Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.3

Question 1.
Find the transpose of each of the following matrices:
(i) (left[begin{array}{c} 5 \ 1 \ hline 2 \ -1 end{array}right])
(ii) (left[begin{array}{cc} 1 & -1 \ 2 & 3 end{array}right])
(iii) (left[begin{array}{ccc} -1 & 5 & 6 \ sqrt{3} & 5 & 6 \ 2 & 3 & -1 end{array}right])
Solution:

Question 2.
If A = (left[begin{array}{ccc}
-1 & 2 & 3 \
5 & 7 & 9 \
-2 & 1 & 1
end{array}right]) + B = (left[begin{array}{ccc}
-4 & 1 & -5 \
1 & 2 & 0 \
1 & 3 & 1
end{array}right]), then verify that
(i) (A+B)’ = A’ + B’
(ii) (A-B)’ = A’ – B’
Solution:

Question 3.
If A’ = (left[begin{array}{cc}
3 & 4 \
-1 & 2 \
0 & 1
end{array}right]) + B = (left[begin{array}{ccc}
-1 & 2 & 1 \
1 & 2 & 3
end{array}right]), then verify that
(i) (A+B)’ = A’ + B’
(ii) (A-B)’ = A’ – B’
Solution:

Question 4.
If A’ = (left[begin{array}{cc}
-2 & 3 \
1 & 2
end{array}right]) and B = (left[begin{array}{cc}
-1 & 0 \
1 & 2
end{array}right]), then find (A + 2B’)
Solution:
A’ = (left[begin{array}{cc}
-2 & 3 \
1 & 2
end{array}right]). ∴ A = (left[begin{array}{cc}
-2 & 1 \
3 & 2
end{array}right]) and B = (left[begin{array}{cc}
-1 & 0 \
1 & 2
end{array}right]).
∴ A + 2B = (left[begin{array}{cc}
-2 & 1 \
3 & 2
end{array}right]) + 2(left[begin{array}{cc}
-1 & 0 \
1 & 2
end{array}right])
= (left[begin{array}{cc} -2 & 1 \ 3 & 2 end{array}right]) + (left[begin{array}{cc} -2 & 0 \ 2 & 4 end{array}right]) = (left[begin{array}{cc} -2-2 & 1+0\ 3+2 & 2+4 end{array}right])
= (left[begin{array}{cc} -4 & 1 \ 5 & 6 end{array}right])
∴(A + 2B)’ = (left[begin{array}{cc} -4 & 1 \ 5 & 6 end{array}right]) = (left[begin{array}{cc} -4 & 5 \ 1 & 6 end{array}right]).

Question 5.
(i) A = (left[begin{array}{c}
1 \
-4 \
3
end{array}right]), B = (left[begin{array}{lll}
-1 & 2 & 1
end{array}right])
(ii) A = (left[begin{array}{c}
0 \
1 \
2
end{array}right]), B = (left[begin{array}{lll}
1 & 5 & 7
end{array}right])
Solution:

Question 6.
If (i) A = (left[begin{array}{cc}
cos alpha & sin alpha \
-sin alpha & cos alpha
end{array}right]), then verify that A’A = I.

(ii) (left[begin{array}{cc}
sin alpha & cos alpha \
-cos alpha & sin alpha
end{array}right]), then verify that A’A = I.
Solution:

Question 7.
(i) Show that the matrix A = (left[begin{array}{ccc} 1 & -1 & 5 \ -1 & 2 & 1 \ 5 & 1 & 3 end{array}right]) is a symmetric matrix.
(ii) Show that the matrix A = (left[begin{array}{ccc} 0 & 1 & -1 \ -1 & 0 & 1 \ 1 & -1 & 0 end{array}right]) is a skew – symmetric matrix.
Solution:
(i) For a symmetric matrix a = a_ij.
Now, A = (left[begin{array}{ccc} 1 & -1 & 5 \ -1 & 2 & 1 \ 5 & 1 & 3 end{array}right])
a₂₁ = – 1 = a₁₂, a₃₁ = 5 = a₁₃
a₃₂ = 1 = a₂₃ a₁₁, a₂₂, a₃₃ are 1, 2, 3 respectively.
Hence, a_ij = a_ji
∴ A is a symmetric matrix.
Alternatively:
A’ = (left[begin{array}{ccc} 1 & -1 & 5 \ -1 & 2 & 1 \ 5 & 1 & 3 end{array}right])‘ = (left[begin{array}{ccc} 1 & -1 & 5 \ -1 & 2 & 1 \ 5 & 1 & 3 end{array}right]) = A.
∴ A’ = A ⇒ A is a symmetric matrix.

(ii) For a skew symmetric matrix

Question 8.
For a matrix A = (left[begin{array}{ll} 1 & 5 \ 6 & 7 end{array}right]), verify that
(i) (A + A) is a symmetric matrix.
(ii) (A – A) is a skew symmetric matrix.
Solution:
(i) A = (left[begin{array}{ll} 1 & 5 \ 6 & 7 end{array}right]) ∴ A’ = (left[begin{array}{ll} 1 & 5 \ 6 & 7 end{array}right])
∴ A + A’ = (left[begin{array}{ll} 1+1 & 5+6 \ 6+5 & 7+7 end{array}right]) = (left[begin{array}{ll} 2 & 11 \ 11 & 14 end{array}right])
Here, a₂₁ = 11 = a₁₂
∴ A + A’ is a symmetric matrix.

(i) A – A’ = (left[begin{array}{ll} 1 & 5 \ 6 & 7 end{array}right]) – (left[begin{array}{ll} 1 & 6 \ 5 & 7 end{array}right])
= (left[begin{array}{ll} 1-1 & 5-6 \ 6-5 & 7-7 end{array}right]) = (left[begin{array}{ll} 0 & -1 \ 1 & 0 end{array}right]).
Here, a₂₁ = 1, a₁₂ = – 1 or – a₁₂ = 1.
∴ a₂₁ = – a₁₂ a₁₁ = a₂₂ = 0
∴ A – A’ is a symmetric matrix.

Question 9.
Find (frac { 1 }{ 2 })(A +A) and (frac { 1 }{ 2 })(A-A), when A = (left[begin{array}{ccc} 0 & a & b \ -a & 0 & c \ -b & -c & 0 end{array}right]).
Solution:

Question 10.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
(i) (left[begin{array}{ll} 3 & 5 \ 1 & -1 end{array}right])
(ii) (left[begin{array}{ccc} 6 & -2 & 2 \ -2 & 3 & -1 \ 2 & -1 & 3 end{array}right])
(iii) (left[begin{array}{ccc} 3 & 3 & -1 \ -2 & -2 & 1 \ -4 & -5 & 2 end{array}right])
(iv) (left[begin{array}{ll} 1 & 5 \ -1 & 2 end{array}right])
Solution:

Choose the correct answer in the following questions:

Question 11.
If A and B are symmetric matrices of the same order, then AB – BA is a _____.
(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix
Solution:
A and B are symmetric matrices.
So, A’ = A and B’ = B.
Now, (AB – BA)’ = (AB)’ – (BA)’ = B’A’ – A’B’
= BA – AB [∵ B’ = B, A’ = A]
⇒ AB – BA is a skew symmetric matrix.
∴ Part (A) is the correct answer.

Question 12.
If A = (left[begin{array}{cc} cos alpha & -sin alpha \ sin alpha & cos alpha end{array}right]) then A + A’= I, if the value of α is
(A) (frac { π }{ 2 })
(B) (frac { π }{ 3 })
(C) π
(D) (frac { 3π }{ 2 })
Solution:
A = (left[begin{array}{cc} cos alpha & -sin alpha \ sin alpha & cos alpha end{array}right]) ⇒ A’ = (left[begin{array}{cc}
cos alpha & sin alpha \
-sin alpha & cos alpha
end{array}right])
∴ A + A’ = (left[begin{array}{cc} cos alpha+cos alpha & -sin alpha+sin alpha \ sin alpha-sin alpha & cos alpha+cos alpha end{array}right])
= (left[begin{array}{cc} 2 cos alpha & 0 \ 0 & 2 cos alpha end{array}right]) = (left[begin{array}{ll} 1 & 0 \ 0 & 1 end{array}right])
2 cos α = 1 ⇒ cos α = (frac { 1 }{ 2 })
α = (frac { π }{ 3 })
∴ Part (B) is the correct answer.