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1

### AIEEE 2004

If $${a_1},{a_2},{a_3},.........,{a_n},......$$ are in G.P., then the value of the determinant

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|,$$ is

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

## Explanation

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$

$$= \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$$

$$= \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$$

$$= 0\left[ \, \right.$$ Apply $$\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$$
2

### AIEEE 2004

Let $$A = \left( {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right).$$ and $$10$$ $$B = \left( {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right)$$. if $$B$$ is

the inverse of matrix $$A$$, then $$\alpha$$ is

A
$$5$$
B
$$-1$$
C
$$2$$
D
$$-2$$

## Explanation

Given that $$10B$$ $$\,\,\, = \left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$

$$\Rightarrow B = {1 \over {10}}\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right]$$

Also since, $$B = {A^{ - 1}} \Rightarrow AB = I$$

$$\Rightarrow {1 \over {10}}\left[ {\matrix{ 1 & { - 1} & 1 \cr 2 & 1 & { - 3} \cr 1 & 1 & 1 \cr } } \right]\left[ {\matrix{ 4 & 2 & 2 \cr { - 5} & 0 & \alpha \cr 1 & { - 2} & 3 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$\Rightarrow {1 \over {10}}\left[ {\matrix{ {10} & 0 & {5 - 2} \cr 0 & {10} & { - 5 + \alpha } \cr 0 & 0 & {5 + \alpha } \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$

$$\Rightarrow {{5 - \alpha } \over {10}} = 0$$

$$\Rightarrow \alpha = 5$$
3

### AIEEE 2004

Let $$A = \left( {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right)$$. The only correct

statement about the matrix $$A$$ is

A
$${A^2} = 1$$
B
$$A=(-1)I,$$ where $$I$$ is a unit matrix
C
$${A^{ - 1}}$$ does not exist
D
$$A$$ is a zero matrix

## Explanation

$$A = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$$

clearly $$\,\,\,A \ne 0.\,$$ Also $$\,\,\left| A \right| = - 1 \ne 0$$

$$\therefore$$ $${A^{ - 1}}\,\,$$ exists, further

$$\left( { - 1} \right)I = \left[ {\matrix{ { - 1} & 0 & 0 \cr 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr } } \right] \ne A$$

Also $${A^2} = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]\left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$$

$$= \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right] = I$$
4

### AIEEE 2003

If the system of linear equations
$$x + 2ay + az = 0;$$ $$x + 3by + bz = 0;\,\,x + 4cy + cz = 0;$$
has a non - zero solution, then $$a, b, c$$.
A
satisfy $$a+2b+3c=0$$
B
are in A.P
C
are in G.P
D
are in H.P.

## Explanation

For homogeneous system of equations to have non zero solution, $$\Delta = 0$$

$$\left| {\matrix{ 1 & {2a} & a \cr 1 & {3b} & b \cr 1 & {4c} & c \cr } } \right| = 0\,{C_2} \to {C_2} - 2{C_3}$$

$$\left| {\matrix{ 1 & 0 & a \cr 1 & b & b \cr 1 & {2c} & c \cr } } \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}$$

$$\left| {\matrix{ 1 & 0 & a \cr 0 & b & {b - a} \cr 0 & {2c - b} & {c - b} \cr } } \right| = 0$$

$$b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0$$

On simplification, $${2 \over b} = {1 \over a} + {1 \over c}$$

$$\therefore$$ $$a,b,c$$ are in Harmonic Progression.

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