Get personalized and accurate responses to your questions with IDNLearn.com. Get the information you need quickly and accurately with our reliable and thorough Q&A platform.
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].
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is committed to providing the best answers. Thank you for visiting, and see you next time for more solutions.