Solving the Matrix Determinant Equation

To find the value of \(x\) from the given matrix determinant equation, we first need to calculate the determinant of the provided 3x3 matrix and then set it equal to zero. The question asks for the value of \(x\) when the determinant is equal to 0.

Determinant Calculation Steps

The given matrix is:

\(\begin{vmatrix}1&1&0\\\ \rm x^2+2x+2&1&0\\\ 2&1&1\end{vmatrix}=0\)

We can calculate the determinant by expanding along any row or column. Expanding along the third column is most efficient because it contains two zero elements, which simplifies the calculation significantly. The formula for expanding a 3x3 determinant along the third column is:

Determinant \(= a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}\)

Where \(a_{ij}\) is the element in the \(i\)-th row and \(j\)-th column, and \(C_{ij}\) is its cofactor. A cofactor \(C_{ij}\) is given by \((-1)^{i+j}M_{ij}\), where \(M_{ij}\) is the minor (the determinant of the submatrix obtained by deleting the \(i\)-th row and \(j\)-th column).

From the given matrix:

• The element \(a_{13} = 0\).
• The element \(a_{23} = 0\).
• The element \(a_{33} = 1\).

So, the determinant simplifies to:

Determinant \(= 0 \times C_{13} + 0 \times C_{23} + 1 \times C_{33}\)

Determinant \(= 1 \times C_{33}\)

Now, we need to find the cofactor \(C_{33}\). This involves finding the minor \(M_{33}\) and multiplying by \((-1)^{3+3}\).

To find \(M_{33}\), we remove the 3rd row and 3rd column from the original matrix:

\(M_{33} = \begin{vmatrix}1&1\\\ \rm x^2+2x+2&1\end{vmatrix}\)

Next, we calculate this 2x2 determinant:

\(M_{33} = (1 \times 1) - (1 \times (x^2+2x+2))\)

\(M_{33} = 1 - (x^2+2x+2)\)

\(M_{33} = 1 - x^2 - 2x - 2\)

\(M_{33} = -x^2 - 2x - 1\)

Since \(C_{33} = (-1)^{3+3}M_{33} = (-1)^6M_{33} = 1 \times M_{33}\), the determinant of the original matrix is:

Determinant \(= -x^2 - 2x - 1\)

Solving for X Value

The problem states that the determinant is equal to 0. Therefore, we set the expression for the determinant to zero:

\(-x^2 - 2x - 1 = 0\)

To solve this quadratic equation, we can multiply the entire equation by -1 to make the leading coefficient positive:

\(-1 \times (-x^2 - 2x - 1) = -1 \times 0\)

\(x^2 + 2x + 1 = 0\)

This equation is a perfect square trinomial, which can be factored as \((a+b)^2 = a^2 + 2ab + b^2\). In this case, \(a=x\) and \(b=1\). So, the equation becomes:

\((x+1)^2 = 0\)

To find the value of \(x\), we take the square root of both sides of the equation:

\(\sqrt{(x+1)^2} = \sqrt{0}\)

\(x+1 = 0\)

Finally, subtract 1 from both sides to isolate \(x\):

\(x = -1\)

Conclusion

The value of \(x\) that satisfies the given matrix determinant equation, making the determinant equal to zero, is \(-1\).