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Exercise 7.2 · Q2

Q.Show that ∣b+cbcb2c2c+acac2a2a+baba2b2∣=0.\begin{vmatrix} b+c & bc & b^2c^2 \\ c+a & ca & c^2a^2 \\ a+b & ab & a^2b^2 \end{vmatrix} = 0.

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✓ Free question

Scaling the rows by a,b,ca,b,c exposes that column 1 is a linear combination of columns 2 and 3, so the determinant is zero.

We use the property that if one column is a linear combination of the other columns, the determinant is 00.

Step 1. Multiply R1,R2,R3R_1, R_2, R_3 by a,b,ca, b, c respectively. This multiplies the determinant by abcabc, so

abc D=∣a(b+c)abcab2c2b(c+a)abca2bc2c(a+b)abca2b2c∣.abc\,D = \begin{vmatrix} a(b+c) & abc & ab^2c^2 \\ b(c+a) & abc & a^2bc^2 \\ c(a+b) & abc & a^2b^2c \end{vmatrix}.

Step 2. Take out abcabc common from C2C_2 and abcabc common from C3C_3:

abc D=(abc)2∣ab+ac1bcab+bc1caac+bc1ab∣.abc\,D = (abc)^2 \begin{vmatrix} ab+ac & 1 & bc \\ ab+bc & 1 & ca \\ ac+bc & 1 & ab \end{vmatrix}.

Step 3. Observe that C1=(ab+bc+ca) C2−C3C_1 = (ab+bc+ca)\,C_2 - C_3, since (ab+bc+ca)−bc=ab+ac(ab+bc+ca) - bc = ab+ac, and similarly for the other rows. So C1C_1 is a linear combination of C2C_2 and C3C_3.

Step 4. A determinant whose one column is a linear combination of the others is 00. Hence abc D=0abc\,D = 0, giving D=0D = 0.

✓Final answer

∣b+cbcb2c2c+acac2a2a+baba2b2∣=0\begin{vmatrix} b+c & bc & b^2c^2 \\ c+a & ca & c^2a^2 \\ a+b & ab & a^2b^2 \end{vmatrix} = 0.

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