You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2
Question 1.
Without expanding the determinant, prove that
Solution:
Question 2.
Show that
Solution:
Question 3.
Prove that
Solution:
LHS
Taking a from C1, b from C2 and c from C3 we get
Expanding along R1 we get
(2c) (abc) (1) [ab + ab] = abc (2c) (2ab)
1 = (abc) (4abc) = 4a2b2c2
= RHS
Question 4.
Solution:
Question 5.
Prove that
Solution:
Question 6.
Show that
Solution:
Question 7.
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0.
Solution:
Question 8.
Solution:
we get – (aα2 + 2bα + c) [ac – b2]
So Δ = 0 ⇒ (aα2 + 2bα + c) (ac -b2) = – 0 = 0
⇒ aα2 + 2bα + c = 0 or ac – b2 = 0
(i.e.) a is a root of ax2 + 2bx + c = 0
or ac = b2
⇒ a, b, c are in G.P.
Question 9.
Prove that
Solution:
Question 10.
If a, b, c are pth, qth and rth terms of an A.P., find the value of
Solution:
We are given a = tp,b = tq and c = tr
Let a be the first term and d be the common difference
Question 11.
Show that is divisible by x4
Solution:
Multiplying R1 by a, R2 by b and R3 by c and
taking out a from C1 b from C2 and c from C3 we get
==
Question 12.
If a, b, c are all positive, and are pth, qth and rth terms of a G.P., show that
Solution:
Question 13.
Find the value of if x, y, z ≠ 1.
Solution:
Expanding the determinant along R1
Question 14.
Solution:
Question 15.
Without expanding, evaluate the following determinants:
Solution:
Question 16.
If A is a square matrix and |A| = 2, find the value of |AAT|.
Solution:
Given |A| = 2
[Property 1: The determinant of a matrix remains unaltered if its rows are changed into columns and columns. That is, |A| = |AT|]
|AT| = |A| = 2
∴ |A AT| = |A| |AT|
= 2 × 2 = 4
Question 17.
If A and B are square matrices of order 3 such that |A| = -1 and |B| = 3, find the value of |3AB|.
Solution:
Given A and B are square matrices of order 3 such that |A| = -1 and |B| = 3
[It A is a square matrix of order n then det ( kA) = |kA| = kn |A|.]
A and B are square matrices of order 3. Therefore,
AB is also a square matrix of order 3.
|3 AB| = 33 |AB|
= 27 |A| |B|
= 27 × – 1 × 3
|3 AB| = – 81
Question 18.
If λ = -2, determine the value of
Solution:
Given λ = -2
∴ 2λ = -4; λ2 = (-2)2; 3λ2 + 1 = 3 (4) + 1 = 13
6λ – 1 = 6(-2) – 1 = -13
expanding along R1
0(0) + 4 (0 + 13) + 1 (-52 + 0) = 52 – 52 = 0
Aliter: The determinant value of a skew-symmetric matrix is zero
Question 19.
Determine the roots of the equation
Solution:
Given the determinant value is 0
⇒ 30(1 + x) (2 – x) = 0
⇒ 1 + x = 0 or 2 – x = 0
⇒ x = -1 or x = 2
So, x = -1 or 2.
Question 20.
Verify that det (AB) = (det A) (det B) for
Solution:
{(-20)(52) (-19) + (10)(38)(—49) + (2)(64)(-17)} – {(-49)(52) (2) + (-17)(38)(-20) + (-19)(64)(10)}
= (19760 – 18620 – 2176) – (-5096 + 12920 – 12160)
= (19760 + 5096 + 12160) – (18620 + 2176 + 12920)
= 37016 – 33716 = 3300 ….(3)
Now (1) × (2) = (3)
(i.e.,) (-33) (-100) = 3300
⇒ det (AB) = (det A), (det B)
Question 21.
Using cofactors of elements of the second row, evaluate |A|, where
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.2 Additional Problems
Question 1.
Solution:
Question 2.
Solution:
Question 3.
Solution:
Question 4.
Solution:
Question 5.
Solution:
Question 6.
Solution:
Question 7.
Solution:
Question 8.
Solution:
Question 9.
Solution: