Menu Close

Question-57619




Question Number 57619 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
Answered by math1967 last updated on 09/Apr/19
 determinant (((q+r),(r+p),(p+q)),((y+z),(z+x),(x+y)))   determinant (((b+c−c−a−a−b),(c+a),(a+b)),((q+r−r−p−p−q),(r+p),(p+q)),((y+z−z−x−x−y),(z+x),(x+y)))C_1 −C_2 −C_3   −2 determinant ((a,(c+a),(a+b)),(p,(r+p),(p+q)),(x,(z+x),(x+y)))  =−2 determinant ((a,(c+a−a),(a+b−a)),(p,(r+p−p),(p+q−p)),(x,(z+x−x),(x+y−x)))C_2 −C_1 ,C_3 −C_1   =−2 determinant ((a,c,b),(p,r,q),(x,z,(y )))=2 determinant ((a,b,c),(p,q,r),(x,y,z))C_2 ⇔C_3
$$\begin{vmatrix}{{q}+{r}}&{{r}+{p}}&{{p}+{q}}\\{{y}+{z}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix} \\ $$$$\begin{vmatrix}{{b}+{c}−{c}−{a}−{a}−{b}}&{{c}+{a}}&{{a}+{b}}\\{{q}+{r}−{r}−{p}−{p}−{q}}&{{r}+{p}}&{{p}+{q}}\\{{y}+{z}−{z}−{x}−{x}−{y}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix}{C}_{\mathrm{1}} −{C}_{\mathrm{2}} −{C}_{\mathrm{3}} \\ $$$$−\mathrm{2}\begin{vmatrix}{{a}}&{{c}+{a}}&{{a}+{b}}\\{{p}}&{{r}+{p}}&{{p}+{q}}\\{{x}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix} \\ $$$$=−\mathrm{2}\begin{vmatrix}{{a}}&{{c}+{a}−{a}}&{{a}+{b}−{a}}\\{{p}}&{{r}+{p}−{p}}&{{p}+{q}−{p}}\\{{x}}&{{z}+{x}−{x}}&{{x}+{y}−{x}}\end{vmatrix}{C}_{\mathrm{2}} −{C}_{\mathrm{1}} ,{C}_{\mathrm{3}} −{C}_{\mathrm{1}} \\ $$$$=−\mathrm{2}\begin{vmatrix}{{a}}&{{c}}&{{b}}\\{{p}}&{{r}}&{{q}}\\{{x}}&{{z}}&{{y}\:}\end{vmatrix}=\mathrm{2}\begin{vmatrix}{{a}}&{{b}}&{{c}}\\{{p}}&{{q}}&{{r}}\\{{x}}&{{y}}&{{z}}\end{vmatrix}{C}_{\mathrm{2}} \Leftrightarrow{C}_{\mathrm{3}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by math1967 last updated on 09/Apr/19
You are welcome sir
$${You}\:{are}\:{welcome}\:{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
another way  ∣b+c    c+a   a+b ∣  ∣q+r    r+p    p+q∣  ∣y+z    z+x    x+y∣  =∣b  c  a∣ +∣b  c  b∣ +∣b   a  a∣+∣b  a b∣      ∣q  r   p∣     ∣q   r   q∣  ∣q    p    p∣   ∣q   p  q∣  +      ∣ y z x∣      ∣y   z   y∣  ∣y   x    x∣   ∣y  x  y∣       ∣c   c  a∣+∣c  c  b ∣+∣c  a  a∣+∣c  a  b∣    ∣r   r   p∣   ∣r   r  q∣    ∣r   p  p∣   ∣r   p   q∣     ∣z   z   x∣ ∣z    z  y∣   ∣z   x   x∣  ∣z   x   y∣    △=△_1 +△_2 +△_3 +△_4 +△_5 +△_6 +△_7 +△_8   now △_2 =△_3 =△_4 =△_5 =△_6 =△_7 =0  reason two identical collumn    so △=△_1 +△_8   now inter change of collumn keep value of   determinant unchanged  So △=2∣a b c∣                    ∣p  q  r∣                     ∣x   y  z∣ proved
$${another}\:{way} \\ $$$$\mid{b}+{c}\:\:\:\:{c}+{a}\:\:\:{a}+{b}\:\mid \\ $$$$\mid{q}+{r}\:\:\:\:{r}+{p}\:\:\:\:{p}+{q}\mid \\ $$$$\mid{y}+{z}\:\:\:\:{z}+{x}\:\:\:\:{x}+{y}\mid \\ $$$$=\mid{b}\:\:{c}\:\:{a}\mid\:+\mid{b}\:\:{c}\:\:{b}\mid\:+\mid{b}\:\:\:{a}\:\:{a}\mid+\mid{b}\:\:{a}\:{b}\mid \\ $$$$\:\:\:\:\mid{q}\:\:{r}\:\:\:{p}\mid\:\:\:\:\:\mid{q}\:\:\:{r}\:\:\:{q}\mid\:\:\mid{q}\:\:\:\:{p}\:\:\:\:{p}\mid\:\:\:\mid{q}\:\:\:{p}\:\:{q}\mid\:\:+ \\ $$$$\:\:\:\:\mid\:{y}\:{z}\:{x}\mid\:\:\:\:\:\:\mid{y}\:\:\:{z}\:\:\:{y}\mid\:\:\mid{y}\:\:\:{x}\:\:\:\:{x}\mid\:\:\:\mid{y}\:\:{x}\:\:{y}\mid \\ $$$$ \\ $$$$\:\:\:\mid{c}\:\:\:{c}\:\:{a}\mid+\mid{c}\:\:{c}\:\:{b}\:\mid+\mid{c}\:\:{a}\:\:{a}\mid+\mid{c}\:\:{a}\:\:{b}\mid \\ $$$$\:\:\mid{r}\:\:\:{r}\:\:\:{p}\mid\:\:\:\mid{r}\:\:\:{r}\:\:{q}\mid\:\:\:\:\mid{r}\:\:\:{p}\:\:{p}\mid\:\:\:\mid{r}\:\:\:{p}\:\:\:{q}\mid \\ $$$$\:\:\:\mid{z}\:\:\:{z}\:\:\:{x}\mid\:\mid{z}\:\:\:\:{z}\:\:{y}\mid\:\:\:\mid{z}\:\:\:{x}\:\:\:{x}\mid\:\:\mid{z}\:\:\:{x}\:\:\:{y}\mid \\ $$$$ \\ $$$$\bigtriangleup=\bigtriangleup_{\mathrm{1}} +\bigtriangleup_{\mathrm{2}} +\bigtriangleup_{\mathrm{3}} +\bigtriangleup_{\mathrm{4}} +\bigtriangleup_{\mathrm{5}} +\bigtriangleup_{\mathrm{6}} +\bigtriangleup_{\mathrm{7}} +\bigtriangleup_{\mathrm{8}} \\ $$$${now}\:\bigtriangleup_{\mathrm{2}} =\bigtriangleup_{\mathrm{3}} =\bigtriangleup_{\mathrm{4}} =\bigtriangleup_{\mathrm{5}} =\bigtriangleup_{\mathrm{6}} =\bigtriangleup_{\mathrm{7}} =\mathrm{0} \\ $$$${reason}\:{two}\:{identical}\:{collumn} \\ $$$$ \\ $$$${so}\:\bigtriangleup=\bigtriangleup_{\mathrm{1}} +\bigtriangleup_{\mathrm{8}} \\ $$$${now}\:{inter}\:{change}\:{of}\:{collumn}\:{keep}\:{value}\:{of}\: \\ $$$${determinant}\:{unchanged} \\ $$$${So}\:\bigtriangleup=\mathrm{2}\mid{a}\:{b}\:{c}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{p}\:\:{q}\:\:{r}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{x}\:\:\:{y}\:\:{z}\mid\:{proved} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *