ISI-2014-18

Question

Let 

$D_1 = det \begin{pmatrix}a & b & c\\x &y & z\\p& q & r\end{pmatrix}$ and $D_2 = det \begin{pmatrix}-x & a & -p\\y &-b & q\\z & -c & r\end{pmatrix}$

Then

• A. $D_1 = D_2$
• B. $D_1 = 2D_2$
• C. $D_1 = -D_2$
• D. $D_2 = 2D_1$

Answer 1

$D_1 = det\begin{pmatrix}a&b&c\\x&y&z\\p&q&r\end{pmatrix}$

$\qquad = a(yr-qz) -b(xr-pz)+c(xq-py)$

$\qquad =ayr-aqz-bxr+bpz+cxq-cpy$

$D_2 = det\begin{pmatrix}-x&a&-p\\y&-b&q\\z&-c&r\end{pmatrix}$

$\qquad = -x(-br+cq) -a(yr-qz)-p(-cy+bz)$

$\qquad =bxr-cxq-ayr+aqz+cpy-bpz = -D_1$

$∴\color{green}{ \text{Correct option is C)}}$ $\color{red}{D_1 = -D_2}$

Comments

one another way of proving it by 1) Interchange R₁ and R₂ 2) multiply by (-1) in R₂ ..3) Transpose the matrix 4) multiply by (-1) in R₁ ..