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

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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 = AR^{p – 1} ;
b = AR^{q – 1}
and c = AR^{r – 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
∴ T_p = ar^p-1 ;
T_p+1 = ar^p ;
T_p+3 = ar^p+2

T_p+5 = ar^p+4 ;
T_p+7 = ar^p+6 ;
T_p+9 = ar^p+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
= \(\left|\begin{array}{ccc}
{ }^x \mathrm{C}_r & { }^x \mathrm{C}_r+{ }^x \mathrm{C}_{r+1} & { }^x \mathrm{C}_r+{ }^x \mathrm{C}_{r+2} \\
y \mathrm{C}_r & { }^y \mathrm{C}_r+{ }^y \mathrm{C}_{r+1} & { }^y \mathrm{C}_r+{ }^y \mathrm{C}_{r+2} \\
z \mathrm{C}_r & z \mathrm{C}_r+{ }^z \mathrm{C}_{r+1} & z \mathrm{C}_r+{ }^z \mathrm{C}_{r+2}
\end{array}\right|\) ;

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

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

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

Question 11.
If Δ_r = \(\left|\begin{array}{ccc}
r-1 & n & 6 \\
(r-1)^2 & 2 n^2 & 4 n-2 \\
(r-1)^3 & 3 n^3 & 3 n^2-3 n
\end{array}\right|\), prove that \(\sum_{i=1}^n\)Δ_r = 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 = \(\frac{n^2(n-1)^2}{4}\)
∴ \(\sum_{i=1}^n\)Δ_r = \(\left|\begin{array}{ccc}
\frac{n(n-1)}{2} & n & 6 \\
\frac{n(n-1)(2 n-1)}{6} & 2 n^2 & 4 n-2 \\
\frac{n^2(n-1)^2}{4} & 3 n^2 & 3 n^2-3 n
\end{array}\right|\) ;
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.