Peer review of ML Aggarwal Class 12 ISC Solutions Chapter 4 Determinants Ex 4.7 can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 4 Determinants Ex 4.7

Typical Problems:

Question 1.
If |x| represents greatest integer less than or equal to the real number x and – 1 ≤ p < 0, 0 ≤ q < 1, 1 ≤ r < 2, then find the value of the determinant:
$$\left|\begin{array}{ccc} {[p]+1} & {[q]} & {[r]} \\ {[p]} & {[q]+1} & {[r]} \\ {[p]} & {[r]} & {[r]+1} \end{array}\right|$$.
Solution:
Let Δ = $$\left|\begin{array}{ccc} {[p]+1} & {[q]} & {[r]} \\ {[p]} & {[q]+1} & {[r]} \\ {[p]} & {[r]} & {[r]+1} \end{array}\right|$$
Since, – 1 ≤ p < 0
[p] = – 1 ≤ p < 0
⇒ [p] = – 1
0 ≤ q < 1
⇒ [q] = 0
and 1 ≤ r < 2
⇒ [r] = 1
∴ Δ = $$\left|\begin{array}{ccc} -1+1 & 0 & 1 \\ -1 & 0+1 & 1 \\ -1 & 0 & 1+1 \end{array}\right|$$
= $$\left|\begin{array}{rrr} 0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{array}\right|$$
Expanding along R1
∴ Δ = 0 – 0 + 1 (0 + 1) = 1

Question 2.
If a > 0, d > 0, find the value of:
$$\left|\begin{array}{ccc} \frac{1}{a} & \frac{1}{a(a+d)} & \frac{1}{(a+d)(a+2 d)} \\ \frac{1}{a+d} & \frac{1}{(a+d)(a+2 d)} & \frac{1}{(a+2 d)(a+3 d)} \\ \frac{1}{a+2 d} & \frac{1}{(a+2 d)(a+3 d)} & \frac{1}{(a+3 d)(a+4 d)} \end{array}\right|$$.
Solution:
Taking $$\frac{1}{a(a+d)(a+2 d)}$$, $$\frac{1}{(a+d)(a+2 d)(a+3 d)}, \frac{1}{(a+2 d)(a+3 d)(a+4 d)}$$
Common from R1, R2 and R3 respectively
∴ Given determinant = $$\frac{1}{a(a+d)^2(a+2 d)^3(a+3 a)^2(a+4 d)}\left|\begin{array}{ccc} (a+d)(a+2 d) & (a+2 d) & a \\ (a+2 d)(a+3 d) & a+3 d & a+d \\ (a+3 d)(a+4 d) & a+4 d & a+2 d \end{array}\right|$$
operate C2 → C2 – C3
= $$\frac{1}{a(a+d)^2(a+2 d)^3(a+3 d)^2(a+4 d)}\left|\begin{array}{ccc} (a+d)(a+2 d) & 2 d & a \\ (a+2 d)(a+3 d) & 2 d & a+d \\ (a+3 d)(a+4 d) & 2 d & a+2 d \end{array}\right|$$
operate R2 → R2 – R1 ;
R3 → R3 – R1
= $$\frac{1}{a(a+d)^2(a+2 d)^3(a+3 d)^2(a+4 d)}\left|\begin{array}{ccc} (a+d)(a+2 d) & 2 d & a \\ (a+2 d) 2 d & 0 & d \\ 2 d(2 a+5 d) & 0 & 2 d \end{array}\right|$$
Expanding along C2 ; we have
= $$\frac{-2 d}{a(a+d)^2(a+2 d)^3(a+3 d)^2(a+4 d)}$$ [4d2 (a + 2d) – 2d2 (2a + 5d)]
= $$\frac{4 d^4}{a(a+d)^2(a+2 d)^3(a+3 d)^2(a+4 d)}$$.

Question 3.
If a, b, c (all positive) are the pth, qth and rth terms respectively of a G.P. , then prove that $$\left|\begin{array}{lll} \log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1 \end{array}\right|$$ = 0.
Solution:
Let A be the first term and R be the common ratio of G.P.
∴ a = ARp – 1 ;
b = ARq – 1
and c = ARr – 1
= $$\mid \begin{array}{lll} \log \mathrm{A}+(p-1) \log \mathrm{R} & p & 1 \\ \log \mathrm{A}+(q-1) \log \mathrm{R} & q & 1 \\ \log \mathrm{A}+(r-1) \log \mathrm{R} & r & 1 \end{array}$$
= $$\left|\begin{array}{lll} \log \mathrm{A} & p & 1 \\ \log \mathrm{A} & q & 1 \\ \log \mathrm{A} & r & 1 \end{array}\right|+\left|\begin{array}{ccc} (p-1) \log \mathrm{R} & p & 1 \\ (q-1) \log \mathrm{R} & q & 1 \\ (r-1) \log \mathrm{R} & r & 1 \end{array}\right|$$ ;
Taking log A common from C1 in Ist det and log R common from C1 in 2nd det.
= log A $$\left|\begin{array}{lll} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{array}\right|$$ + log R $$\left|\begin{array}{ccc} p-1 & p & 1 \\ q-1 & q & 1 \\ r-1 & r & 1 \end{array}\right|$$
= log A . 0 + log R $$\left|\begin{array}{ccc} p-1 & p & 1 \\ q-1 & q & 1 \\ r-1 & r & 1 \end{array}\right|$$
[∵ C1 and C3 are identical]
operate C1 → C1 + C3
= log R $$\left|\begin{array}{lll} p & p & 1 \\ q & q & 1 \\ r & r & 1 \end{array}\right|$$
= log R . 0 = 0
[∵ C1 and C2 are identical]

Question 4.
If T be the pth term of a G.P. of positive terms, then evaluate $$\left|\begin{array}{ccc} \log \mathrm{T}_{p+1} & \log \mathrm{T}_{p+3} & \log \mathrm{T}_{p+5} \\ \log \mathrm{T}_{p+3} & \log \mathrm{T}_{p+5} & \log \mathrm{T}_{p+7} \\ \log \mathrm{T}_{p+5} & \log \mathrm{T}_{p+7} & \log \mathrm{T}_{p+9} \end{array}\right|$$.
Solution:
Let r be the common ratio and a be the first term of G.P. series
∴ Tp = arp-1 ;
Tp+1 = arp ;
Tp+3 = arp+2

Tp+5 = arp+4 ;
Tp+7 = arp+6 ;
Tp+9 = arp+8 ;
operate R2 → R2 – R1 ;
R3 → R3 – R1

∴ Given det = $$\left|\begin{array}{ccc} \log \mathrm{T}_{p+1} & \log \mathrm{T}_{p+3} & \log \mathrm{T}_{p+5} \\ \log \left(\frac{\mathrm{T}_{p+3}}{\mathrm{~T}_{p+1}}\right) & \log \left(\frac{\mathrm{T}_{p+5}}{\mathrm{~T}_{p+3}}\right) & \log \left(\frac{\mathrm{T}_{p+7}}{\mathrm{~T}_{p+5}}\right) \\ \log \left(\frac{\mathrm{T}_{p+5}}{\mathrm{~T}_{p+1}}\right) & \log \left(\frac{\mathrm{T}_{p+7}}{\mathrm{~T}_{p+3}}\right) & \log \left(\frac{\mathrm{T}_{p+9}}{\mathrm{~T}_{p+5}}\right) \end{array}\right|$$

= $$\left|\begin{array}{ccc} \log \mathrm{T}_{p+1} & \log \mathrm{T}_{p+3} & \log \mathrm{T}_{p+5} \\ \log r^2 & \log r^2 & \log r^2 \\ \log r^4 & \log r^4 & \log r^4 \end{array}\right|$$

$$\frac{T_{p+3}}{T_{p+1}}=\frac{a r^{p+2}}{a r^p}=r^2 ; \frac{T_{p+5}}{T_{p+3}}=\frac{a r^{p+4}}{a r^{p+2}}=r^2 ; \frac{T_{p+7}}{T_{p+5}}=\frac{a r^{p+6}}{a r^{p+4}}=r^2$$
Also $$\frac{T_{p+5}}{T_{p+1}}=\frac{a r^{p+4}}{a r^p}=r^4 ; \frac{T_{p+7}}{T_{p+3}}=\frac{a r^{p+6}}{a r^{p+2}}=r^4 ; \frac{T_{p+9}}{T_{p+5}}=\frac{a r^{p+8}}{a r^{p+4}}=r^4$$

Taking log r2 common from R2 and log r common from R3
= $$\left|\begin{array}{ccc} \log \mathrm{T}_{p+1} & \log \mathrm{T}_{p+3} & \log \mathrm{T}_{p+5} \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right|$$ = 0
[∵ R2 and R3 are identical]

Question 5.
Prove that $$\left|\begin{array}{ccc} {cosec} x & 1 & 0 \\ 1 & 2 {cosec} x & 1 \\ 0 & 1 & 2 {cosec} x \end{array}\right|$$ ≥ 1 for all x ∈ (0, π).
Solution:
Let ∆ = $$\left|\begin{array}{ccc} {cosec} x & 1 & 0 \\ 1 & 2 {cosec} x & 1 \\ 0 & 1 & 2 {cosec} x \end{array}\right|$$ ;
expanding along R1 ; we have
= cosec x (4 cosec2 x – 1) – 1 (2 cosec x – 0) + 0
= 4 cosec3 x – 3 cosec x
= cosec x (4 cosec2 x – 3)
when x ∈ (0, π)
⇒ cosec x ≥ 1
also cosec2 x ≥ 4
⇒ 4 cosec2 x ≥ 4
⇒ 4 cosec2 x – 3 ≥ 4 – 3 = 1
∴ Δ = cosec x (4 cosec2 x – 3) ≥ 1 x ∈ (0, π).

Question 6.
Solve the equation: $$\left|\begin{array}{ccc} 4 x & 6 x+2 & 8 x+1 \\ 6 x+2 & 9 x+3 & 12 x \\ 8 x+1 & 12 x & 16 x+2 \end{array}\right|$$ = 0
Solution:
Given, $$\left|\begin{array}{ccc} 4 x & 6 x+2 & 8 x+1 \\ 6 x+2 & 9 x+3 & 12 x \\ 8 x+1 & 12 x & 16 x+2 \end{array}\right|$$ = 0
Operating R2 → R2 – $$\frac{3}{2}$$ R1 ;
R3 → R3 – 2R1
$$\left|\begin{array}{ccc} 4 x & 6 x+2 & 8 x+1 \\ 2 & 0 & -3 / 2 \\ 1 & -4 & 0 \end{array}\right|$$ = 0
expanding along R1 ;
4x (0 – 6) – (6x + 2) (0 + $$\frac{3}{2}$$) + (8x + 1) (- 8) = 0
⇒ – 24x – 9x – 3 – 64x – 8 = 0
⇒ – 97x – 11 = 0
⇒ x = $$-\frac{11}{97}$$.

Question 7.
If a, b, c are the sides of a triangle ABC with vertices A (x1, y1), B (x2, y2) and C (x3, y3), show that the equation of the internal bisector of ∠A is $$\left|\begin{array}{ccc} x & y & 1 \\ x_1 & y_1 & 1 \\ b x_2+c x_3 & b y_2+c y_3 & b+c \end{array}\right|$$ = 0.
Solution:
Let AD be the internal bisector of ∠A that divides BC in the ratio c : b
i.e., BD : DC :: c : b
∴ Coordinates of point D are $$\left(\frac{c x_3+b x_2}{c+b}, \frac{c y_3+b y_2}{c+b}\right)$$

Let P (x, y) be any point on AD.
Then points P, A and D lies on same line AD.
i.e. P, A and D are collinear.
∴ Area of Δ PAD = 0
$$\frac{1}{2}\left|\begin{array}{ccc} x & y & 1 \\ x_1 & y_1 & 1 \\ \frac{b x_2+c x_3}{b+c} & \frac{b y+c y_3}{b+c} & 1 \end{array}\right|$$ = 0
[Multiply R3 by b + c]
⇒ $$\left|\begin{array}{ccc} x & y & 1 \\ x_1 & y_1 & 1 \\ b x_2+c x_3 & b y_2+c y_3 & b+c \end{array}\right|$$ = 0 [Hence proved]

Question 8.
Prove that $$\left|\begin{array}{lll} { }^x \mathrm{C}_r & { }^x \mathrm{C}_{r+1} & { }^x \mathrm{C}_{r+2} \\ { }^y \mathrm{C}_r & { }^y \mathrm{C}_{r+1} & { }^y \mathrm{C}_{r+2} \\ { }^z \mathrm{C}_r & { }^z \mathrm{C}_{r+1} & { }^z \mathrm{C}_{r+2} \end{array}\right|$$ = $$\left|\begin{array}{lll} { }^x \mathrm{C}_r & { }^{x+1} \mathrm{C}_{r+1} & { }^{x+2} \mathrm{C}_{r+2} \\ { }^y \mathrm{C}_r & { }^{y+1} \mathrm{C}_{r+1} & { }^{y+2} \mathrm{C}_{r+2} \\ { }^z \mathrm{C}_r & { }^{z+1} \mathrm{C}_{r+1} & { }^{z+2} \mathrm{C}_{r+2} \end{array}\right|$$.
Solution:
L.H.S. = $$\left|\begin{array}{lll} { }^x \mathrm{C}_r & { }^x \mathrm{C}_{r+1} & { }^x \mathrm{C}_{r+2} \\ { }^y \mathrm{C}_r & { }^y \mathrm{C}_{r+1} & { }^y \mathrm{C}_{r+2} \\ { }^z \mathrm{C}_r & { }^z \mathrm{C}_{r+1} & { }^z \mathrm{C}_{r+2} \end{array}\right|$$
Operate C2 → C2 + C1 ;
C3 → C3 + C1
= $$\left|\begin{array}{ccc} { }^x \mathrm{C}_r & { }^x \mathrm{C}_r+{ }^x \mathrm{C}_{r+1} & { }^x \mathrm{C}_r+{ }^x \mathrm{C}_{r+2} \\ y \mathrm{C}_r & { }^y \mathrm{C}_r+{ }^y \mathrm{C}_{r+1} & { }^y \mathrm{C}_r+{ }^y \mathrm{C}_{r+2} \\ z \mathrm{C}_r & z \mathrm{C}_r+{ }^z \mathrm{C}_{r+1} & z \mathrm{C}_r+{ }^z \mathrm{C}_{r+2} \end{array}\right|$$ ;

= $$\left|\begin{array}{ccc} { }^x \mathrm{C}_r & { }^{x+1} \mathrm{C}_{r+1} & { }^{x+1} \mathrm{C}_{r+2} \\ { }^y \mathrm{C}_r & { }^{y+1} \mathrm{C}_{r+1} & { }^{y+1} \mathrm{C}_{r+2} \\ { }^z \mathrm{C}_r & { }^{z+1} \mathrm{C}_{r+1} & { }^{z+1} \mathrm{C}_{r+2} \end{array}\right|$$
[∵xCr + xCr+1 = x+1Cr+1
and xCr + xCr+2 = x+1Cr+2]

Question 9.
If Δr = $$\left|\begin{array}{ccc} r & x & \frac{n(n+1)}{2} \\ 2 r-1 & y & n^2 \\ 3 r-2 & z & \frac{n(3 n-1)}{2} \end{array}\right|$$, prove that $$\sum_{i=1}^n$$Δr = 0.
Solution:
Δr = $$\left|\begin{array}{ccc} r & x & \frac{n(n+1)}{2} \\ 2 r-1 & y & n^2 \\ 3 r-2 & z & \frac{n(3 n-1)}{2} \end{array}\right|$$

Question 10.
If Δr = $$\left|\begin{array}{ccc} 2^{r-1} & 2.3^{r-1} & 4.5^{r-1} \\ x & y & z \\ 2^n-1 & 3^n-1 & 5^n-1 \end{array}\right|$$, prove that $$\sum_{i=1}^n$$Δr = 0.
Solution:
Δr = $$\left|\begin{array}{ccc} 2^{r-1} & 2.3^{r-1} & 4.5^{r-1} \\ x & y & z \\ 2^n-1 & 3^n-1 & 5^n-1 \end{array}\right|$$

Question 11.
If Δr = $$\left|\begin{array}{ccc} r-1 & n & 6 \\ (r-1)^2 & 2 n^2 & 4 n-2 \\ (r-1)^3 & 3 n^3 & 3 n^2-3 n \end{array}\right|$$, prove that $$\sum_{i=1}^n$$Δr = 0.
Solution:
$$\sum_{i=1}^n$$ (r – 1) = Sum of first (n – 1) natural numbers
= $$\frac{(n-1) n}{2}$$
$$\sum_{i=1}^n$$ (r – 1)2 = Sum of squares of first (n – 1) natural numbers = $$\frac{n(n-1)(2 n-1)}{6}$$
$$\sum_{i=1}^n$$ (r – 1)3 = Sum of squares of first (n – 1) natural numbers = $$\frac{n^2(n-1)^2}{4}$$
∴ $$\sum_{i=1}^n$$Δr = $$\left|\begin{array}{ccc} \frac{n(n-1)}{2} & n & 6 \\ \frac{n(n-1)(2 n-1)}{6} & 2 n^2 & 4 n-2 \\ \frac{n^2(n-1)^2}{4} & 3 n^2 & 3 n^2-3 n \end{array}\right|$$ ;
Taking $$\frac{n(n-1)}{12}$$ common from c1 and n common from C2 ; we have
= $$\frac{n^2(n-1)}{12}$$ $$\left|\begin{array}{ccc} 6 & 1 & 6 \\ (2 n-1) 2 & 2 n & 4 n-2 \\ 3 n(n-1) & 3 n & 3 n^2-3 n \end{array}\right|$$
= $$\frac{n^2(n-1)}{12}$$ . 0 = 0
[∵ C1 and C3 are identical]

Question 12.
The determinant $$\left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0 \end{array}\right|$$ is equal to zero if
(i) a, b, c are in A.P.
(ii) a, b, c are in G.P.
(iii) a, b, c are in H.P.
(iv) α is a root of axsup>2 + bx + c = 0
(v) x – α is a factor of axsup>2 + bx + c . pick out the correct alternatives.
Solution:
Let ∆ = $$\left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0 \end{array}\right|$$
operate R3 → R3 – αR1 – R2
= $$\left|\begin{array}{ccc} a & b & a \alpha+b \\ b & c & b \alpha+c \\ 0 & 0 & -a \alpha^2-2 b \alpha-c \end{array}\right|$$
Expanding along C3 ; we have
= (- aα2 – 2bα – c) (ac – b2)
Now ∆ = 0 if (- aα2 – 2bα – c) (ac – b2) = 0
if either aα2 + 2bα + c = 0 or ac – b2 = 0
if α be a root of ax2 + 2bx + c = 0 or ac – b2 = 0
i.e. x – α is a factor of ax2 + 2bx + c or a, b, c in G.P.
Thus Ans. (ii) and (v).

Question 13.
If the matrix $$\left[\begin{array}{ccr} \cos x & \sin x & 2 \\ \sin x & \cos x & -3 \\ 0 & 0 & 1 \end{array}\right]$$ is singular, find the values of x.
Solution:
Let A = $$\left[\begin{array}{ccr} \cos x & \sin x & 2 \\ \sin x & \cos x & -3 \\ 0 & 0 & 1 \end{array}\right]$$
∴ |A| = $$\left|\begin{array}{ccr} \cos x & \sin x & 2 \\ \sin x & \cos x & -3 \\ 0 & 0 & 1 \end{array}\right|$$ ;
expanding along R3
= cos2 x – sin2 x
= cos 2x
Since A is given to be singular
∴ |A| = 0
⇒ cos 2x = 0
⇒ 2x = (2n + 1) $$\frac{\pi}{2}$$ ∀ n ∈ I
⇒ x = (2n + 1) $$\frac{\pi}{4}$$ ∀ n ∈ I

Question 14.
If $$\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} p & -q \\ q & p \end{array}\right]$$, find the values of p and q.
Solution:
Given eqn. be
$$\left[\begin{array}{cc} 1 & -\tan x \\ \tan x & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} p & -q \\ q & p \end{array}\right]$$ …………(1)
Let A = $$\left[\begin{array}{cc} 1 & \tan x \\ -\tan x & 1 \end{array}\right]$$
∴ |A| = 1 + tan2 x
= sec2 x
Here, A11 = 1;
A12 = tan x ;
A21 = – tan x;
A22 = 1

Thus their corresponding elements are equal.
∴ p = cos 2x
and q = sin 2x

Question 15.
If A = diagonal [a1, a2, a3] prove that A-1 = diagonal $$\left[a_1^{-1}, a_2^{-1}, a_3^{-1}\right]$$.
Solution:
Given A = $$\left[\begin{array}{ccc} a_1 & 0 & 0 \\ 0 & a_2 & 0 \\ 0 & 0 & a_3 \end{array}\right]$$ = diag [a1, a2, a3]
Here |D| = a1 $$\mid \begin{array}{cc} a_2 & 0 \\ 0 & a_3 \end{array}$$
= a1a2a3 ≠ 0
∴ D-1 exists and
D-1 = $$\frac{1}{|D|}$$ adj D ………….(1)
Cofactors of 1st row are ;
a2a3 ; 0 ; 0
Cofactors of 2nd row are ;
0 ; a1a3 ; 0
Cofactors of 3rd row are ;
0 ; 0 ; a1a2
∴ adj D = $$\left[\begin{array}{ccc} a_2 a_3 & 0 & 0 \\ 0 & a_1 a_3 & 0 \\ 0 & 0 & a_1 a_2 \end{array}\right]$$
∴ From (1) ; we have
D-1 = $$\frac{1}{a_1 a_2 a_3}\left[\begin{array}{ccc} a_2 a_3 & 0 & 0 \\ 0 & a_1 a_3 & 0 \\ 0 & 0 & a_1 a_2 \end{array}\right]$$
Therefore, D-1 = $$\left[\begin{array}{ccc} 1 / a_1 & 0 & 0 \\ 0 & 1 / a_2 & 0 \\ 0 & 0 & 1 / a_3 \end{array}\right]$$
= diag $$\left[a_1^{-1}, a_2^{-1}, a_3^{-1}\right]$$

Question 16.
Obtain the inverse of the matrices $$\left[\begin{array}{lll} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{array}\right]$$ and $$\left[\begin{array}{lll} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{array}\right]$$. Hence find the inverse of the matrix $$\left[\begin{array}{ccc} 1+p q & p & 0 \\ q & 1+p q & p \\ 0 & q & 1 \end{array}\right]$$.
Solution:
Let A = $$\left[\begin{array}{lll} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{array}\right]$$
and B = $$\left[\begin{array}{lll} 1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1 \end{array}\right]$$
Here, |A| = $$\left|\begin{array}{lll} 1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1 \end{array}\right|$$ ;
expanding along c1
= 1 (1 – 0)
= 1 ≠ 0

Question 17.
Find the product of matrices A = $$\left[\begin{array}{rrr} 2 & 3 & 4 \\ 5 & 4 & -6 \\ 3 & -2 & -2 \end{array}\right]$$ and B = $$\left[\begin{array}{rrr} 20 & 2 & 34 \\ 8 & 16 & -32 \\ 22 & -13 & 7 \end{array}\right]$$. Hence, solve the system of equations.
$$\frac{2}{x}+\frac{3}{y}+\frac{4}{z}$$ = – 3, $$\frac{5}{x}+\frac{4}{y}-\frac{6}{z}$$ = 4, $$\frac{3}{x}-\frac{2}{y}-\frac{2}{z}$$ = 6.
Solution:
Given A = $$\left[\begin{array}{rrr} 2 & 3 & 4 \\ 5 & 4 & -6 \\ 3 & -2 & -2 \end{array}\right]$$
and B = $$\left[\begin{array}{rrr} 20 & 2 & 34 \\ 8 & 16 & -32 \\ 22 & -13 & 7 \end{array}\right]$$

AB = $$\left[\begin{array}{rrr} 2 & 3 & 4 \\ 5 & 4 & -6 \\ 3 & -2 & -2 \end{array}\right]\left[\begin{array}{rrr} 20 & 2 & 34 \\ 8 & 16 & -32 \\ 22 & -13 & 7 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 40+24+88 & 4+48-52 & 68-96+28 \\ 100+32-132 & 10+64+78 & 170-128-42 \\ 60-16-44 & 6-32+26 & 102+64-14 \end{array}\right]$$
= $$\left[\begin{array}{rrr} 152 & 0 & 0 \\ 0 & 152 & 0 \\ 0 & 0 & 152 \end{array}\right]$$

⇒ AB = 152 $$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$$
= 152 I2

⇒ A ($$\frac{1}{152}$$ B) = I3
⇒ A-1 = $$\frac{1}{152}$$ B ………..(1)
[by def. of inverse]
putting $$\frac{1}{x}$$ = u;
$$\frac{1}{y}$$ = v ;
and $$\frac{1}{z}$$ = w
in given system of eqn’s we have

2u + 3v + 4w = – 3
5u + 4v – 6w = 4
3u – 2v – 2w = 6
Given system of eqns in equivalent to AX = C
where A = $$\left[\begin{array}{rrr} 2 & 3 & 4 \\ 5 & 4 & -6 \\ 3 & -2 & -2 \end{array}\right]$$ ;
X = $$\left[\begin{array}{l} u \\ v \\ w \end{array}\right]$$ ;
C = $$\left[\begin{array}{r} -3 \\ 4 \\ 6 \end{array}\right]$$
Here |A| = 2 (- 8 – 12) – 3 (- 10 + 18) + 4 (- 10 – 12)
= – 40 – 24 – 88
= – 152 ≠ 0
∴ A-1 exists and given system has unique solution.
∴ X = A-1C
=$$\frac{1}{152}$$ BC [ising (1)]
⇒ X = $$\frac{1}{152}\left[\begin{array}{rrr} 20 & 2 & 34 \\ 8 & 16 & -32 \\ 22 & -13 & 7 \end{array}\right]\left[\begin{array}{r} -3 \\ 4 \\ 6 \end{array}\right]$$
= $$\frac{1}{152}\left[\begin{array}{l} -60+8+204 \\ -23+64-192 \\ -66-52+42 \end{array}\right]$$

⇒ $$\left[\begin{array}{l} u \\ v \\ w \end{array}\right]$$ = $$\frac{1}{152}\left[\begin{array}{r} 152 \\ -152 \\ -76 \end{array}\right]$$
= $$\left[\begin{array}{r} 1 \\ -1 \\ -1 / 2 \end{array}\right]$$
i.e. u = 1 ; v = – 1 ; w = – $$\frac{1}{2}$$
⇒ $$\frac{1}{x}$$ = 1 ;
$$\frac{1}{y}$$ = – 1 ;
$$\frac{1}{z}$$ = – $$\frac{1}{2}$$
Thus, x = 1 ; y = – 1 and z = – 2.