JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 10)
Suppose the vectors x1, x2 and x3 are the
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if
$${x_1} = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, $${x_2} = \left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right]$$, $${x_3} = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right]$$
$${b_1} = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$, $${b_2} = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$ and $${b_3} = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$,
then the determinant of A is equal to :
solutions of the system of linear equations,
Ax = b when the vector b on the right side is equal to b1, b2 and b3 respectively. if
$${x_1} = \left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right]$$, $${x_2} = \left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right]$$, $${x_3} = \left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right]$$
$${b_1} = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$, $${b_2} = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$ and $${b_3} = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$,
then the determinant of A is equal to :
$${3 \over 2}$$
4
2
$${1 \over 2}$$
Explanation
Let A = $$\left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_4}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right]$$
For Ax1 = b1 :
$$ \Rightarrow $$ $$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$
$$ \therefore $$ $${a_1} + {a_2} + {a_3} = 1$$ ....(1)
$${a_4} + {a_5} + {a_6} = 0$$ ......(2)
$${a_7} + {a_8} + {a_9} = 0$$ .....(3)
For Ax2 = b2 :
$$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$
$$ \therefore $$ $$2{a_2} + {a_3} = 0$$ .....(4)
$$2{a_5} + {a_6} = 2$$ ....(5)
$$2{a_8} + {a_9} = 0$$ ....(6)
For Ax3 = b3 :
$$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$
$$ \therefore $$ $${{a_3} = 0}$$
$${{a_6} = 0}$$
$${{a_9} = 2}$$
Putting value of $${{a_3}}$$ in equation (4), we get
$${{a_2}}$$ = 0
Putting value of $${{a_6}}$$ in equation (5), we get
$${{a_5}}$$ = 1
Putting value of $${{a_9}}$$ in equation (6), we get
$${{a_8}}$$ = -1
Putting value of $${{a_2}}$$ and $${{a_3}}$$ in equation (1), we get
$${{a_1}}$$ = 1
Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
$${{a_4}}$$ = -1
Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
$${{a_4}}$$ = -1
Putting value of $${{a_8}}$$ and $${{a_9}}$$ in equation (6), we get
$${{a_7}}$$ = -1
$$ \therefore $$ A = $$\left[ {\matrix{ 1 & 0 & 0 \cr { - 1} & 1 & 0 \cr { - 1} & { - 1} & 2 \cr } } \right]$$
So, |A| = 2(1) = 2
For Ax1 = b1 :
$$ \Rightarrow $$ $$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 1 \cr 1 \cr 1 \cr } } \right] = \left[ {\matrix{ 1 \cr 0 \cr 0 \cr } } \right]$$
$$ \therefore $$ $${a_1} + {a_2} + {a_3} = 1$$ ....(1)
$${a_4} + {a_5} + {a_6} = 0$$ ......(2)
$${a_7} + {a_8} + {a_9} = 0$$ .....(3)
For Ax2 = b2 :
$$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 2 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 2 \cr 0 \cr } } \right]$$
$$ \therefore $$ $$2{a_2} + {a_3} = 0$$ .....(4)
$$2{a_5} + {a_6} = 2$$ ....(5)
$$2{a_8} + {a_9} = 0$$ ....(6)
For Ax3 = b3 :
$$\left[ {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right]\left[ {\matrix{ 0 \cr 0 \cr 1 \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 2 \cr } } \right]$$
$$ \therefore $$ $${{a_3} = 0}$$
$${{a_6} = 0}$$
$${{a_9} = 2}$$
Putting value of $${{a_3}}$$ in equation (4), we get
$${{a_2}}$$ = 0
Putting value of $${{a_6}}$$ in equation (5), we get
$${{a_5}}$$ = 1
Putting value of $${{a_9}}$$ in equation (6), we get
$${{a_8}}$$ = -1
Putting value of $${{a_2}}$$ and $${{a_3}}$$ in equation (1), we get
$${{a_1}}$$ = 1
Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
$${{a_4}}$$ = -1
Putting value of $${{a_5}}$$ and $${{a_6}}$$ in equation (6), we get
$${{a_4}}$$ = -1
Putting value of $${{a_8}}$$ and $${{a_9}}$$ in equation (6), we get
$${{a_7}}$$ = -1
$$ \therefore $$ A = $$\left[ {\matrix{ 1 & 0 & 0 \cr { - 1} & 1 & 0 \cr { - 1} & { - 1} & 2 \cr } } \right]$$
So, |A| = 2(1) = 2
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