Students can download 11th Business Maths Chapter 1 Matrices and Determinants Ex 1.5 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 1.5

Samacheer Kalvi 11th Business Maths Matrices and Determinants Ex 1.5 Text Book Back Questions and Answers

Choose the Correct Answer.

Question 1.
The value of x if \(\left|\begin{array}{lll}
0 & 1 & 0 \\
x & 2 & x \\
1 & 3 & x
\end{array}\right|=0\) is
(a) 0, -1
(b) 0, 1
(c) -1, 1
(d) -1, -1
Answer:
(b) 0, 1
Hint:
0 – 1[x2 – x] + 0 = 0
⇒ x2 – x = 0
⇒ x(x – 1) = 0
⇒ x = 0 (or) x = 1

Question 2.
The value of \(\left|\begin{array}{lll}
2 x+y & x & y \\
2 y+z & y & z \\
2 z+x & z & x
\end{array}\right|\) is
(a) xyz
(b) x + y + z
(c) 2x + 2y + 2z
(d) 0
Answer:
(d) 0
Hint:
= \(\left|\begin{array}{lll}
2 x & x & y \\
2 y & y & z \\
2 z & z & x
\end{array}\right|\) C1 → C1 – C3
= 0 (C1 and C2 are proportional)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 3.
The cofactor of -7 in the determinant \(\left|\begin{array}{rrr}
2 & -3 & 5 \\
6 & 0 & 4 \\
1 & 5 & -7
\end{array}\right|\) is
(a) -18
(b) 18
(c) -7
(d) 7
Answer:
(b) 18
Hint:
A cofactor of -7 = \(\left|\begin{array}{rr}
2 & -3 \\
6 & 0
\end{array}\right|\)
= 0 + 18
= 18

Question 4.
If Δ = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
2 & 3 & 1
\end{array}\right|\) then \(\left|\begin{array}{lll}
3 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right|\) is
(a) Δ
(b) -Δ
(c) 3Δ
(d) -3Δ
Answer:
(b) -Δ
Hint:
\(\left|\begin{array}{lll}
3 & 1 & 2 \\
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right|=-\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 1 & 2 \\
2 & 3 & 1
\end{array}\right|\) R1 ↔ R2
= -Δ

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 5.
The value of the determinant \(\left|\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right|^{2}\) is
(a) abc
(b) 0
(c) a2b2c2
(d) -abc
Answer:
(c) a2b2c2
Hint:
\(a^{2} b^{2} c^{2}\left|\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|\)
= a2b2c2 × 12
= a2b2c2

Question 6.
If A is square matrix of order 3 then |kA| is:
(a) k|A|
(b) -k|A|
(c) k3|A|
(d) -k3|A|
Answer:
(c) k3|A|

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 7.
adj (AB) is equal to:
(a) adj A adj B
(b) adj AT adj BT
(c) adj B adj A
(d) adj BT adj AT
Answer:
(c) adj B adj A

Question 8.
The inverse matrix of \(\left(\begin{array}{cc}
\frac{4}{5} & \frac{5}{12} \\
\frac{2}{5} & \frac{1}{2}
\end{array}\right)\) is
(a) \(\frac{7}{30}\left(\begin{array}{cc}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)
(b) \(\frac{7}{30}\left(\begin{array}{cc}
\frac{1}{2} & \frac{-5}{12} \\
\frac{-2}{5} & \frac{1}{5}
\end{array}\right)\)
(c) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)
(d) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{-5}{12} \\
\frac{-2}{5} & \frac{4}{5}
\end{array}\right)\)
Answer:
(c) \(\frac{30}{7}\left(\begin{array}{rr}
\frac{1}{2} & \frac{5}{12} \\
\frac{2}{5} & \frac{4}{5}
\end{array}\right)\)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 9.
If A = \(\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\) such that ad – bc ≠ 0 then A-1 is:
(a) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & b \\
-c & a
\end{array}\right]\)
(b) \(\frac{1}{a d-b c}\left[\begin{array}{ll}
d & b \\
c & a
\end{array}\right]\)
(c) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
(d) \(\frac{1}{a d-b c}\left[\begin{array}{ll}
d & -b \\
c & a
\end{array}\right]\)
Answer:
(c) \(\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
Hint:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5 Q9

Question 10.
The number of Hawkins-Simon conditions for the viability of input-output analysis is:
(a) 1
(b) 3
(c) 4
(d) 2
Answer:
(d) 2

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 11.
The inventor of input-output analysis is:
(a) Sir Francis Galton
(b) Fisher
(c) Prof. Wassily W. Leontief
(d) Arthur Cayley
Answer:
(c) Prof. Wassily W. Leontief

Question 12.
Which of the following matrix has no inverse?
(a) \(\left(\begin{array}{rr}
-1 & 1 \\
1 & -4
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\)
(c) \(\left(\begin{array}{cc}
\cos a & \sin a \\
-\sin a & \cos a
\end{array}\right)\)
(d) \(\left(\begin{array}{rr}
\sin a & \sin a \\
-\cos a & \cos a
\end{array}\right)\)
Answer:
(b) \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\)
Hint:
So \(\left(\begin{array}{rr}
2 & -1 \\
-4 & 2
\end{array}\right)\) has no inverse.

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 13.
Inverse of \(\left(\begin{array}{ll}
3 & 1 \\
5 & 2
\end{array}\right)\) is:
(a) \(\left(\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
-2 & 5 \\
1 & -3
\end{array}\right)\)
(c) \(\left(\begin{array}{rr}
3 & -1 \\
-5 & -3
\end{array}\right)\)
(d) \(\left(\begin{array}{rr}
-3 & 5 \\
1 & -2
\end{array}\right)\)
Answer:
(a) \(\left(\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right)\)
Hint:
Let A = \(\left(\begin{array}{ll}
3 & 1 \\
5 & 2
\end{array}\right)\)
|A| = [6 – 5] = 1
adj A = \(\left[\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right]\)
∴ A-1 = \(\left[\begin{array}{rr}
2 & -1 \\
-5 & 3
\end{array}\right]\)

Question 14.
If A = \(\left(\begin{array}{rr}
-1 & 2 \\
1 & -4
\end{array}\right)\) then A (adj A) is:
(a) \(\left(\begin{array}{ll}
-4 & -2 \\
-1 & -1
\end{array}\right)\)
(b) \(\left(\begin{array}{rr}
4 & -2 \\
-1 & 1
\end{array}\right)\)
(c) \(\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)
(d) \(\left(\begin{array}{ll}
0 & 2 \\
2 & 0
\end{array}\right)\)
Answer:
(c) \(\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)
Hint:
A = \(\left(\begin{array}{rr}
-1 & 2 \\
1 & -4
\end{array}\right)\)
|A| = 4 – 2 = 2
We know that A (adj A) = |A| I
⇒ 2 \(\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right)\)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 15.
If A and B non-singular matrix then, which of the following is incorrect?
(a) A2 = I implies A-1 = A
(b) I-1 = I
(c) If AX = B then X = B-1A
(d) If A is square matrix of order 3 then |adj A| = |A|2
Answer:
(c) If AX = B then X = B-1A
Hint:
If AX = B then X = A-1B so, X = B-1A is incorrect.

Question 16.
The value of \(\left|\begin{array}{rrr}
5 & 5 & 5 \\
4 x & 4 y & 4 z \\
-3 x & -3 y & -3 z
\end{array}\right|\) is:
(a) 5
(b) 4
(c) 0
(d) -3
Answer:
(c) 0
Hint:
= 4 × (-3) \(\left|\begin{array}{lll}
5 & 5 & 5 \\
x & y & z \\
x & y & z
\end{array}\right|\)
[Take out 4 from R2 and -3 from R3]
= 0 (∵ R2 ≡ R3)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 17.
If A is an invertible matrix of order 2 then det (A-1) be equal
(a) det (A)
(b) \(\frac{1}{{det}(A)}\)
(c) 1
(d) 0
Answer:
(b) \(\frac{1}{{det}(A)}\)
Hint:
AA-1 = I
|AA-1| = |I|
|A| |A-1| = 1
|A-1| = \(\frac{1}{|\mathrm{A}|}\)
det A-1 = \(\frac{1}{\det (A)}\)

Question 18.
If A is 3 × 3 matrix and |A| = 4 then |A-1| is equal to:
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{16}\)
(c) 2
(d) 4
Answer:
(a) \(\frac{1}{4}\)
Hint:
|A-1| = \(\frac{1}{|A|}=\frac{1}{4}\)

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 19.
If A is a square matrix of order 3 and |A| = 3 then |adj A| is equal to:
(a) 81
(b) 27
(c) 3
(d) 9
Answer:
(d) 9
Hint:
|adj A| = |A|2 = 32 = 9

Question 20.
The value of \(\left|\begin{array}{ccc}
x & x^{2}-y z & 1 \\
y & y^{2}-z x & 1 \\
z & z^{2}-x y & 1
\end{array}\right|\) is:
(a) 1
(b) 0
(c) -1
(d) -xyz
Answer:
(b) 0
Hint:
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5 Q20
Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5 Q20.1

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 21.
If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), then |2A| is equal to:
(a) 4 cos 2θ
(b) 4
(c) 2
(d) 1
Answer:
(b) 4
Hint:
|2A| = 22 |A|
= 4 \(\left|\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right|\)
= 4 [cos2θ + sin2θ]
= 4 × 1
= 4

Question 22.
If Δ = \(\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\) and Aij is cofactor of aij, then value of Δ is given by:
(a) a11A31 + a12A32 + a13A33
(b) a11A11 + a12A21 + a13A31
(c) a21A11 + a22A12 + a23A13
(d) a11A11 + a21A21 + a31A31
Answer:
(d) a11A11 + a21A21 + a31A31

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 23.
If \(\left|\begin{array}{ll}
x & 2 \\
8 & 5
\end{array}\right|=0\) then the value of x is:
(a) \(\frac{-5}{6}\)
(b) \(\frac{5}{6}\)
(c) \(\frac{-16}{5}\)
(d) \(\frac{16}{5}\)
Answer:
(d) \(\frac{16}{5}\)
Hint:
\(\left|\begin{array}{ll}
x & 2 \\
8 & 5
\end{array}\right|=0\)
5x – 16 = 0
⇒ x = \(\frac{16}{5}\)

Question 24.
If \(\left|\begin{array}{ll}
4 & 3 \\
3 & 1
\end{array}\right|\) = -5 then the value of \(\left|\begin{array}{rr}
20 & 15 \\
15 & 5
\end{array}\right|\) is:
(a) -5
(b) -125
(c) -25
(4) 0
Answer:
(b) -125
Hint:
\(\left|\begin{array}{rr}
20 & 15 \\
15 & 5
\end{array}\right|\)
= 5 × 5 \(\left|\begin{array}{ll}
4 & 3 \\
3 & 1
\end{array}\right|\)
= 5 × 5 × (-5)
= -125

Samacheer Kalvi 11th Business Maths Guide Chapter 1 Matrices and Determinants Ex 1.5

Question 25.
If any three rows or columns of a determinant are identical then the value of the determinant is:
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(a) 0