Solving Determinant Equation for Alpha

The problem asks us to find the equation whose root is \(\alpha\) given a determinant \(\Delta(a, b, c, \alpha)\) and the condition \(\Delta(7, 4, 2, \alpha) = 0\).

The given determinant is: \[ \Delta(a, b, c, \alpha) = \left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0 \end{array}\right| \]

We are given that \(\Delta(7, 4, 2, \alpha) = 0\). This means we substitute \(a=7\), \(b=4\), and \(c=2\) into the determinant: \[ \Delta(7, 4, 2, \alpha) = \left|\begin{array}{ccc} 7 & 4 & 7 \alpha+4 \\ 4 & 2 & 4 \alpha+2 \\ 7 \alpha+4 & 4 \alpha+2 & 0 \end{array}\right| \]

Now, we need to evaluate this determinant and set it equal to zero. We can expand the determinant along the third row for simplicity, as it contains a zero element. The formula for expanding a \(3 \times 3\) determinant along the third row is: \[ \text{det} = (a \alpha+b) \times C_{31} + (b \alpha+c) \times C_{32} + 0 \times C_{33} \] where \(C_{ij}\) are the cofactors.

Let's calculate the cofactors \(C_{31}\) and \(C_{32}\) for the substituted values \(a=7, b=4, c=2\):

• Cofactor \(C_{31}\): This is \( (-1)^{3+1} \) times the determinant of the minor obtained by removing the 3rd row and 1st column. \[ C_{31} = (-1)^4 \left|\begin{array}{cc} 4 & 7 \alpha+4 \\ 2 & 4 \alpha+2 \end{array}\right| = 1 \times [(4)(4 \alpha+2) - (2)(7 \alpha+4)] \] \[ C_{31} = (16 \alpha + 8) - (14 \alpha + 8) = 16 \alpha + 8 - 14 \alpha - 8 = 2 \alpha \]
• Cofactor \(C_{32}\): This is \( (-1)^{3+2} \) times the determinant of the minor obtained by removing the 3rd row and 2nd column. \[ C_{32} = (-1)^5 \left|\begin{array}{cc} 7 & 7 \alpha+4 \\ 4 & 4 \alpha+2 \end{array}\right| = -1 \times [(7)(4 \alpha+2) - (4)(7 \alpha+4)] \] \[ C_{32} = -1 \times [(28 \alpha + 14) - (28 \alpha + 16)] = -1 \times [28 \alpha + 14 - 28 \alpha - 16] = -1 \times [-2] = 2 \]

Now substitute these cofactors back into the determinant expansion formula: \[ \Delta = (7 \alpha+4) \times C_{31} + (4 \alpha+2) \times C_{32} \] \[ \Delta = (7 \alpha+4)(2 \alpha) + (4 \alpha+2)(2) \] \[ \Delta = (14 \alpha^2 + 8 \alpha) + (8 \alpha + 4) \] \[ \Delta = 14 \alpha^2 + 16 \alpha + 4 \]

We are given that \(\Delta(7, 4, 2, \alpha) = 0\). So, we set the calculated determinant equal to zero: \[ 14 \alpha^2 + 16 \alpha + 4 = 0 \]

This is a quadratic equation in terms of \(\alpha\). We can simplify it by dividing the entire equation by 2: \[ \frac{14 \alpha^2}{2} + \frac{16 \alpha}{2} + \frac{4}{2} = \frac{0}{2} \] \[ 7 \alpha^2 + 8 \alpha + 2 = 0 \]

The question asks for the equation of which \(\alpha\) is a root. This quadratic equation \(7 \alpha^2 + 8 \alpha + 2 = 0\) shows the relationship that \(\alpha\) must satisfy. If we replace \(\alpha\) with a general variable like \(x\), the equation becomes \(7x^2 + 8x + 2 = 0\).

Comparing this with the given options:

Our derived equation \(7x^2 + 8x + 2 = 0\) matches Option 3. Therefore, \(\alpha\) is a root of the equation \(7x^2 + 8x + 2 = 0\).

Revision Table: Key Steps

Additional Information: Determinants and Roots

A determinant is a scalar value that can be computed from the elements of a square matrix. Determinants have many applications in linear algebra, including solving systems of linear equations, finding matrix inverses, and calculating areas and volumes.

• Determinant Expansion: For a \(3 \times 3\) matrix \(A = \left|\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|\), the determinant can be calculated by expanding along any row or column using cofactors. The cofactor \(C_{ij}\) of an element \(a_{ij}\) is given by \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the determinant of the minor matrix (the matrix obtained by removing the \(i\)-th row and \(j\)-th column).
• Determinant = 0: If the determinant of a matrix is zero, it means the matrix is singular. In the context of a system of linear equations represented by the matrix, a zero determinant implies that the system either has no unique solution or has infinitely many solutions. In this problem, setting the determinant involving \(\alpha\) to zero gives an equation that \(\alpha\) must satisfy for this specific condition to hold.
• Roots of an Equation: The roots of an equation are the values of the variable that satisfy the equation. In this case, \(\alpha\) is a root of the resulting quadratic equation because it is the value that makes the determinant equal to zero.

This problem combined the concept of evaluating a determinant with solving an algebraic equation (a quadratic equation in this case) derived from setting the determinant to zero.