Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.2 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 1 Matrices and Determinants Ex 1.2

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.2 Text Book Back Questions and Answers

Question 1.
Find the adjoint of the matrix A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 4
\end{array}\right]\)
Adj A = \(\left[\begin{array}{rr}
4 & -3 \\
-1 & 2
\end{array}\right]\)

Question 2.
If A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\) then verify that A(adj A) = |A| I and also find A-1.
Solution:
Given A = \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\)
|A| = \(\left|\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right|\)
= \(1\left|\begin{array}{ll}
4 & 3 \\
3 & 4
\end{array}\right|-3\left|\begin{array}{ll}
1 & 3 \\
1 & 4
\end{array}\right|+3\left|\begin{array}{ll}
1 & 4 \\
1 & 3
\end{array}\right|\)
= 1[16 – 9] – 3[4 – 3] + 3[3 – 4]
= 1(7) – 3(1) + 3(-1)
= 7 – 3 – 3
= 1
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q2
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q2.1
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q2.2
From (1) and (2), A(Adj A) = |A| I

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2

Question 3.
Find the inverse of each of the following matrices:
(i) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
(ii) \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 3
\end{array}\right]\)
(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
(iv) \(\left[\begin{array}{rrr}
-3 & -5 & 4 \\
-2 & 3 & -1 \\
1 & -4 & -6
\end{array}\right]\)
Solution:
(i) \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3

(ii) \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 3
\end{array}\right]\)
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.1

(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.2
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.3
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.4

(iv) \(\left[\begin{array}{rrr}
-3 & -5 & 4 \\
-2 & 3 & -1 \\
1 & -4 & -6
\end{array}\right]\)
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.5
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q3.6

Question 4.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
1 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 4 \\
1 & -2
\end{array}\right]\), then verify adj(AB) = (adj B) (adj A).
Solution:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q4
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q4.1
From (1) and (2), adj (AB) = (adj B) (adj A)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2

Question 5.
If A = \(\left[\begin{array}{rrr}
2 & -2 & 2 \\
2 & 3 & 0 \\
9 & 1 & 5
\end{array}\right]\) then, show that (adj A) A = O.
Solution:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q5
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q5.1

Question 6.
If A = \(\left[\begin{array}{rrr}
-1 & 2 & -2 \\
4 & -3 & 4 \\
4 & -4 & 5
\end{array}\right]\) then, show that the inverse of A is A itself.
Solution:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q6
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q6.1
∴ A-1 = A
Hence proved.

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2

Question 7.
If A-1 = \(\left[\begin{array}{rrr}
1 & 0 & 3 \\
2 & 1 & -1 \\
1 & -1 & 1
\end{array}\right]\) then, find A.
Solution:
Given A-1 = \(\left[\begin{array}{rrr}
1 & 0 & 3 \\
2 & 1 & -1 \\
1 & -1 & 1
\end{array}\right]\)
We know that (A-1)-1 = A
So we have to find inverse of A-1
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q7
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q7.1

Question 8.
Show that the matrices A = \(\left[\begin{array}{lll}
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{4}{5} & \frac{-2}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{3}{5} & \frac{-1}{5} \\
\frac{-1}{5} & \frac{-2}{5} & \frac{4}{5}
\end{array}\right]\) are inverses of each other.
Solution:
To prove that A and B are inverses of each other.
We have to prove that AB = BA = I.
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q8
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q8.1
Thus AB = BA = I
Hence A and B are inverses of each other.

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2

Question 9.
If A = \(\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right]\), then verify that (AB)-1 = B-1A-1
Solution:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q9
Now we will find B-1A-1
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q9.1
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q9.2
From (1) and (2), (AB)-1 = B-1A-1

Question 10.
Find λ if the matrix \(\left[\begin{array}{rrr}
1 & 1 & 3 \\
2 & \lambda & 4 \\
9 & 7 & 11
\end{array}\right]\) has no inverse.
Solution:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q10
1[11λ – 28] – 1[22 – 36] + 3[14 – 9λ] = 0
11λ – 28 + 14 + 42 – 27λ = 0
-16λ + 28 = 0
-16λ = -28
λ = \(\frac{-28}{-16}=\frac{7}{4}\)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2

Question 11.
If X = \(\left[\begin{array}{rrr}
8 & -1 & -3 \\
-5 & 1 & 2 \\
10 & -1 & -4
\end{array}\right]\) and Y = \(\left[\begin{array}{rrr}
2 & 1 & -1 \\
0 & 2 & 1 \\
5 & p & q
\end{array}\right]\) then, find p, q if Y = X-1
Solution:
Given that Y is the inverse of X.
∴ XY = I
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.2 Q11
6 – 3p = 0 and -9 – 3q = 0
6 = 3p and -9 = 3q
∴ p = 2; q = -3