Discover the best answers to your questions with the help of IDNLearn.com. Our Q&A platform offers reliable and thorough answers to ensure you have the information you need to succeed in any situation.
Sagot :
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].
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
