Samacheer Kalvi 11th Maths Guide Chapter 7 Matrices and Determinants Ex 7.5

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, A² = 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 = A² + B², 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 = a² + 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)
a² + b² + 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^TA
(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)^TA^T
= AA^T
∴ AA^T is symmetric.

(3) A^TA
(A^TA)^T = A^T(A^T)^T
= A^TA
∴ A^TA is symmetric.

(4) A – A^T
(A – A^T)^T = A^T – (A^T)^T
= A^TA
∴ 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) (a² + 1)²
(3) a² – 1
(4) (a² – 1)²
Answer:
(4) (a² – 1)²

Explaination:

= (a² + x²) (y² + a²) – (ax + ay) (ax + ay)
= a² y² + a⁴ + x² y² + a² x² – ((ax)² + (ay)² + 2 (ax)(ay))
= a² x² + a² y² + a⁴ + (xy)² – a² x² – a² y² – 2a² xy
= a⁴ + (1)² – 2a²(1)
= a⁴ – 2a² + 1
|AA^T| = (a² – 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 (x₁, y₁) = (x, – 2) ,
(x₂, y₂) = (5, 2) and (x₃, y₃) = (8, 8)
The condition for the three points (x₁, y₁), (x₂, y₂) and (x₃, y₃) 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 + α² + βγ = 0
(2) 1 – α² – βγ = 0
(3) 1 – α² + βγ = 0
(4) 1 + α² – βγ = 0
Answer:
(1) 1 + α² + βγ = 0

Explaination:

– α² – βγ = 1
α² + βγ + 1 = 0

Question 15.

(1) Δ
(2) kΔ
(3) 3 kΔ
(4) k³ Δ
Answer:
(4) k³ 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) a² + b² + c²
Answer:
(3) 0

Explaination:

= 0 – a(0 – bc) – b (ac – 0)
= abc – abc = 0

Question 18.
If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in geometric progression with the same common ratio, then the points (x₁, y₁), (x₂, y₂), (x₃, y₃) 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 x₁, x₂ , x₃, as well as y₁, y₂, y₃, are in geometric progression with the same common ratio.
∴ x₁ = a, x₂ = ar, x₃ = ar² ,
y₁ = b, y₂ = br, y₃ = br²
(x₁, y₁) = (a, b),
(x₂, y₂) = (ar ,br)
and (x₃, y₃) = (ar², br²)
Area of the triangle whose vertices are
(x₁, y₁),(x₂, y₂) and (x₃ y₃) is

= \(\frac{1}{2}\)ab[1(r – r²) – 1 (r – r²) + 1 (r³ – r³)]
= \(\frac{1}{2}\)ab[r – r² – r + r² + 0]
= \(\frac{1}{2}\)ab × 0 = 0
∴ The points (x₁, y₁),(x₂, y₂) and (x₃ y₃) 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) b³
(4) ab + bc
Answer:
(3) b³

Explaination:

(a – 6) (b² – ac) = 0
Since a ≠ 6 , we have a – 6 ≠ 0
∴ b² – ac = 0
b² = ac
b³ = 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^Tis of order 1 × n
∴ C^T A is of order (1 × n) × (n × n) = (1 × n)
C^TAC 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^TAC is a skew – symmetric matrix.
A matrix of order 1 is skew – symmetric if A is zero matrix.
Since C^TAC is a skew – symmetric and its order is 1.
∴ C^TAC 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. ^Then 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^TB = 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.