JEE Advance - Mathematics (2012 - Paper 2 Offline - No. 19)
If the ad joint of a 3 $$\times$$ 3 matrix P is $$\left[ {\matrix{
1 & 4 & 4 \cr
2 & 1 & 7 \cr
1 & 1 & 3 \cr
} } \right]$$, then the possible value(s) of the determinant of P is(are)
$$-$$2
$$-$$1
1
2
Explanation
Concept Involved If $$\left| {{A_{n \times n}}} \right| = \Delta $$, then $$\left| {adj\,A} \right| = {\Delta ^{n - 1}}$$
Here, $${P_{3 \times 3}} = \left[ {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right]$$
$$ \Rightarrow \left| {adj\,P} \right| = {\left| P \right|^2}$$
$$\therefore$$ $$\left| {adj\,P} \right| = \left| {\matrix{ 1 & 4 & 4 \cr 2 & 1 & 7 \cr 1 & 1 & 3 \cr } } \right|$$
$$ = 1(3 - 7) - 4(6 - 7) + 4(2 - 1)$$
$$ = - 4 + 4 + 4 = 4$$
$$ \Rightarrow \left| P \right| = \pm \,2$$
Comments (0)
