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Exercise C · Q5

Q.Evaluate using properties of determinants:

(i) ∣b−cc−aa−bc−aa−bb−ca−bb−cc−a∣\begin{vmatrix} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{vmatrix}
(ii) ∣x+yy+zx+zzxy111∣\begin{vmatrix} x+y & y+z & x+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}
(iii) ∣x+yy+zx+zzxy111∣\begin{vmatrix} x+y & y+z & x+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}
(iv) ∣265240219240225198219198181∣\begin{vmatrix} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{vmatrix}
(v) ∣abca+2xb+2yc+2zxyz∣\begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}.
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Each determinant is 00 because in every case a row (or column) is a linear combination of the others.

A determinant is 00 if one row/column is a scalar multiple of, or a linear combination of, the other rows/columns (equivalently, applying Ri→Ri+λRjR_i\to R_i+\lambda R_j produces a zero or proportional row).

  1. (i) ∣b−cc−aa−bc−aa−bb−ca−bb−cc−a∣\begin{vmatrix}b-c&c-a&a-b\\c-a&a-b&b-c\\a-b&b-c&c-a\end{vmatrix}. Apply C1→C1+C2+C3C_1\to C_1+C_2+C_3. Each entry of the new C1C_1 is (b−c)+(c−a)+(a−b)=0(b-c)+(c-a)+(a-b)=0, so C1C_1 is all zeros ⇒\Rightarrow determinant =0=0.
  2. (ii) ∣x+yy+zx+zzxy111∣\begin{vmatrix}x+y&y+z&x+z\\z&x&y\\1&1&1\end{vmatrix}. Apply R1→R1+R2R_1\to R_1+R_2: the entries become (x+y+z, x+y+z, x+y+z)=(x+y+z)R3(x+y+z,\ x+y+z,\ x+y+z)=(x+y+z)R_3. Since R1R_1 is now a multiple of R3R_3, determinant =0=0.
  3. (iii) Identical to (ii): by the same operation R1→R1+R2=(x+y+z)R3R_1\to R_1+R_2=(x+y+z)R_3, so determinant =0=0. …

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