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Exercise 7.2 · Q3

Q.Prove that ∣a2bcac+c2a2+abb2acabb2+bcc2∣=4a2b2c2.\begin{vmatrix} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix} = 4a^2b^2c^2.

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Factor a,b,ca,b,c from the three columns, then reduce the remaining 3×33\times3 determinant with column and row operations to obtain 4abc4abc; multiplying back gives 4a2b2c24a^2b^2c^2.

We use that a common factor of a column can be taken outside, and that adding a multiple of one column/row to another does not change the value.

Step 1. Factor: C1=a(a, a+b, b)TC_1 = a(a,\,a+b,\,b)^T, C2=b(c, b, b+c)TC_2 = b(c,\,b,\,b+c)^T, C3=c(a+c, a, c)TC_3 = c(a+c,\,a,\,c)^T. Taking a,b,ca, b, c out:

D=abc∣aca+ca+bbabb+cc∣.D = abc\begin{vmatrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}.

Step 2. Apply C3→C3−C1−C2C_3 \to C_3 - C_1 - C_2: the column becomes (0, −2b, −2b)T(0,\,-2b,\,-2b)^T, giving

D=abc∣ac0a+bb−2bbb+c−2b∣.D = abc\begin{vmatrix} a & c & 0 \\ a+b & b & -2b \\ b & b+c & -2b \end{vmatrix}.

Step 3. Apply R3→R3−R2R_3 \to R_3 - R_2: bottom row becomes (−a, c, 0)(-a,\,c,\,0). Then R1→R1+R3R_1 \to R_1 + R_3: top row becomes (0, 2c, 0)(0,\,2c,\,0):

D=abc∣02c0a+bb−2b−ac0∣.D = abc\begin{vmatrix} 0 & 2c & 0 \\ a+b & b & -2b \\ -a & c & 0 \end{vmatrix}.

Step 4. Expand along the first row (only the 2c2c entry survives): the inner determinant =2c⋅(−1)1+2∣a+b−2b−a0∣=2c⋅(−1)(−2ab)=4abc= 2c\cdot(-1)^{1+2}\begin{vmatrix} a+b & -2b \\ -a & 0 \end{vmatrix} = 2c\cdot(-1)(-2ab) = 4abc.

Step 5. Therefore D=abc⋅4abc=4a2b2c2D = abc \cdot 4abc = 4a^2b^2c^2.

✓Final answer

∣a2bcac+c2a2+abb2acabb2+bcc2∣=4a2b2c2\begin{vmatrix} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix} = 4a^2b^2c^2.

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