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### AIEEE 2006

If $$A$$ and $$B$$ are square matrices of size $$n\, \times \,n$$ such that
$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right),$$ then which of the following will be always true?
A
$$A=B$$
B
$$AB=BA$$
C
either of $$A$$ or $$B$$ is a zero matrix
D
either of $$A$$ or $$B$$ is identity matrix

## Explanation

$${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$$

$${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$$

$$\Rightarrow AB = BA$$
2

### AIEEE 2005

If $${a^2} + {b^2} + {c^2} = - 2$$ and

f$$\left( x \right) = \left| {\matrix{ {1 + {a^2}x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {\left( {1 + {a^2}} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}} \right)x} \cr {\left( {1 + {a^2}} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|,$$

then f$$(x)$$ is a polynomial of degree

A
$$1$$
B
$$0$$
C
$$3$$
D
$$2$$

## Explanation

Applying, $${C_1} \to {C_1} + {C_2} + {C_3}\,\,\,$$ we get

$$f\left( x \right) = \left| {\matrix{ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$$

$$= \left| {\matrix{ 1 & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr 1 & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$$

$$\left[ \, \right.$$ As given that $${a^2} + {b^2} + {c^2} = - 2$$ $$\left. {} \right]$$

$$\therefore$$ $${a^2} + {b^2} + {c^2} + 2 = 0$$

Applying $${R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3}$$

$$\therefore$$ $$f\left( x \right) = \left| {\matrix{ 0 & {x - 1} & 0 \cr 0 & {1 - x} & {x - 1} \cr 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$$

$$f\left( x \right) = {\left( {x - 1} \right)^2}$$

Hence degree $$=2.$$
3

### AIEEE 2005

If $${a_1},{a_2},{a_3},........,{a_n},.....$$ are in G.P., then the determinant $$\Delta \left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$\$
is equal to
A
$$1$$
B
$$0$$
C
$$4$$
D
$$2$$

## Explanation

As $$\,\,\,\,{a_1},{a_2},{a_3},.........$$ are in $$G.P.$$

$$\therefore$$ Using $${a_n} = a{r^{n - 1}},\,\,\,$$ we get the given determinant,

as $$\,\,\,\,\,\,\,\left| {\matrix{ {\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \cr {\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \cr {\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \cr } } \right|$$

Operating $${C_3} - {C_2}$$ and $${C_2} - {C_1}$$ and using

$$\log m - \log n = \log {m \over n}\,\,\,\,$$ we get

$$= \left| {\matrix{ {\log a{r^{n - 1}}} & {\log r} & {\log r} \cr {\log a{r^{n + 2}}} & {\log r} & {\log r} \cr {\log a{r^{n + 5}}} & {\log r} & {\log r} \cr } } \right|$$

$$=0$$ (two columns being identical)
4

### AIEEE 2005

The system of equations

$$\matrix{ {\alpha \,x + y + z = \alpha - 1} \cr {x + \alpha y + z = \alpha - 1} \cr {x + y + \alpha \,z = \alpha - 1} \cr }$$

has infinite solutions, if $$\alpha$$ is

A
$$-2$$
B
either $$-2$$ or $$1$$
C
not $$-2$$
D
$$1$$

## Explanation

$$ax + y + z = \alpha - 1$$

$$x + \alpha \,y + z = \alpha - 1;$$

$$x + y + z\alpha = \alpha - 1$$

$$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & 1 & \alpha \cr } } \right|$$

$$= \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)$$

$$= \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)$$

For infinite solutions, $$\Delta = 0$$

$$\Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0$$

$$\Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0$$

$$\Rightarrow \alpha = - 2,1;$$

But $$\alpha \ne 1.\,\,\,$$ $$\therefore$$ $$\,\,\alpha = - 2$$

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