What are the values of x that satisfy the equation \(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 0\;?\)
-2 ± √6
Finding the Values of x in a Determinant Equation
We are asked to find the values of \(x\) that satisfy the given equation involving the sum of two 3x3 determinants:
\(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| + \left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 0\;\) To solve this, we need to calculate the value of each determinant and then solve the resulting equation for \(x\).
Calculating the First Determinant
Let's calculate the first determinant: \(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right|\). We can expand this determinant along the first row:
\(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| = x \cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right| - 0 \cdot \left| {\begin{array}{*{20}{c}} {2x}&1\\ 1&1 \end{array}} \right| + 2 \cdot \left| {\begin{array}{*{20}{c}} {2x}&2\\ 1&1 \end{array}} \right|\)
Now, calculate the 2x2 determinants:
- \( \left| {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right| = (2 \cdot 1) - (1 \cdot 1) = 2 - 1 = 1 \)
- \( \left| {\begin{array}{*{20}{c}} {2x}&1\\ 1&1 \end{array}} \right| = (2x \cdot 1) - (1 \cdot 1) = 2x - 1 \)
- \( \left| {\begin{array}{*{20}{c}} {2x}&2\\ 1&1 \end{array}} \right| = (2x \cdot 1) - (2 \cdot 1) = 2x - 2 \)
Substitute these values back into the expansion:
\(\left| {\begin{array}{*{20}{c}} x&0&2\\ {2x}&2&1\\ 1&1&1 \end{array}} \right| = x(1) - 0(2x - 1) + 2(2x - 2) = x - 0 + 4x - 4 = 5x - 4\)
Calculating the Second Determinant
Next, let's calculate the second determinant: \(\left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right|\). We can expand this determinant along the first row:
\(\left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 3x \cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right| - 0 \cdot \left| {\begin{array}{*{20}{c}} {{x^2}}&1\\ 0&1 \end{array}} \right| + 2 \cdot \left| {\begin{array}{*{20}{c}} {{x^2}}&2\\ 0&1 \end{array}} \right|\)
Now, calculate the 2x2 determinants:
- \( \left| {\begin{array}{*{20}{c}} 2&1\\ 1&1 \end{array}} \right| = (2 \cdot 1) - (1 \cdot 1) = 2 - 1 = 1 \)
- \( \left| {\begin{array}{*{20}{c}} {{x^2}}&1\\ 0&1 \end{array}} \right| = (x^2 \cdot 1) - (1 \cdot 0) = x^2 - 0 = x^2 \)
- \( \left| {\begin{array}{*{20}{c}} {{x^2}}&2\\ 0&1 \end{array}} \right| = (x^2 \cdot 1) - (2 \cdot 0) = x^2 - 0 = x^2 \)
Substitute these values back into the expansion:
\(\left| {\begin{array}{*{20}{c}} {3x}&0&2\\ {{x^2}}&2&1\\ 0&1&1 \end{array}} \right| = 3x(1) - 0(x^2) + 2(x^2) = 3x - 0 + 2x^2 = 2x^2 + 3x\)
Forming and Solving the Equation
The original equation is the sum of the two determinants equal to zero:
\((5x - 4) + (2x^2 + 3x) = 0\)
Combine like terms to form a quadratic equation:
\(2x^2 + (5x + 3x) - 4 = 0\)
\(2x^2 + 8x - 4 = 0\)
We can simplify this equation by dividing the entire equation by 2:
\(x^2 + 4x - 2 = 0\)
This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a=1\), \(b=4\), and \(c=-2\). We can solve for \(x\) using the quadratic formula:
\(x = \frac{{ - b \pm \sqrt{{b^2 - 4ac}} }}{{2a}}\)
Substitute the values of \(a\), \(b\), and \(c\) into the formula:
\(x = \frac{{ - 4 \pm \sqrt{{4^2 - 4(1)(-2)}} }}{{2(1)}}\)
\(x = \frac{{ - 4 \pm \sqrt{{16 + 8}} }}{2}\)
\(x = \frac{{ - 4 \pm \sqrt{{24}} }}{2}\)
Simplify the square root term \(\sqrt{24}\):
\(\sqrt{24} = \sqrt{4 \cdot 6} = \sqrt{4} \cdot \sqrt{6} = 2\sqrt{6}\)
Substitute the simplified radical back into the equation for \(x\):
\(x = \frac{{ - 4 \pm 2\sqrt{{6}} }}{2}\)
Factor out a 2 from the numerator and cancel it with the denominator:
\(x = \frac{{2(-2 \pm \sqrt{{6}})}}{2}\)
\(x = -2 \pm \sqrt{{6}}\)
Thus, the values of \(x\) that satisfy the equation are \(x = -2 + \sqrt{6}\) and \(x = -2 - \sqrt{6}\).
The values of x are \(-2 \pm \sqrt{6}\).
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The determinant \(\left| {\begin{array}{*{20}{c}} {xp + y}&x&y\\ {yp + z}&y&z\\ 0&{xp + y}&{yp + z} \end{array}} \right| = 0,\) if
