AP Inter 2nd Year Maths Exercise 4f Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 4 Determinants Exercise 4f Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Determinants Solutions Exercise 4f

I.

Question 1.
Prove that the determinant \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\) is independent of θ.
Solution:
∆ = \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\)
= x(-x² – 1) – sin θ(-x sin θ – cos θ) + cos θ(-sin θ + x cos θ)
= -x³ – x + x sin²θ – sin θ cos θ sin θ cos θ + x cos²θ
= -x³ – x + x(sin²θ + cos²θ) = -x³ – x + x(1) = -x³
Thus, ∆ is independent of θ.

Question 2.
Evaluate \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|\)
Expanding along C₃, we get
∆ = -sin α(-sin αsin²β – cos²β sin α) + cos α(cosα(cos²β + cosαsin²β)
= sin²α(sin²β + cos²β) + cos²α(cos²β + sin²β)
= sin²α(1) + cos²α(1) = 1

Question 3.
Evaluate \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\)
Solution:
∆ = \(\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|\) [R₂ → R₂ – R₁ and R₃ → R₃ – R₁]
= 1 (xy – 0) [Expanding along C₁]
= xy

II.

Question 1.
If A^-1 = \(\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\), find (AB)^-1
Solution:
We know that (AB)^-1 = B^-1A^-1.
Given that B = \(\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
∴ |B| = 1(3) – 2(-1) – 2(2) = 3 + 2 – 4 = 5 – 4 = 1
Now, B₁₁ = 3; B₁₂ = 1; B₁₃ = 2
B₂₁ = 2; B₂₂ = 1; B₂₃ = 2
B₃₁ = 6; B₃₂ = 2; B₃₃ = 5

Question 2.
Evaluate \(\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|\)
Solution:
∆ = \(\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|\)

Question 3.
Show that \(\left|\begin{array}{ccc}
y+z & x & x \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) = 4xyz
Solution:
L.H.S = \(\left|\begin{array}{ccc}
y+z & x & x \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) = \(\left|\begin{array}{ccc}
0 & -2 z & -2 y \\
y & z+x & y \\
z & z & x+y
\end{array}\right|\) (∵ R₁ → (R₁ – (R₂ + R₃))
= -2\(\left|\begin{array}{ccc}
0 & \mathrm{z} & \mathrm{y} \\
\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{z} & \mathrm{x}+\mathrm{y}
\end{array}\right|\) = -2\(\left|\begin{array}{lll}
0 & z & y \\
y & x & 0 \\
z & 0 & x
\end{array}\right|\) (∵ R₂ → R₂ – R₁ ; R₃ → R₃ – R₁)
= -2[0(x² – 0) – z(xy – 0) + y(0 – xz)]
= -2[0 – xyz – xyz] = -2(-2 xyz) = 4xyz
= R.H.S

Question 4.
Show that \(\left|\begin{array}{ccc}
\mathrm{a} & \mathrm{~b} & \mathrm{c} \\
\mathrm{~b} & \mathrm{c} & \mathrm{a} \\
\mathrm{c} & \mathrm{a} & \mathrm{~b}
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)²
Solution:
Let ∆ = \(\left|\begin{array}{ccc}
\mathrm{a} & \mathrm{~b} & \mathrm{c} \\
\mathrm{~b} & \mathrm{c} & \mathrm{a} \\
\mathrm{c} & \mathrm{a} & \mathrm{~b}
\end{array}\right|\) = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ + b³ + abc + abc – c³
= -(a³+ b³ + c³ – 3abc)
⇒ ∆² = (a³+ b³ + c³ – 3abc)² ………. (1)

III.

Question 1.
Let A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\). Verify that (i) |adj A|^-1 = adj (A^-1)
(ii) (A^-1)^-1 = A
Solution:
Given that A = \(\left[\begin{array}{lll}
1 & 2 & 1 \\
2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]\)
⇒|A| = 1(15 – 1) – 2(10 – 1) + 1(2 – 3) = 14 – 18 – 1 = -5
Cofactor matrix of A:
Now, A₁₁ = 14; A₁₂ = -9; A₁₃ = -1
A₂₁ = -9; A₂₂ = 4; A₂₃ = 1
A₃₁ = -1; A₃₂ = 1; A₃₃ = -1
Hence, adj A = \(\left[\begin{array}{ccc}
14 & -9 & -1 \\
-9 & 4 & 1 \\
-1 & 1 & -1
\end{array}\right]\) …………. (1)
∴ A^-1 = \(\frac{1}{\mathrm{~A}}\) (adj A) = –\(\frac{1}{5}\left[\begin{array}{ccc}
14 & -9 & -1 \\
-9 & 4 & 1 \\
-1 & 1 & -1
\end{array}\right]\) = \(\frac{1}{5}\left[\begin{array}{ccc}
-14 & 9 & 1 \\
9 & -4 & -1 \\
1 & -1 & 1
\end{array}\right]\) …………… (2)
(i) Using (1) & (2) we prove the result [adj A]^-1 = adj (A^-1)
(i) |adjA| = 14(-4 – 1) + 9(9 + 1) – 1(-9 + 4)
= 14(-5) + 9(10) + (-1)(-5)= -70 + 90 + 5 = 25
From (1) Cofactor matrix of AdjA
A₁₁ = -5; A₁₂ = -10; A₁₃ = -5
A₂₁ = -10; A₂₂ = -15; A₂₃ = -5
A₃₁ = -5; A₃₂ = -5; A₃₃ = -25

Hence, (A^-1)^-1 = A is proved.

Question 2.
Solve the system of equations \(\frac{2}{x}+\frac{3}{y}+\frac{10}{z}\) = 4, \(\frac{4}{x}-\frac{6}{y}+\frac{5}{z}\) = 1, \(\frac{6}{x}+\frac{9}{y}-\frac{20}{z}\) = 2
Solution:
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, and \(\frac{1}{z}\) = r
∴ The given system of equations is
2p + 3q + 10r = 4
4p – 6q + 5r = 1
6p + 9q – 20r = 2
This system can be written in the form of AX = B, where
A = \(\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\), X = \(\left[\begin{array}{l}
\mathrm{p} \\
\mathrm{q} \\
\mathrm{r}
\end{array}\right]\) B = \(\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]\)
∴ |A = 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)
= 150 + 330 + 720 = 1200
Thus, A is non-singular
∴ A^-1 exists.
Here, A₁₁ = 75; A₁₂ = 110; A₁₃ = 72
A₂₁ = 150; A₂₂ = -100; A₂₃ = 0
A₃₁ = 75; A₃₂ = 30; A₃₃ = -24