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1

### AIEEE 2006

Let $$A = \left( {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right)$$ and $$B = \left( {\matrix{ a & 0 \cr 0 & b \cr } } \right),a,b \in N.$$ Then
A
there cannot exist any $$B$$ such that $$AB=BA$$
B
there exist more then one but finite number of $$B'$$s such that $$AB=BA$$
C
there exists exactly one $$B$$ such that $$AB=BA$$
D
there exist infinitely many $$B'$$s such that $$AB=BA$$

## Explanation

$$A = \left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right]\,\,\,\,B = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]$$

$$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$

$$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$

Hence, $$AB=BA$$ only when $$a=b$$

$$\therefore$$ There can be infinitely many $$B's$$

for which $$AB=BA$$
2

### 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

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

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

$$\Rightarrow AB = BA$$
3

### 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 $$=2.$$
4

### 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)

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