TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I.
Question 1.
Find the determinants of the following matrices.

(i) $\left[\begin{array}{cc} 2 & 1 \\ 1 & -5 \end{array}\right]$

Answer:
Let A = $\left[\begin{array}{cc} 2 & 1 \\ 1 & -5 \end{array}\right]$ then determinant A
= det A = |A| = 2(-5) – 1(1)
= -10 – 1
= -11

(ii) $\left[\begin{array}{cc} 4 & 5 \\ -6 & 2 \end{array}\right]$

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

(iii) $\left[\begin{array}{cc} \mathrm{i} & 0 \\ 0 & -\mathrm{i} \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{cc} \mathrm{i} & 0 \\ 0 & -\mathrm{i} \end{array}\right]$ then
det A = i(-1) – 0 = -i² = 1 (∵ i² = -1)

(iv) $\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$
Answer:

(v) $\left|\begin{array}{rrr} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{array}\right|$
Answer:
Let A = $\left[\begin{array}{rrr} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{array}\right]$
Then det A = 1$\left|\begin{array}{rr} -1 & 4 \\ 7 & 6 \end{array}\right|$ – 4$\left|\begin{array}{ll} -2 & 4 \\ -3 & 6 \end{array}\right|$ + 2$\left|\begin{array}{rr} 2 & -1 \\ -3 & 7 \end{array}\right|$
= 1(-6 – 28) – 4(12 + 12)+ 2(14 – 3)
= 1 (- 34) – 4(24) + 2(11)
= -34 – 96 + 22
= -108

(vi) $\left[\begin{array}{rrr} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{rrr} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{array}\right]$
Then det A = 2$\left|\begin{array}{rr} -3 & 1 \\ 2 & 1 \end{array}\right|$ + 1$\left|\begin{array}{ll} 4 & 1 \\ 1 & 1 \end{array}\right|$ + 4$\left|\begin{array}{rr} 4 & -3 \\ 1 & 2 \end{array}\right|$
= 2(- 3 – 2)+ 1(4 – 1) + 4(8 + 3)
= 2(-5) + 3 + 4(11)
= – 10 + 3 + 44
= 37

(vii) $\left[\begin{array}{rrr} 1 & 2 & -3 \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{rrr} 1 & 2 & -3 \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{array}\right]$
Then det A = 1$\left|\begin{array}{rr} -1 & 7 \\ 4 & -6 \end{array}\right|$ – 2$\left|\begin{array}{rr} 4 & 7 \\ 2 & -6 \end{array}\right|$ – 3$\left|\begin{array}{rr} 4 & -1 \\ 2 & 4 \end{array}\right|$
= 1(6 – 28) – 2(- 24 – 14) – 3(16 + 2)
= -22 + 76 – 54 = 0
[Note : Since R₁ and R₂ are proportional, we have det A = 0.]

(viii) $\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]$
Then det A = a$\left|\begin{array}{ll} b & f \\ f & c \end{array}\right|$ – h$\left|\begin{array}{ll} \mathrm{h} & \mathrm{f} \\ \mathrm{g} & \mathrm{c} \end{array}\right|$ – g$\left|\begin{array}{ll} h & b \\ g & f \end{array}\right|$
= a(bc – f²) – h(ch – fg) + g(fh – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

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

Question 2.
If A = $\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right]$ and det A = 45 then find x.

Answer:
det A = 45
⇒ $\left|\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right|$ = 45
⇒ 1(3x + 24) = 45
⇒ 3x = 21
⇒ x = 7

II.
Question 1.
Show that $\left|\begin{array}{lll} \mathrm{b c} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c a} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{a b} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|$ = (a – b)(b – c)(c – a).

Answer:
Operating R₂ – R₁, R₃ – R₁, on the given determinant
LHS = $\left|\begin{array}{ccc} b c & b+c & 1 \\ c(a-b) & a-b & 0 \\ b(a-c) & a-c & 0 \end{array}\right|$
= (a – b)(a – c)$\left|\begin{array}{ccc} b c & b+c & 1 \\ c & 1 & 0 \\ b & 1 & 0 \end{array}\right|$
= (a – b)(a – c)(1)(c – b)
= (a – b)(b – c)(c – a) (exponding on 3rd column)
= RHS

Question 2.
Show that $\left|\begin{array}{ccc} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{array}\right|$ = a² + b² + c² – 3abc (Mar. 2008; May 2007)

Answer:

= (a + b + c) [(c – b) (c – a) – (a – b) (b – a)]
= (a + b + c) [c² – bc – ac + ab + a² – 2ab + b²]
= (a + b + c) [a² + b² + c² – ab – bc – ca]
= a² + b² + c² – 3abc

Question 3.
Show that $\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z} & \mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{y} \\ \mathrm{z} & \mathrm{z} & \mathrm{x}+\mathrm{y} \end{array}\right|$ = 4xyz.
Answer:
R₁ – (R₂ + R₃) on the given determinant gives

= 2[z(xy) – y(-xz)]
= 2[2xyz] = 4xyz = RHS

Question 4.
If $\left|\begin{array}{ccc} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{array}\right|$ = 0 and $\left|\begin{array}{ccc} \mathrm{a} & \mathrm{a}^2 & 1 \\ \mathrm{b} & \mathrm{b}^2 & 1 \\ \mathrm{c} & \mathrm{c}^2 & 1 \end{array}\right|$ ≠ 0, then show that abc = -1. (Mar. ’14)

Answer:

⇒ abc + 1 = 0
⇒ abc = -1

Question 5.
Without expanding the determinant, prove that
(i) $\left|\begin{array}{lll} \mathrm{a} & \mathrm{a}^2 & \mathrm{b c} \\ \mathrm{b} & \mathrm{b}^2 & \mathrm{c a} \\ \mathrm{c} & \mathrm{c}^2 & \mathrm{a b} \end{array}\right|=\left|\begin{array}{ccc} 1 & \mathrm{a}^2 & \mathrm{a}^3 \\ 1 & \mathrm{b}^2 & \mathrm{b}^3 \\ 1 & \mathrm{c}^2 & \mathrm{c}^3 \end{array}\right|$

Answer:

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

(iii) $\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|$ (Board Model Paper)
Answer:

(∵ R₂ – R₁; R₃ – R₁)
= (b – a) (c² – a²) – (c – a) (b² – a²)
= (b – a) (c – a) (c + a) – (c – a) (b – a) (b + a)
= (b – a) (c – a) (c + a – b – a)
= (b – a) (c – a) (c – b)
= (a – b) (b – c) (c – a)
LHS = RHS

Question 6.
If Δ₁ = $\left|\begin{array}{ccc} \mathrm{a}_1^2+\mathrm{b}_1+c_1 & \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_2+c_2 & \mathrm{a}_1 \mathrm{a}_3+\mathrm{b}_3+c_3 \\ \mathrm{b}_1 \mathrm{b}_2+c_1 & \mathrm{b}_2^2+\mathrm{c}_2 & \mathrm{b}_2 \mathrm{b}_3+\mathrm{c}_3 \\ \mathrm{c}_3 c_1 & \mathrm{c}_3 \mathrm{c}_2 & \mathrm{c}_3^2 \end{array}\right|$ and Δ₂ = $\left|\begin{array}{lll} \mathrm{a}_1 & \mathrm{b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{b}_2 & \mathrm{c}_2 \\ \mathrm{a}_3 & \mathrm{b}_3 & \mathrm{c}_3 \end{array}\right|$, then find the value of $\frac{\Delta_1}{\Delta_2}$.

Answer:

Question 7.
If Δ₁ = $\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{array}\right|$ and Δ₂ = $\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$ and Δ₁ = Δ₂ then show that cos²α + cos²β + cos²γ = 1.

Answer:
Given $\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{array}\right|$
= (1 – cos²γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos²γ – cos²α + cos β cos α cos γ + cos α cos β cos γ – cos²β
= 1 – (cos²α + cos²β + cos²γ) + 2 cos α cos β cos γ

Δ₂ = $\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$
= – cos α (0 – cos γ cos β) + cos β (cos α cos γ)
= cos α cos β cos γ + cos α cos β cos γ
= 2cos α cos β cos γ
Also given Δ₁ = Δ₂
⇒ 1 – (cos²α + cos²β + cos²γ) + 2 cos α cos β cos γ
= 2 cos α cos β cos γ
⇒ 1 – (cos²α + cos²β + cos²γ) = 0
∴ cos²α + cos²β + cos²γ = 1

III.
Question 1.
Show that
$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|$ = 2(a + b + c)³

Answer:

Question 2.
Show that $\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|^2$ = $\left|\begin{array}{ccc} 2 b c-a^2 & c^2 & b^2 \\ c^2 & 2 a c-b^2 & a^2 \\ b^2 & a^2 & 2 a b-c^2 \end{array}\right|$ = (a³ + b³ + c³ – 3abc)². (May 2014, Mar. 01′)

Answer:
Let Δ = $\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$ = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= – (a³ + b³ + c³ – 3abc)
⇒ Δ² = (a³ + b³ + c³ – 3abc)² …………..(1)

From (1) and (2) the result is proved.

Question 3.
Show that $\left|\begin{array}{ccc} a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$ = (a – 1)³. (March 2007)

Answer:
Apply operations R₁ – R₂ and R₂ – R₃ we get

Question 4.
Show that $\left|\begin{array}{ccc} \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{a}^2 & \mathrm{b}^2 & \mathrm{c}^2 \\ \mathrm{a}^3 & \mathrm{b}^3 & \mathrm{c}^3 \end{array}\right|$ = abc(a – b)(b – c)(c – a)

Answer:
LHS = abc$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right|$
= abc$\left|\begin{array}{ccc} 0 & 0 & 1 \\ a-b & b-c & c \\ a^2-b^2 & b^2-c^2 & c^2 \end{array}\right|$ (Use operations C₁ – C₂, C₂ – C₃)
= abc [(a – b) (b² – c²) – (b – c) (a² – b²)]
= abc [(a – b) (b – c) (b + c) – (b – c) (a – b) (a + b)]
= abc (a – b) (b – c) [b + c – a – b]
= abc (a – b) (b – c) (c – a)

Question 5.
Show that $\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right|$ = 4(a + b)(b + c)(c + a)

Answer:

= 0 (∵ R₁ & R₃ are similar)
∴ (a + b) is a factor of Δ.
Similarly putting b + c = 0 and c + a = 0 we shall find that b + c and c + a are also factors of Δ.
∵ Δ is a 3rd degree expression in a, b, c.

Let Δ = k (a + b) (b + c) (c + a)
Where k ≠ 0 is a scalar.
Put a = 1, b = 1, c = 1 then
= k(1 + 1) (1 + 1) (1 + 1)
= 8k -2(4 – 4) – 2(-4 – 4) + 2(4 + 4)
= 8k
⇒ -16 + 16 = 8k ⇒ k = 4
Δ = 4(a + b) (b + c) (c + a)
Here $\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right|$ = 4(a + b)(b + c)(c + a)

Question 6.
Show that $\left|\begin{array}{lll} \mathrm{a}-\mathrm{b} & \mathrm{b}-\mathrm{c} & \mathrm{c}-\mathrm{a} \\ \mathrm{b}-\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{a}-\mathrm{b} \\ \mathrm{c}-\mathrm{a} & \mathrm{a}-\mathrm{b} & \mathrm{b}-\mathrm{c} \end{array}\right|$ = 0
Answer:
R₁ + (R₂ + R₃) given
$\left|\begin{array}{ccc} 0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$
= 0 (∵ If one row or column elements of a square matrix are zeroes then the value of the determinant of that matrix is equal to zero)
= RHS.

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

Answer:
Make operations R₂ – R₁, R₃ – R₁ then the given determinant.

Question 8.
Show that $\left|\begin{array}{lll} x & a & a \\ a & x & a \\ a & a & x \end{array}\right|$ = (x + 2a)(x – a)².

Answer: