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Exercise 7.2 · Q8

Q.If ∣abaα+bbcbα+caα+bbα+c0∣=0,\begin{vmatrix} a & b & a\alpha+b \\ b & c & b\alpha+c \\ a\alpha+b & b\alpha+c & 0 \end{vmatrix} = 0, prove that a,b,ca, b, c are in G.P. or α\alpha is a root of ax2+2bx+c=0ax^2 + 2bx + c = 0.

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The column operation C3→C3−αC1−C2C_3 \to C_3 - \alpha C_1 - C_2 collapses the determinant to −(aα2+2bα+c)(ac−b2)-(a\alpha^2+2b\alpha+c)(ac-b^2); setting it to 00 gives the two stated alternatives.

We use that a column operation of the form C3→C3−αC1−C2C_3 \to C_3 - \alpha C_1 - C_2 leaves the value unchanged.

Step 1. Apply C3→C3−αC1−C2C_3 \to C_3 - \alpha C_1 - C_2:

  • Row 1: (aα+b)−αa−b=0(a\alpha+b) - \alpha a - b = 0
  • Row 2: (bα+c)−αb−c=0(b\alpha+c) - \alpha b - c = 0
  • Row 3: 0−α(aα+b)−(bα+c)=−(aα2+2bα+c)0 - \alpha(a\alpha+b) - (b\alpha+c) = -(a\alpha^2 + 2b\alpha + c)

Step 2. The determinant becomes

∣ab0bc0aα+bbα+c−(aα2+2bα+c)∣.\begin{vmatrix} a & b & 0 \\ b & c & 0 \\ a\alpha+b & b\alpha+c & -(a\alpha^2+2b\alpha+c) \end{vmatrix}.

Step 3. Expand along the third column (only the (3,3)(3,3) entry is nonzero):

=−(aα2+2bα+c)∣abbc∣=−(aα2+2bα+c)(ac−b2).= -(a\alpha^2+2b\alpha+c)\begin{vmatrix} a & b \\ b & c \end{vmatrix} = -(a\alpha^2+2b\alpha+c)(ac - b^2). …

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