Determinants Matrices Question 176
Question: Consider the system of linear equations $ a_1x+b_1y+c_1z+d_1=0, $
$ a_2x+b_2y+c_2z+d_2=0, $
$ a_3x+b_3y+c_3z+d_3=0, $ Let us denote by $ \Delta (a,b,c) $ the determinant $ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix} $ , if $ \Delta (a,b,c)#0, $ then the value of x in the unique solution of the above equations is
Options:
A) $ \frac{\Delta (b,c,d)}{\Delta (a,b,c)} $
B) $ \frac{-\Delta (b,c,d)}{\Delta (a,b,c)} $
C) $ \frac{\Delta (a,c,d)}{\Delta (a,b,c)} $
D) $ -\frac{\Delta (a,b,d)}{\Delta (a,b,c)} $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] From the given system of equations, $ x=\frac{D_1}{D}, $
$ y=\frac{D_2}{D}, $
$ z=\frac{D_3}{D} $ where, $ D=\Delta (a,b,c); $
$ D_1=\Delta (d,b,c) $
$ D_2=\Delta (a,d,c); $
$ D_1=\Delta (a,b,d) $ Now, $ x=\frac{\Delta (d,b,c)}{\Delta (a,b,c)} $ where, $ \Delta (d,b,c)= \begin{vmatrix} -d_1 & b_1 & c_1 \\ -d_2 & b_2 & c_2 \\ -d_3 & b_3 & c_3 \\ \end{vmatrix} $
$ =- \begin{vmatrix} b_1 & -d_1 & c_1 \\ b_2 & -d_2 & c_2 \\ b_3 & -d_3 & c_3 \\ \end{vmatrix}=+ \begin{vmatrix} b_1 & c_1 & -d_1 \\ b_2 & c_2 & -d_2 \\ b_3 & c_3 & -d_3 \\ \end{vmatrix} $
$ =\Delta (b,c,d) $ Hence, $ x=\frac{\Delta (b,c,d)}{\Delta (a,b,c)} $
BETA
