JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 4)
The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2 $$ \times $$ 2 matrix such that the trace of A is 3 and the trace of A3 is $$-$$18, then the value of the determinant of A is .............
Answer
5
Explanation
Let a square matrix 'A' of order 2 $$ \times $$ 2, such that tr(A) = 3, is
$$A = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
So, $${A^2} = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ = \left[ {\matrix{ {{x^2} + yz} & {xy + 3y - xy} \cr {xz + 3z - xz} & {yz + {{(3 - x)}^2}} \cr } } \right]$$
$$ \therefore $$ $${A^3} = \left[ {\matrix{ {{x^2} + yz} & {3y} \cr { + 3z} & {yz + 9 + {x^2} - 6x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ \because $$ $${t_r}({A^3}) = {x^3} + xyz + 3yz + 3yz + 3yz - xyz + 27 - 9x + 3{x^2} - {x^3} - 18x + 6{x^2}$$
$$ = 9yz + 27 - 27x + 9{x^2} = - 18$$ (given)
$$ \Rightarrow yz + 3 - 3x + {x^2} = - 2$$
$$ \Rightarrow 3x - {x^2} - yz = 5$$
$$ \because $$ $$|A|\, = \,\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ = 3x - {x^2} - yz = 5$$
$$A = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
So, $${A^2} = \left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ = \left[ {\matrix{ {{x^2} + yz} & {xy + 3y - xy} \cr {xz + 3z - xz} & {yz + {{(3 - x)}^2}} \cr } } \right]$$
$$ \therefore $$ $${A^3} = \left[ {\matrix{ {{x^2} + yz} & {3y} \cr { + 3z} & {yz + 9 + {x^2} - 6x} \cr } } \right]\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ \because $$ $${t_r}({A^3}) = {x^3} + xyz + 3yz + 3yz + 3yz - xyz + 27 - 9x + 3{x^2} - {x^3} - 18x + 6{x^2}$$
$$ = 9yz + 27 - 27x + 9{x^2} = - 18$$ (given)
$$ \Rightarrow yz + 3 - 3x + {x^2} = - 2$$
$$ \Rightarrow 3x - {x^2} - yz = 5$$
$$ \because $$ $$|A|\, = \,\left[ {\matrix{ x & y \cr z & {3 - x} \cr } } \right]$$
$$ = 3x - {x^2} - yz = 5$$
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