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1

### AIEEE 2011

Let $$A$$ and $$B$$ be two symmetric matrices of order $$3$$.
Statement - 1: $$A(BA)$$ and $$(AB)$$$$A$$ are symmetric matrices.
Statement - 2: $$AB$$ is symmetric matrix if matrix multiplication of $$A$$ with $$B$$ is commutative.
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

$$\therefore$$ $$\,\,\,\,\,A' = A,B' = B$$

Now $$\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$$

$$= \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$$

Similarly $$\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$$

So, $$A\left( {BA} \right)\,\,\,\,$$ and $$A\left( {BA} \right)\,\,\,\,$$ are symmetric matrices.

Again $$\left( {AB} \right)' = B'A' = BA$$

Now if $$BA=AB$$, then $$AB$$ is symmetric matrix.
2

### AIEEE 2010

The number of $$3 \times 3$$ non-singular matrices, with four entries as $$1$$ and all other entries as $$0$$, is
A
$$5$$
B
$$6$$
C
at least $$7$$
D
less than $$4$$

## Explanation

$$\left[ {\matrix{ 1 & {...} & {...} \cr {...} & 1 & {...} \cr {...} & {...} & 1 \cr } } \right]\,\,$$ are $$6$$ non-singular matrices because $$6$$

blanks will be filled by $$5$$ zeros and $$1$$ one.

Similarly, $$\left[ {\matrix{ {...} & {...} & 1 \cr {...} & 1 & {...} \cr 1 & {...} & {...} \cr } } \right]\,\,$$ are $$6$$ non-singular matrices.

So, required cases are more than $$7,$$ non-singular $$3 \times 3$$ matrices.
3

### AIEEE 2010

Consider the system of linear equations; $$\matrix{ {{x_1} + 2{x_2} + {x_3} = 3} \cr {2{x_1} + 3{x_2} + {x_3} = 3} \cr {3{x_1} + 5{x_2} + 2{x_3} = 1} \cr }$$\$
The system has
A
exactly $$3$$ solutions
B
a unique solution
C
no solution
D
infinitenumber of solutions

## Explanation

$$D = \left| {\matrix{ 1 & 2 & 1 \cr 2 & 3 & 1 \cr 3 & 5 & 2 \cr } } \right| = 0$$

$${D_1}\left| {\matrix{ 3 & 2 & 1 \cr 3 & 3 & 1 \cr 1 & 5 & 2 \cr } } \right| \ne 0$$

$$\Rightarrow$$ Given system, does not have any solution.

$$\Rightarrow$$ No solution
4

### AIEEE 2010

Let $$A$$ be a $$\,2 \times 2$$ matrix with non-zero entries and let $${A^2} = I,$$
where $$I$$ is $$2 \times 2$$ identity matrix. Define
$$Tr$$$$(A)=$$ sum of diagonal elements of $$A$$ and $$\left| A \right| =$$ determinant of matrix $$A$$.
Statement- 1: $$Tr$$$$(A)=0$$.
Statement- 2: $$\left| A \right| = 1$$ .
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

Let $$A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$$ where $$a,b,c,d$$ $$\ne 0$$

$${A^2} = \left( {\matrix{ a & b \cr c & d \cr } } \right)\left( {\matrix{ a & b \cr c & d \cr } } \right)$$

$$\Rightarrow {A^2} = \left( {\matrix{ {{a^2} + bc} & {ab + bd} \cr {ac + cd} & {bc + {d^2}} \cr } } \right)$$

$$\Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$$

$$ab + bd = ac + cd = 0$$

$$c \ne 0\,\,\,\,\,b \ne 0$$

$$\Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$$

$$\left| A \right| = ad - bc = - {a^2} - bc = - 1$$

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