Let $$A$$ be a $$\,2 \times 2$$ matrix
Statement - 1 : $$adj\left( {adj\,A} \right) = A$$
Statement - 2 :$$\left| {adj\,A} \right| = \left| A \right|$$
A
statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.
B
statement - 1 is true, statement - 2 is false.
C
statement - 1 is false, statement -2 is true
D
statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.
Explanation
We know that $$\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$$b
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|$$
$$\therefore$$ $$\,\,\,\,\,\,$$ Both the statements are true and statement $$-2$$ is a correct explanation for statement - $$1.$$
3
AIEEE 2008
MCQ (Single Correct Answer)
Let $$A$$ be a square matrix all of whose entries are integers.
Then which one of the following is true?
A
If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are not necessarily integers
B
If det $$A \ne \pm 1,$$ then $${A^{ - 1}}$$ exists and all its entries are non integers
C
If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ exists but all its entries are integers
D
If det $$A = \pm 1,$$ then $${A^{ - 1}}$$ need not exists
Explanation
As all entries of square matrix $$A$$ are integers, therefore all co-factors should also be integers.
If det $$A = \pm 1\,\,$$ then $${A^{ - 1}}\,\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers.
4
AIEEE 2008
MCQ (Single Correct Answer)
Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to
A
$$2$$
B
$$-1$$
C
$$0$$
D
$$1$$
Explanation
The given equations are
$$\matrix{
{ - x + cy + bz = 0} \cr
{cx - y + az = 0} \cr
{bx + ay - z = 0} \cr
} $$
As $$x,y,z$$ are not all zero
$$\therefore$$ The above system should not have unique (zero) solution
$$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{
{ - 1} & c & b \cr
c & { - 1} & a \cr
b & a & { - 1} \cr
} } \right| = 0$$
