# AP Inter 1st Year Maths 1A Solutions Chapter 3 మాత్రికలు Ex 3(d)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Chapter 3 మాత్రికలు Exercise 3(d) will help students to clear their doubts quickly.

## AP Inter 1st Year Maths 1A Solutions Chapter 3 మాత్రికలు Exercise 3(d)

I.

Question 1.
కింది మాత్రికల నిర్ధారకాలు కనుక్కోండి.
(i) $$\left[\begin{array}{cc} 2 & 1 \\ 1 & -5 \end{array}\right]$$
Solution:
det A = ad – bc
= 2(-5) – 1(1)
= -10 – 1
= -11

(ii) $$\left[\begin{array}{cc} 4 & 5 \\ -6 & 2 \end{array}\right]$$
Solution:
det A = 4(2) – (-6)(5)
= 8 + 30
= 38

(ii) $$\left[\begin{array}{cc} \mathbf{i} & 0 \\ 0 & -\mathbf{i} \end{array}\right]$$
Solution:
det A = -i2 – 0
= 1 – 0
= 1

(iv) $$\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
Solution:
det A = 0(0 – 1) – 1(0 – 1) + 1(1 – 0)
= 1 + 1
= 2

(v) $$\left[\begin{array}{ccc} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{array}\right]$$
Solution:
det A = 1(-6 – 28) – 4(12 + 12) + 2(14 – 3)
= -34 – 96 + 22
= -108

(vi) $$\left[\begin{array}{ccc} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{array}\right]$$
Solution:
det A = 2(-3 – 2) + 1(4 – 1) + 4(8 + 3)
= -10 + 3 + 44
= 37

(vii) $$\left[\begin{array}{ccc} 1 & 2 & -3 \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{array}\right]$$
Solution:
det A = 0,
∵ R1 మరియు R3 ల అనురూప మూలకాలు సమాన నిష్పత్తిలో ఉన్నవి.

(viii) $$\left[\begin{array}{lll} \mathbf{a} & \mathbf{h} & \mathbf{g} \\ \mathbf{h} & \mathbf{b} & \mathbf{f} \\ \mathbf{g} & \mathbf{f} & \mathbf{c} \end{array}\right]$$
Solution:
det A = a(bc – f2) – h(ch – fg) + g(hf – bg)
= abc – af2 – ch2 + fgh + fgh – bg2
= abc + 2fgh – af2 – bg2 – ch2

(ix) $$\left[\begin{array}{lll} \mathbf{a} & \boldsymbol{b} & \mathbf{c} \\ \mathbf{b} & \mathbf{c} & \mathbf{a} \\ \mathbf{c} & \mathbf{a} & \mathbf{b} \end{array}\right]$$
Solution:
det A = a(bc – a2) – b(b2 – ac) + c(ab – c2)
= abc – a3 – b3 + abc + abc – c3
= 3abc – a3 – b3 – c3

(x) $$\left[\begin{array}{ccc} \mathbf{1}^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{array}\right]$$
Solution:
det A = $$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$$
= 1(225 – 256) – 4(100 – 144) + 9(64 – 81)
= -31 + 176 – 153
= -184 + 176
= -8

Question 2.
A = $$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right]$$ అయి, det A = 45 అయితే x విలువ కనుక్కోండి. [Mar. ’07, ’03]
Solution:
det A = 45
⇒ $$\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right|$$ = 45
⇒ 3x + 24 = 45
⇒ 3x – 45 + 24 = 0
⇒ 3x – 21 = 0
⇒ x = 7

II.

Question 1.
$$\left|\begin{array}{lll} b c & b+c & 1 \\ c a & c+a & 1 \\ a b & a+b & 1 \end{array}\right|$$ = (a – b) (b – c) (c – a) అని చూపండి.
Solution:

Question 2.
$$\left|\begin{array}{ccc} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{array}\right|$$ = a3 + b3 + c3 – 3abc అని చూపండి. [May ’13, ’07; Mar. ’08]
Solution:

= (a + b + c) [(-ac + b2) – (-c2 + ab) + (-bc + a2))
= (a + b + c) (-ac + b2 + c2 – ab – bc + a2)
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a3 + b3 + c3 – 3abc

Question 3.
$$\left|\begin{array}{ccc} \mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\ \mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\ \mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y} \end{array}\right|$$ = 4xyz అని చూపండి.
Solution:
L.H.S. = $$\left|\begin{array}{ccc} \mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\ \mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\ \mathbf{z} & \mathbf{z} & \mathbf{x}+\mathbf{y} \end{array}\right|$$
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) (zx + yz + x2 + xy – yz) – x(xy + y2 – yz) + x(yz – z2 – zx)
= (y + z) (zx + x2 + xy) – x(xy + y2 – yz) + x(yz – z2 – zx)
= xyz + x2y + xy2 + xz2 + x2z + xyz – x2y – xy2 + xyz + xyz – xz2 – x2z
= 4xyz

Question 4.
$$\left|\begin{array}{ccc} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{array}\right|$$ = 0, $$\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|$$ ≠ 0 అయితే abc = -1 అని చూపండి.
సూచన: ఒక చతురస్ర మాత్రికలో ఏదైనా ఒక అడ్డు వరుస (లేదా నిలువు వరుస) లోని ప్రతీ మూలకం రెండు సంఖ్యల మొత్తంగా ఉంటే, ఆ మాత్రిక నిర్ధారకాన్ని రెండు మాత్రికల నిర్ధారకాలు మొత్తంగా వ్రాయవచ్చు.
Solution:

∴ 1 + abc = 0
⇒ abc = -1

Question 5.
నిర్ధారకాన్ని విస్తరించండి.
(i) $$\left|\begin{array}{ccc} a & a^2 & b c \\ b & b^2 & c a \\ c & c^2 & a b \end{array}\right|=\left|\begin{array}{ccc} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{array}\right|$$
Solution:

(ii) $$\begin{array}{ccc} a x & b y & c z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{array}=\left|\begin{array}{ccc} a & b & c \\ x & y & z \\ y z & z x & x y \end{array}\right|$$
Solution:

(iii) $$\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|$$ అని నిరూపించండి.
Solution:

Question 6.
Δ1 = $$\left|\begin{array}{ccc} a_1^2+b_1+c_1 & a_1 a_2+b_2+c_2 & a_1 a_3+b_3+c_3 \\ b_1 b_2+c_1 & b_2^2+c_2 & b_2 b_3+c_3 \\ c_3 c_1 & c_3 c_2 & c_3^2 \end{array}\right|$$ మరియు Δ2 = $$\left|\begin{array}{lll} a_1 & b_2 & c_2 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|$$, అయితే $$\frac{\Delta_1}{\Delta_2}$$ విలువ కనుక్కోండి.
Solution:

Question 7.
Δ1 = $$\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \alpha & 1 \end{array}\right|$$, Δ2 = $$\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$$, Δ1 = Δ2 అయితే cos2α + cos2β + cos2γ = 1 అని చూపండి.
Solution:
Δ1 = $$\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \alpha & 1 \end{array}\right|$$
= 1(1 – cos2γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos2γ – cos2α + cos α cos β cos γ + cos α cos β cos γ – cos2β
= 1 – cos2γ – cos2α – cos2β + 2 cos α cos β cos γ
Δ2 = $$\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$$
= 0(0 – cos2γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ
ఇచ్చినది Δ1 = Δ2
1 – cos2γ – cos2α – cos2β + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos2α – cos2β – cos2γ = 0
1 = cos2α + cos2β + cos2γ

III.

Question 1.
$$\left|\begin{array}{ccc} \mathbf{a}+\mathbf{b}+\mathbf{2 c} & \mathbf{a} & \mathbf{b} \\ \mathbf{c} & \mathbf{b}+\mathbf{c}+\mathbf{2 a} & \mathbf{b} \\ \mathbf{c} & \mathbf{a} & \mathbf{c}+\mathbf{a}+\mathbf{2 b} \end{array}\right|$$ = 2(a + b + c)3 అని చూపండి.
Solution:

Question 2.
$$\left|\begin{array}{lll} \mathbf{a} & \mathbf{b} & \mathbf{c} \\ \mathbf{b} & \mathbf{c} & \mathbf{a} \\ \mathbf{c} & \mathbf{a} & \mathbf{b} \end{array}\right|^2$$ = $$\left|\begin{array}{ccc} 2 b c-a^2 & c^2 & b^2 \\ c^2 & 2 a c-b^2 & a^2 \\ b^2 & a^2 & 2 a b-c^2 \end{array}\right|$$ = (a3 + b3 + c3 – 3abc)2 అని చూపండి. [Mar. ’01]
Solution:

Question 3.
$$\left|\begin{array}{ccc} a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$$ = (a – 1)3 అని చూపండి. [Mar. ’13, ’07]
Solution:

= (a – 1)2 [0(6 – 3) – 0[3(a + 1) – 3) + 1(a + 1 – 2)]
= (a – 1)2 (a – 1)
= (a – 1)3
= R.H.S.

Question 4.
$$\left|\begin{array}{ccc} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{array}\right|$$ = abc (a – b) (b – c) (c – a) అని చూపండి. [May ’06]
Solution:

= abc (a – b) (b – c) [0(c2 – c(b + c) – 0(c2 – c(a + b) + 1(b + c – a – b)]
= abc (a – b) (b – c) (c – a)

Question 5.
$$\left|\begin{array}{ccc} -2 a & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\ \mathbf{a}+\mathbf{b} & -2 \mathbf{b} & \mathbf{b}+\mathbf{c} \\ \mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c} \end{array}\right|$$ = 4(a + b) (b + c) (c + a) అని చూపండి.
Solution:

= 0 [∵ R1 = R3]
∴ (a + b), Δ కు కారణరాశి.
ఇదేవిధంగా b + c, c + a లు కూడా Δ కు కారణరాశులు.
∴ Δ అనునది a, b, c లలో 3వ పరిమాణం.
Δ = k(a + b) (b + c) (c + a), k శూన్యేతర సంఖ్య.
a = 1, b = 1, c = 1 అయిన
$$\left|\begin{array}{ccc} -2 & 2 & 2 \\ 2 & -2 & 2 \\ 2 & 2 & -2 \end{array}\right|$$ = k(1 + 1) (1 + 1) (1 + 1)
⇒ -2(4 – 4) – 2(-4 – 4) + 2(4 + 4) = 8k
⇒ 16 + 16 = 8k
⇒ k = 4
∴ Δ = 4(a + b) (b + c) (c + a)
∴ $$\left|\begin{array}{ccc} -2 a & \mathbf{a}+\mathbf{b} & \mathbf{c}+\mathbf{a} \\ \mathbf{a}+\mathbf{b} & -2 \mathbf{b} & \mathbf{b}+\mathbf{c} \\ \mathbf{c}+\mathbf{a} & \mathbf{c}+\mathbf{b} & -2 \mathbf{c} \end{array}\right|$$ = 4(a + b) (b + c) (c + a).

Question 6.
$$\left|\begin{array}{ccc} \mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} \\ \mathbf{b}-\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} \\ \mathbf{c}-\mathbf{a} & \mathbf{a}-\mathbf{b} & \mathbf{b}-\mathbf{c} \end{array}\right|$$ = 0 అని చూపండి.
Solution:

Question 7.
$$\left|\begin{array}{lll} 1 & a & a^2-b c \\ 1 & b & b^2-c a \\ 1 & c & c^2-a b \end{array}\right|$$ = 0 అని చూపండి.
Solution:

Question 8.
$$\left|\begin{array}{lll} \mathbf{x} & \mathbf{a} & \mathbf{a} \\ \mathbf{a} & \mathbf{x} & \mathbf{a} \\ \mathbf{a} & \mathbf{a} & \mathbf{x} \end{array}\right|$$ = (x + 2a) (x – a)2 అని చూపండి.
Solution:

= (x + 2a) [1(x – a)2 – a(0(x – a) – 0)] + a[0 – 0(x – a)]
= (x + 2a) (x – a)2
= R.H.S.