Understanding the Determinant Expansion and Finding 'e'

The problem asks us to find the value of the constant term 'e' in the polynomial expansion of a given 3x3 determinant. The determinant is given as:

\(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x \end{array}\right|\)

This determinant is stated to be equal to a polynomial of degree 4:

\(ax^4 + bx^3 + cx^2 + dx + e\)

Here, 'e' is the constant term, which means it's the value of the polynomial when \(x=0\). Similarly, the value of the determinant when \(x=0\) will give us the constant term 'e'.

Step-by-Step Calculation of 'e'

To find the value of 'e', we substitute \(x=0\) into the determinant:

The determinant becomes:

\(\left|\begin{array}{ccc} 0^2+3(0) & 0-1 & 0+3 \\ 0+1 & -2(0) & 0-4 \\ 0-3 & 0+4 & 3(0) \end{array}\right|\)

This simplifies to:

\(\left|\begin{array}{ccc} 0 & -1 & 3 \\ 1 & 0 & -4 \\ -3 & 4 & 0 \end{array}\right|\)

Now, we need to calculate the value of this 3x3 determinant. We can expand along the first row for simplicity:

Determinant value \( = 0 \cdot C_{11} + (-1) \cdot C_{12} + 3 \cdot C_{13}\)

Where \(C_{ij}\) is the cofactor of the element in the \(i\)-th row and \(j\)-th column.

• Cofactor \(C_{11}\): \( (-1)^{1+1} \cdot \left|\begin{array}{cc} 0 & -4 \\ 4 & 0 \end{array}\right| = 1 \cdot (0 \cdot 0 - (-4) \cdot 4) = 1 \cdot (0 + 16) = 16 \)
• Cofactor \(C_{12}\): \( (-1)^{1+2} \cdot \left|\begin{array}{cc} 1 & -4 \\ -3 & 0 \end{array}\right| = -1 \cdot (1 \cdot 0 - (-4) \cdot (-3)) = -1 \cdot (0 - 12) = -1 \cdot (-12) = 12 \)
• Cofactor \(C_{13}\): \( (-1)^{1+3} \cdot \left|\begin{array}{cc} 1 & 0 \\ -3 & 4 \end{array}\right| = 1 \cdot (1 \cdot 4 - 0 \cdot (-3)) = 1 \cdot (4 - 0) = 4 \)

Now, substitute these cofactor values back into the determinant expansion formula:

Determinant value \( = 0 \cdot (16) + (-1) \cdot (12) + 3 \cdot (4) \)

Determinant value \( = 0 - 12 + 12 \)

Determinant value \( = 0 \)

So, the value of the determinant when \(x=0\) is 0.

Since the determinant is equal to \(ax^4 + bx^3 + cx^2 + dx + e\), substituting \(x=0\) into the polynomial gives:

\(e = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e\)

\(e = 0 + 0 + 0 + 0 + e\)

Wait, that's not right. Substituting \(x=0\) into the polynomial expression \(ax^4 + bx^3 + cx^2 + dx + e\) yields the value of the expression at \(x=0\), which is exactly 'e'. So, the value of the determinant at \(x=0\) is equal to 'e'.

Value of determinant at \(x=0\) = \(e\)

\(0 = e\)

Thus, the value of 'e' is 0.

Summary of Finding Constant Term

The constant term 'e' in a polynomial \(P(x) = ax^4 + bx^3 + cx^2 + dx + e\) is simply the value of the polynomial when \(x=0\), i.e., \(P(0) = e\).

In this problem, the polynomial is the expansion of the determinant. Therefore, the constant term 'e' is the value of the determinant when \(x=0\).

We calculated the determinant with \(x=0\):

Expanding this determinant gave us a value of 0.

Therefore, \(e = 0\).

Revision Table: Determinant and Polynomials

Additional Information: Properties of Determinants

• Determinants are only defined for square matrices.
• The determinant of a 3x3 matrix can be calculated using the cofactor expansion method or the Sarrus' rule.
• If a row or column of a determinant consists entirely of zeros, the value of the determinant is zero. In our case, when \(x=0\), no row or column is entirely zero, so we must calculate the determinant.
• The degree of the polynomial resulting from expanding a determinant where entries are linear or quadratic functions of \(x\) can be estimated, but calculating the value at \(x=0\) directly finds the constant term 'e'.