Q1 Evaluate the determinant
A= 2 4
−5. −1
Solution
= (2 × (−1)) − (4 × (−5))
= −2 + 20 = 18
Q2. Evaluate the determinants
A=3 −1 −2
0 0 −1
3. −5 0
Solution
= 3(0 − 5) + 1(0 + 3) − 2(0 − 0) = −15 + 3 − 0 = −12
Q3.Find values of 𝑥, if determinant of matrix A is equal to determinant of B
A=2 4.
5. 1
B= 2𝑥 4
6. 𝑥
Solution
Determinant of A
=2-20=-18
Determinant of B
=2x^2-24
Determinant of A=Determinant of B
-18=2x^2-24
-18+24=2x^2
6=2x^2
x=+-√3
Q4 Find area of the triangle with vertices at the point given in each of the following:
(1, 0), (6, 0), (4, 3)
Solution
We know, area of triangle
1/2[𝑥1 𝑦1 1
𝑥2. 𝑦2. 1
𝑥3. 𝑦3 1]
Area of triangle =
1/2[1 0 1
6 0 1
4 3 1]
=
1/2[1(0 − 3) − 0(6 − 4) + 1(18 − 0)]
=1/2(15) = 7.5 square units
Q5 Find equation of line joining (1, 2) and (3, 6) using determinants.
Solution
Let, 𝑃(𝑥, 𝑦) be any point lie on the line joining 𝐴(1, 2) and 𝐵(3, 6). Hence, the points 𝐴, 𝐵
and 𝑃 will be collinear and area of triangle 𝐴𝐵𝐶 will be zero.
Therefore, area of triangle =
1/2[1 2 1
3 6 1
𝑥 𝑦 1]=0
1/2[1(6 − 𝑦) − 2(3 − 𝑥) + 1(3𝑦 − 6𝑥)] = 0
⇒ 6 − 𝑦 − 6 + 2𝑥 + 3𝑦 − 6𝑥 = 0
⇒ −4𝑥 + 2𝑦 = 0
⇒ 2𝑥 = 𝑦
Hence the equation of line joining the points (1, 2) and (3, 6) is 2𝑥 − 𝑦 = 0
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