JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 3)
Let S be the set of all column matrices $$\left[ {\matrix{
{{b_1}} \cr
{{b_2}} \cr
{{b_3}} \cr
} } \right]$$ such that $${b_1},{b_2},{b_3} \in R$$ and the system of equations (in real variables)
$$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $$\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr {{b_3}} \cr } } \right]$$$$ \in $$S?
$$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $$
has at least one solution. Then, which of the following system(s) (in real variables) has (have) at least one solution for each $$\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr {{b_3}} \cr } } \right]$$$$ \in $$S?
$$x + 2y + 3z = {b_1}$$, $$\,4y + 5z = {b_2}$$ and $$x + 2y + 6z = {b_3}$$
$$x + y + 3z = {b_1}$$, $$5x + 2y + 6z = {b_2}$$ and $$ - 2x - y - 3z = {b_3}$$
$$ - x + 2y - 5z = {b_1}$$, $$\,2x - 4y + 10z = {b_2}$$ and $$x - 2y + 5z = {b_3}$$
$$x + 2y + 5z = {b_1}$$, $$2x + 3z = {b_2}$$ and $$x + 4y - 5z = {b_3}$$
Explanation
We have,
$$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $$
has at least one solution.
$$ \therefore $$ $$D = \left| {\matrix{ { - 1} & 2 & 5 \cr 2 & { - 4} & 3 \cr 1 & { - 2} & 2 \cr } } \right|$$
and $${D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow {D_1} = \left| {\matrix{ {{b_1}} & 2 & 5 \cr {{b_2}} & { - 4} & 3 \cr {{b_3}} & { - 2} & 2 \cr } } \right|$$
$$ = - 2{b_1} - 14{b_2} + 26{b_3} = 0$$
$$ \Rightarrow {b_1} + 7{b_2} = 13{b_3}$$ ....(i)
(a) $$ \Rightarrow D = \left| {\matrix{ 1 & 2 & 3 \cr 0 & 4 & 5 \cr 1 & 2 & 6 \cr } } \right| = 1(24 - 10) + 1(10 - 12)$$
$$ = 14 - 2 = 12 \ne 0$$
Here, D $$ \ne $$ 0 $$ \Rightarrow $$ unique solution for any b1, b2, b3.
(b) $$D = \left| {\matrix{ 1 & 1 & 3 \cr 5 & 2 & 6 \cr { - 2} & { - 1} & { - 3} \cr } } \right|$$
$$ = 1( - 6 + 6) - 1( - 15 + 12) + 3( - 5 + 4) = 0$$
For at least one solution
$${D_1} = {D_2} = {D_3} = 0$$
Now, $${D_1} = \left| {\matrix{ {{b_1}} & 1 & 3 \cr {{b_2}} & 2 & 6 \cr {{b_3}} & { - 1} & { - 3} \cr } } \right|$$
$$ = {b_1}( - 6 + 6) - {b_2}( - 3 + 3) + {b_3}(6 - 6) = 0$$
$${D_2} = \left| {\matrix{ 1 & {{b_1}} & 3 \cr 5 & {{b_2}} & 6 \cr { - 2} & {{b_3}} & { - 3} \cr } } \right|$$
$$ = {b_1}( - 15 + 12) + {b_2}( - 3 + 6) - {b_3}(6 - 15)$$
$$ = 3{b_1} + 3{b_2} + 9{b_3} = 0 \Rightarrow {b_1} + {b_2} + 3{b_3} = 0$$
not satisfies the Eq. (i)
$$ \therefore $$ It has no solution.
(c) $$D = \left| {\matrix{ { - 1} & 2 & { - 5} \cr 2 & { - 4} & {10} \cr 1 & { - 2} & 5 \cr } } \right|$$
= $$ - 1( - 20 + 20) - 2(10 - 10) - 5( - 4 + 4) = 0$$
Here, b2 = $$-$$2b1 and b3 = $$-$$b1 satisfies the Eq. (i)
Planes are parallel.
(d) $$D = \left| {\matrix{ 1 & 2 & 5 \cr 2 & 0 & 3 \cr 1 & 4 & { - 5} \cr } } \right|$$
$$ = 1(0 - 12) - 2( - 10 - 3) + 5(8 - 0) = 54$$
$$D \ne 0$$
$$ \therefore $$ It has unique solutino for any b1, b2, b3.
$$\eqalign{ & - x + 2y + 5z = {b_1} \cr & 2x - 4y + 3z = {b_2} \cr & x - 2y + 2z = {b_3} \cr} $$
has at least one solution.
$$ \therefore $$ $$D = \left| {\matrix{ { - 1} & 2 & 5 \cr 2 & { - 4} & 3 \cr 1 & { - 2} & 2 \cr } } \right|$$
and $${D_1} = {D_2} = {D_3} = 0$$
$$ \Rightarrow {D_1} = \left| {\matrix{ {{b_1}} & 2 & 5 \cr {{b_2}} & { - 4} & 3 \cr {{b_3}} & { - 2} & 2 \cr } } \right|$$
$$ = - 2{b_1} - 14{b_2} + 26{b_3} = 0$$
$$ \Rightarrow {b_1} + 7{b_2} = 13{b_3}$$ ....(i)
(a) $$ \Rightarrow D = \left| {\matrix{ 1 & 2 & 3 \cr 0 & 4 & 5 \cr 1 & 2 & 6 \cr } } \right| = 1(24 - 10) + 1(10 - 12)$$
$$ = 14 - 2 = 12 \ne 0$$
Here, D $$ \ne $$ 0 $$ \Rightarrow $$ unique solution for any b1, b2, b3.
(b) $$D = \left| {\matrix{ 1 & 1 & 3 \cr 5 & 2 & 6 \cr { - 2} & { - 1} & { - 3} \cr } } \right|$$
$$ = 1( - 6 + 6) - 1( - 15 + 12) + 3( - 5 + 4) = 0$$
For at least one solution
$${D_1} = {D_2} = {D_3} = 0$$
Now, $${D_1} = \left| {\matrix{ {{b_1}} & 1 & 3 \cr {{b_2}} & 2 & 6 \cr {{b_3}} & { - 1} & { - 3} \cr } } \right|$$
$$ = {b_1}( - 6 + 6) - {b_2}( - 3 + 3) + {b_3}(6 - 6) = 0$$
$${D_2} = \left| {\matrix{ 1 & {{b_1}} & 3 \cr 5 & {{b_2}} & 6 \cr { - 2} & {{b_3}} & { - 3} \cr } } \right|$$
$$ = {b_1}( - 15 + 12) + {b_2}( - 3 + 6) - {b_3}(6 - 15)$$
$$ = 3{b_1} + 3{b_2} + 9{b_3} = 0 \Rightarrow {b_1} + {b_2} + 3{b_3} = 0$$
not satisfies the Eq. (i)
$$ \therefore $$ It has no solution.
(c) $$D = \left| {\matrix{ { - 1} & 2 & { - 5} \cr 2 & { - 4} & {10} \cr 1 & { - 2} & 5 \cr } } \right|$$
= $$ - 1( - 20 + 20) - 2(10 - 10) - 5( - 4 + 4) = 0$$
Here, b2 = $$-$$2b1 and b3 = $$-$$b1 satisfies the Eq. (i)
Planes are parallel.
(d) $$D = \left| {\matrix{ 1 & 2 & 5 \cr 2 & 0 & 3 \cr 1 & 4 & { - 5} \cr } } \right|$$
$$ = 1(0 - 12) - 2( - 10 - 3) + 5(8 - 0) = 54$$
$$D \ne 0$$
$$ \therefore $$ It has unique solutino for any b1, b2, b3.
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