Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 7 Matrices and Determinants Ex 7.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.

If a_{ij} = \(\) (3i – 2j) and A = [a_{ij}]_{3 × 2} is

(1)

(2)

(3)

(4)

Answer:

(2)

Explaination:

Question 2.

What must be the matrix X, if

(1)

(2)

(3)

(4)

Answer:

(1)

Explaination:

Question 3.

Which one of the following is not true about the matrix \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \) ?.

(1) a scalar matrix

(2) a diagonal matrix

(3) an upper triangular matrix

(4) a lower triangular matrix

Answer:

(1) a scalar matrix

Explaination:

Let A = \(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)

(1) a scalar matrix – not true

(2) a diagonal matrix – true

(3) an upper triangular matrix – true

(4) a lower triangular matrix – true

[(1) A square matrix A = [a_{ij}]_{m × n} is called a diagonal matrix if a_{ij} = 0 whenever i ≠ j

(2) A diagonal matrix whose entries along the principle diagonal are equal is called a scalar matrix.

(3) A square matrix is said to be an upper triangular matrix if all the elements below the main diagonal are zero.

(4) A square matrix is said to be a lower triangular matrix if all the elements above the main diagonal are zero.]

Question 4.

If A and B are two matrices such that A + B and AB are both defined, then

(1) A and B are two matrices not necessarily of same order

(2) A and B are square matrices of same order

(3) Number of columns of A is equal to the number of rows of B

(4) A = B

Answer:

(2) A and B are square matrices of same order

Explaination:

Given A and B are two matrices such that A + B and AB are both defined.

A + B defined means A and B are same order.

AB defined means , Number of columns of A equal to Number of rows of B.

A + B and AB are simultaneously defined.

Therefore A and B are of same order.

Question 5.

If A = \(\left[ \begin{matrix} λ & 1 \\ -1 & -λ \end{matrix} \right] \), then for what value of λ, A^{2} = 0 ?

(1) 0

(2) ± 1

(3) – 1

(4) 1

Answer:

(2) ± 1

Explaination:

Question 6.

If A = \(\left[ \begin{matrix} 1 & -1 \\ 2 & -1 \end{matrix} \right]\), B = \(\left[ \begin{matrix} a & 1 \\ b & -1 \end{matrix} \right]\) and (A + B)^{2} = A^{2} + B^{2}, then the values of a and b are

(1) a = 4, b = 1

(2) a = 1, b = 4

(3) a = 0, b = 4

(4) a = 2, b = 4

Answer:

(2) a = 1, b = 4

Explaination:

L .2

Equating the like entrices

(a – 1)^{2} = a^{2} + b – 1 ………. (3)

o = a – 1 ………. (4)

2a+ 2 + ab + b = ab – b ………. (5)

4 = b ………. (6)

a = 1, b = 4

Question 7.

If A = \(\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{matrix} \right] \)

the equation AA^{T} = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b)

is equal to

(1) (2, -1)

(2) (- 2, 1)

(3) (2, 1)

(4) (- 2, – 1)

Answer:

(4) (- 2, – 1)

Explaination:

Equating the like entries

a + 2b + 4 = 0 …………. (1)

2a – 2b + 2 = 0

a – b + 1 = 0 …………. (2)

a^{2} + b^{2} + 4 = 9 …………. (3)

3b = – 3 ⇒ b = -1

Substituting in equation (1) we get

a + 2 × – 1 + 4 = 0

a – 2 + 4 = 0

a = – 2

Substituting the values a = – 2 , b = – 1 in equation (3)

we have

(- 2)^{2} + (-1)^{2} + 4 = 9

4 + 1 + 4 = 9

9 = 9

∴ The required value of the ordered pair (a, b) is

(a, b) = (- 2, – 1)

Question 8.

If A is a square matrix, then which of the following is not symmetric?

(1) A + A^{T}

(2) AA^{T}

(3) A^{T}A

(4) AA^{T}

Answer:

(4) AA^{T}

Explaination:

Given A is a square matrix.

A square matrix A is symmetric if A^{T} = A

(1) A + A^{T}

(A + A^{T})^{T} = A^{T} + (A^{T})^{T}

= A^{T} + A = A + A^{T}

∴ A + A^{T} is symmetric.

(2) AA^{T}

(AA^{T})^{T} = (A^{T})^{T}A^{T}

= AA^{T}

∴ AA^{T} is symmetric.

(3) A^{T}A

(A^{T}A)^{T} = A^{T}(A^{T})^{T}

= A^{T}A

∴ A^{T}A is symmetric.

(4) A – A^{T}

(A – A^{T})^{T} = A^{T} – (A^{T})^{T}

= A^{T}A

∴ A – A^{T} is not symmetric.

Question 9.

If A and B are symmetric matrices of order n , where (A ≠ B) , then

(1) A + B is skew – symmetric

(2) A + B is symmetric

(3) A + B is a diagonal matrix

(4) A + B is a zero matrix

Answer:

(2) A + B is symmetric

Explaination:

Given A and B are symmetric matrices of order n.

∴ A^{T} = A and B^{T} = B

A matrix A is skew symmetric if A^{T} = – A

(1)(A + B)^{T} = A^{T} + B^{T} = A + B

A + B is not skew symmetric.

(2)(A + B)^{T} = A^{T}+B^{T} = A + B

∴ A + B is symmetric.

(3) A + B is a diagonal matrix is incorrect.

(4) A + B is a zero matrix is incorrect.

Question 10.

If A = \(\left[ \begin{matrix} a & x \\ y & a \end{matrix} \right] \) and if xy = 1, then det (AA^{T}) is equal to

(1) (a – 1)^{2}

(2) (a^{2} + 1)^{2}

(3) a^{2} – 1

(4) (a^{2} – 1)^{2}

Answer:

(4) (a^{2} – 1)^{2}

Explaination:

= (a^{2} + x^{2}) (y^{2} + a^{2}) – (ax + ay) (ax + ay)

= a^{2} y^{2} + a^{4} + x^{2} y^{2} + a^{2} x^{2} – ((ax)^{2} + (ay)^{2} + 2 (ax)(ay))

= a^{2} x^{2} + a^{2} y^{2} + a^{4} + (xy)^{2} – a^{2} x^{2} – a^{2} y^{2} – 2a^{2} xy

= a^{4} + (1)^{2} – 2a^{2}(1)

= a^{4} – 2a^{2} + 1

|AA^{T}| = (a^{2} – 1)^{2}

Question 11.

The value of x, for which the matrix A = \(\left[ \begin{matrix} { e }^{ x-2 } & { e }^{ 7+x } \\ { e }^{ 2+x } & { e }^{ 2x+3 } \end{matrix} \right] \) is singular

(1) 9

(2) 8

(3) 7

(4) 6

Answer:

(2) 8

Explaination:

Question 12.

If the points (x, – 2), (5, 2), (8, 8) are collinear , then x is equal to

(1) – 3

(2) \(\frac{1}{3}\)

(3) 1

(4) 3

Answer:

(4) 3

Explaination:

Let the given points be (x_{1}, y_{1}) = (x, – 2) ,

(x_{2}, y_{2}) = (5, 2) and (x_{3}, y_{3}) = (8, 8)

The condition for the three points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) to be collinear is

\(\frac{1}{2}\) [x(2 – 8) + 2(5 – 8) + 1(40 – 16)] = 0

x × – 6 + 2 × – 3 + 1 × 24 = 0

– 6x – 6 + 24 = 0

– 6x + 18 = 0

6x = 18 ⇒ x = \(\frac{18}{6}\) = 3

x = 3

Question 13.

If \(\left| \begin{matrix} 2a & { x }_{ 1 } & { y }_{ 1 } \\ 2b & { x }_{ 2 } & { y }_{ 2 } \\ 2c & { x }_{ 3 } & { y }_{ 3 } \end{matrix} \right| \) = \(\frac{\text { abc }}{2}\) ≠ 0, then the area of the triangle whose vertices are

(1) \(\frac{1}{4}\)

(2) \(\frac{1}{4}\)abc

(3) \(\frac{1}{8}\)

(4) \(\frac{1}{8}\)abc

Answer:

(3) \(\frac{1}{8}\)

Explaination:

Question 14.

If the square of the matrix \(\left[ \begin{matrix} α & β \\ γ & -α \end{matrix} \right] \) is the unit matrix of order 2, then α, β, and γ should

(1) 1 + α^{2} + βγ = 0

(2) 1 – α^{2} – βγ = 0

(3) 1 – α^{2} + βγ = 0

(4) 1 + α^{2} – βγ = 0

Answer:

(1) 1 + α^{2} + βγ = 0

Explaination:

– α^{2} – βγ = 1

α^{2} + βγ + 1 = 0

Question 15.

(1) Δ

(2) kΔ

(3) 3 kΔ

(4) k^{3} Δ

Answer:

(4) k^{3} Δ

Explaination:

Question 16.

A root of the equation \(\left| \begin{matrix} 3-x & -6 & 3 \\ -6 & 3-x & 3 \\ 3 & 3 & -6-x \end{matrix} \right| \) = 0 is

(1) 6

(2) 3

(3) 0

(4) -6

Answer:

(3) 0

Explaination:

0 = 0

x = 0 satisfies the given equation.

Hence the root of the given equation is x = 0.

Question 17.

The value of the determinant of A = \(\left[ \begin{matrix} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{matrix} \right] \) is

(1) -2 abc

(2) abc

(3) 0

(4) a^{2} + b^{2} + c^{2}

Answer:

(3) 0

Explaination:

= 0 – a(0 – bc) – b (ac – 0)

= abc – abc = 0

Question 18.

If x_{1}, x_{2}, x_{3} as well as y_{1}, y_{2}, y_{3} are in geometric progression with the same common ratio, then the points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) are

(1) vertices of an equilateral triangle

(2) vertices of a right angled triangle

(3) vertices of a right angled isosceles triangle

(4) collinear

Answer:

(4) collinear

Explaination:

Given x_{1}, x_{2} , x_{3}, as well as y_{1}, y_{2}, y_{3}, are in geometric progression with the same common ratio.

∴ x_{1} = a, x_{2} = ar, x_{3} = ar^{2} ,

y_{1} = b, y_{2} = br, y_{3} = br^{2}

(x_{1}, y_{1}) = (a, b),

(x_{2}, y_{2}) = (ar ,br)

and (x_{3}, y_{3}) = (ar^{2}, br^{2})

Area of the triangle whose vertices are

(x_{1}, y_{1}),(x_{2}, y_{2}) and (x_{3} y_{3}) is

= \(\frac{1}{2}\)ab[1(r – r^{2}) – 1 (r – r^{2}) + 1 (r^{3} – r^{3})]

= \(\frac{1}{2}\)ab[r – r^{2} – r + r^{2} + 0]

= \(\frac{1}{2}\)ab × 0 = 0

∴ The points (x_{1}, y_{1}),(x_{2}, y_{2}) and (x_{3} y_{3}) are collinear.

Question 19.

If denotes the greatest integer less than or equal to the real number under consideration and – 1 ≤ x< 0, 0 ≤ y < 1, 1 ≤ z < 2 , then the value of the determinant

(1)

(2)

(3)

(4)

Answer:

(1)

Explaination:

Question 20.

If a ≠ b, b, c satisfy \(\left| \begin{matrix} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{matrix} \right| \) = 0 then abc =

(1) a + b + c

(2) 0

(3) b^{3}

(4) ab + bc

Answer:

(3) b^{3}

Explaination:

(a – 6) (b^{2} – ac) = 0

Since a ≠ 6 , we have a – 6 ≠ 0

∴ b^{2} – ac = 0

b^{2} = ac

b^{3} = abc

Question 21.

If A = \(\left| \begin{matrix} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{matrix} \right| \) and B = \(\left| \begin{matrix} -2 & 4 & 2 \\ 6 & 2 & 0 \\ -2 & 4 & 8 \end{matrix} \right| \), then B is given by

(1) B = 4A

(2) B = – 4A

(3) B = – A

(4) B = 6A

Answer:

(2) B = – 4A

Explaination:

Question 22.

IfA skew-symmetric of order n and C is a column matrix of order n × 1, then C_{T} AC is

(1) an identity matrix of order n

(2) an identity matrix of order I

(3) a zero matrix of order 1

(4) an identity matrix of order 2

Answer:

(3) a zero matrix of order 1

Explaination:

Given A is a skew symmetric matrix of order n.

∴ A^{T} = – A

C is a column matrix of order n × 1

C^{T}is of order 1 × n

∴ C^{T} A is of order (1 × n) × (n × n) = (1 × n)

C^{T}AC is of order (1 × n) × (n × 1) = (1 × 1)

Since A is a skew – symmetric matrix, we have

A^{T} = -A

(C^{T} AC)^{T}= C^{T} A^{T} (C^{T})^{T} = C^{T} (-A) C

= -C^{T} AC

∴ C^{T}AC is a skew – symmetric matrix.

A matrix of order 1 is skew – symmetric if A is zero matrix.

Since C^{T}AC is a skew – symmetric and its order is 1.

∴ C^{T}AC is a zero matrix.

Question 23.

The matrix A satisfying the equation

(1)

(2)

(3)

(4)

Answer:

(3)

Explaination:

x + 3z = 1 ——– (1)

y + 3t = 1 ——- (2)

z = 0

t = – 1

(1) ⇒ x + 3 × 0 = 1 ⇒ x = 1

(2) ⇒ y + 3 × – 1 = 1 ⇒ y = 4

∴ A = \(\left[ \begin{matrix} 1 & 4 \\ 0 & -1 \end{matrix} \right] \)

Question 24.

If A + 1 = \(\left[ \begin{matrix} 3 & -2 \\ 4 & 1 \end{matrix} \right]\), then (A + I) (A – I) is equal to

(1)

(2)

(3)

(4)

Answer:

(1)

Explaination:

Question 25.

Let A and B be two symmetrh matrices of same order. ^{T}hen which one of the following statement is not true?

(1) A + B is a symmetric matrix

(2) AB is a symmetric matrix

(3) AB = (BA)^{T}

(4) A^{T}B = MI^{T}

Answer:

(2) AB is a symmetric matrix

Explaination:

Given A and B are two symmetric matrices of the same order.

A = A^{T} , B = B^{T}

(1)(A+B)^{T} = A^{T} + B^{T} = A + B

A + B is symmetric.

(2) (AB)^{T} = B^{T} A^{T} BA

Thus (AB)^{T} ≠ AB

Hence , AB is not symmetric.

(3) AB = (BA)^{T}

= A^{T} B^{T} = AB

Statement is true.

(4) A^{T} B = AB^{T} Since A^{T} = A

B = B^{T}

Statement is true.