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1

MCQ (Single Correct Answer)

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.

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

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.

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

MCQ (Single Correct Answer)

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 $$4$$

$$\left[ {\matrix{
1 & {...} & {...} \cr
{...} & 1 & {...} \cr
{...} & {...} & 1 \cr
} } \right]\,\,$$ 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.

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

MCQ (Single Correct Answer)

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

The system has

A

exactly $$3$$ solutions

B

a unique solution

C

no solution

D

infinitenumber of solutions

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

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

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

$$ \Rightarrow $$ No solution

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

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

$$ \Rightarrow $$ No solution

4

MCQ (Single Correct Answer)

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:** $$\left| A \right| = 1$$ .

where $$I$$ is $$2 \times 2$$ identity matrix. Define

$$Tr$$$$(A)=$$ sum of diagonal elements of $$A$$ and $$\left| A \right| = $$ determinant of matrix $$A$$.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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