ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.5

Practicing ML Aggarwal Class 12 Solutions ISC Chapter 3 Matrices Ex 3.5 is the ultimate need for students who intend to score good marks in examinations.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 3 Matrices Ex 3.5

Very short answer/objective questions (1 to 2):

Question 1.
Use elementary row operation
R₁ → R₁ – 2R₂ in the following matrix equation :
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right]=\left[\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
5 & -3 & 1
\end{array}\right]\), where A is a 3 × 3 matrix.
Solution:
Given \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right]=\left[\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
5 & -3 & 1
\end{array}\right]\) A
operate R₁ → R₁ – 2R₂
\(\left[\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 2
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
1 & 0 & 0 \\
5 & -3 & 1
\end{array}\right]\)

Question 1 (old).
Use elemenatry column operation C₂ → C₂ + 2C₁ in the following matrix equation:
\(\left[\begin{array}{ll}
2 & 1 \\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
3 & 1 \\
2 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 1
\end{array}\right]\)
Solution:
Given \(\left[\begin{array}{ll}
2 & 1 \\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
3 & 1 \\
2 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 1
\end{array}\right]\)
Operate C₂ → C₂ + 2C₁
\(\left[\begin{array}{ll}
2 & 5 \\
2 & 4
\end{array}\right]=\left[\begin{array}{ll}
3 & 1 \\
2 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
-1 & -1
\end{array}\right]\)

Question 2.
Use elementary column operation
C₃ → C₃ – 2C₁ in the following matrix equation:
\(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
1 & 3 & 1
\end{array}\right]=A\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\), where A is a 3 × 3 matrix.
Solution:
Given \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
1 & 3 & 1
\end{array}\right]=\mathrm{A}\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Operate C₃ → C₃ – 2C₁
\(\left[\begin{array}{rrr}
1 & -1 & 0 \\
2 & 1 & -1 \\
1 & 3 & -1
\end{array}\right]=\mathrm{A}\left[\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & -2 \\
0 & 0 & 1
\end{array}\right]\)

Question 3.
Find the inverse of the following (2 to 4) matrices, if they exist, by using elementary operations:
(i) \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\) (NCERT)
(ii) \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\) (NCERT)
(iii) \(\left[\begin{array}{rr}
3 & 10 \\
2 & 7
\end{array}\right]\) (NCERT)
Solution:
(i) (By elementary row transformations)
We know that,
A = I₂ A
∴ \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) A ;
R₂ → R₂ – 2 R₁

\(\left[\begin{array}{ll}
2 & 3 \\
1 & 1
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right]\) A ;
R₁ ↔ R₂

\(\left[\begin{array}{ll}
1 & 1 \\
2 & 3
\end{array}\right]=\left[\begin{array}{rr}
-2 & 1 \\
1 & 0
\end{array}\right]\) A ;
R₂ → R₂ – 2 R₁

\(\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-2 & 1 \\
5 & -2
\end{array}\right]\) A ;
R₁ → R₁ – R₂

\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-7 & 3 \\
5 & -2
\end{array}\right]\) A
Hence A^-1 = \(\left[\begin{array}{rr}
-7 & 3 \\
5 & -2
\end{array}\right]\)
(By elementary column transformations)
we know that,
A = I₂ A

(ii) Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\).
Then A = I₂ A
⇒ \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Operate R₂ → R₂ – 2R₁

⇒ \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right]\) A
Operate R₁ → R₁ – 3R₂

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
7 & -3 \\
-2 & 1
\end{array}\right]\) A
⇒ I₂ = BA
where B = \(\left[\begin{array}{rr}
7 & -3 \\
-2 & 1
\end{array}\right]\)
∴ A^-1 = B
= \(\left[\begin{array}{rr}
7 & -3 \\
-2 & 1
\end{array}\right]\).

(iii) Let A = \(\)
We know that A = IA
⇒ \(\left[\begin{array}{rr}
3 & 10 \\
2 & 7
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) A;
operate R₁ → R₁ – R₂

⇒ \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]=\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]\) A;
operate R₂ → R₂ – 2R₁

⇒ \(\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
1 & -1 \\
-2 & 3
\end{array}\right]\) A;
operate R₁ → R₁ – 3R₂

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
7 & -10 \\
-2 & 3
\end{array}\right]\) A;
Hence A^-1 = \(\left[\begin{array}{rr}
7 & -10 \\
-2 & 3
\end{array}\right]\)

Question 4.
(i) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]\) (NCERT)
(iii) \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Solution:
(i) By elementary row transformations
we know that,
A = I₂ A
⇒ \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) A ;
operate : R₂ → R₂ – 2R₁

⇒ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right]\) A ;
operate : R₁ → R₁ + \(\frac{1}{5}\)R₂

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 5
\end{array}\right]=\left[\begin{array}{rr}
3 / 5 & 1 / 5 \\
-2 & 1
\end{array}\right]\) ;
R₂ → \(\frac{1}{5}\)R₂

\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\) A
Hence A^-1 = \(\left[\begin{array}{rr}
3 / 5 & 1 / 5 \\
-2 / 5 & 1 / 5
\end{array}\right]\)
= \(\frac{1}{5}\left[\begin{array}{rr}
3 & 1 \\
-2 & 1
\end{array}\right]\)

(ii) \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]\)

∴ A^-1 = \(\frac{1}{2}\left[\begin{array}{ll}
2 & 1 \\
4 & 3
\end{array}\right]\) [by def. of inverse]

(iii) Let A = \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]\)
Then A = I₂A
⇒ \(\left[\begin{array}{rr}
10 & -2 \\
-5 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) A
Operate R₁ ↔ R₂

\(\left[\begin{array}{rr}
-5 & 1 \\
10 & -2
\end{array}\right]=\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) A
Operate R₂ → R₂ + 2R₁

\(\left[\begin{array}{rr}
-5 & 1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 1 \\
1 & 2
\end{array}\right]\) A
Here all entries in the 2nd row of L.H.S. of the matrix eqn. are zeroes.
∴ A is not invertible.
i.e., A^-1 does not exists.
[Note : Here |A| = \(\left|\begin{array}{rr}
10 & -2 \\
-5 & +1
\end{array}\right|\)
= + 10 – 10 = 0
∴ A is singular.
∴ A is not invertible.]

Question 5.
(i) \(\left[\begin{array}{rrr}
3 & 0 & -1 \\
2 & 3 & 0 \\
0 & 4 & 1
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
1 & 2 & 3 \\
2 & 5 & 7 \\
-2 & -4 & -5
\end{array}\right]\)
(iii) \(\left[\begin{array}{rrr}
2 & -1 & 4 \\
4 & 0 & 2 \\
3 & -2 & 7
\end{array}\right]\)
Solution:
(i) Let A = \(\left[\begin{array}{rrr}
3 & 0 & -1 \\
2 & 3 & 0 \\
0 & 4 & 1
\end{array}\right]\)
We know that A =IA

Hence A^-1 = \(\left[\begin{array}{rrr}
3 & -4 & 3 \\
-2 & 3 & -2 \\
8 & -12 & 9
\end{array}\right]\).

(ii) We know that A = I₃A

∴ A^-1 = \(\left[\begin{array}{rrr}
3 & -2 & -1 \\
-4 & 1 & -1 \\
2 & 0 & 1
\end{array}\right]\) [by def. of inverse]

(iii) Let A = \(\left[\begin{array}{rrr}
2 & -1 & 4 \\
4 & 0 & 2 \\
3 & -2 & 7
\end{array}\right]\)
We know that A = IA

Question 6.
Using elementary row operations, find the inverse of the following matrix:
(i) A = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-5 & 3 & 1 \\
-3 & 2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
Solution:
Given A = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-5 & 3 & 1 \\
-3 & 2 & 3
\end{array}\right]\)
Then A = I₃A
⇒ \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-5 & 3 & 1 \\
-3 & 2 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) A
operate R₃ → R₃ + R₁

⇒ I₃ = BA
where B = \(\left[\begin{array}{rrr}
-7 & -9 & 10 \\
-12 & -15 & 17 \\
1 & 1 & -1
\end{array}\right]\)
∴ A^-1 = B
= \(\left[\begin{array}{rrr}
-7 & -9 & 10 \\
-12 & -15 & 17 \\
1 & 1 & -1
\end{array}\right]\)

(ii) Let A = \(\left[\begin{array}{rrr}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right]\)
Since A = I₂ A

⇒ I₃ = BA
Thus by def. of inverse,
B = A^-1
= \(\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\)