x2 determinants (minors).
[tex]\[ \left|\begin{array}{cc} 1 & -1 \\ -1 & x + 1 \end{array}\right| = (1) \cdot (x + 1) - (-1) \cdot (-1) = (x + 1) - 1 = x \][/tex]
[tex]\[ \left|\begin{array}{cc} 0 & -1 \\ 0 & x + 1 \end{array}\right| = 0 \cdot (x + 1) - (-1) \cdot 0 = 0 \][/tex]
[tex]\[ \left|\begin{array}{cc} 0 & 1 \\ 0 & -1 \end{array}\right| = 0 \cdot (-1) - (1) \cdot 0 = 0 \][/tex]
Step 3: Substitute these minors back into the original determinant equation.
[tex]\[ \Delta = (5x^2 + 18x - 8) \cdot x - x \cdot 0 + x^2 \cdot 0 = (5x^2 + 18x - 8) \cdot x \][/tex]
[tex]\[ \Delta = 5x^3 + 18x^2 - 8x \][/tex]
Step 4: Set the determinant equal to zero and solve for [tex]\( x \)[/tex].
[tex]\[ 5x^3 + 18x^2 - 8x = 0 \][/tex]
Factor out [tex]\( x \)[/tex]:
[tex]\[ x(5x^2 + 18x - 8) = 0 \][/tex]
This gives us one solution:
[tex]\[ x = 0 \][/tex]
To find the other solutions, solve the quadratic equation [tex]\( 5x^2 + 18x - 8 = 0 \)[/tex].
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 18 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 5 \cdot (-8)}}{2 \cdot 5} \][/tex]
[tex]\[ x = \frac{-18 \pm \sqrt{324 + 160}}{10} \][/tex]
[tex]\[ x = \frac{-18 \pm \sqrt{484}}{10} \][/tex]
[tex]\[ x = \frac{-18 \pm 22}{10} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{-18 + 22}{10} = \frac{4}{10} = \frac{2}{5} \][/tex]
[tex]\[ x = \frac{-18 - 22}{10} = \frac{-40}{10} = -4 \][/tex]
Step 5: Compile all the solutions.
The values of [tex]\( x \)[/tex] that make the determinant zero are:
[tex]\[ x = -4, \quad x = 0, \quad x = \frac{2}{5} \][/tex]
So, the determinant of the matrix is [tex]\( 5x^3 + 18x^2 - 8x \)[/tex], and the solutions to the equation [tex]\( 5x^3 + 18x^2 - 8x = 0 \)[/tex] are [tex]\( x = -4, \ x = 0,\)[/tex] and [tex]\( x = \frac{2}{5} \)[/tex].
