2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.2

Students can Download Basic Maths Exercise 1.2 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.2

Part – A

2nd PUC Basic Maths Matrices and Determinants One Mark Questions and Answers

Question 1.
If A = \(\left[ \begin{matrix} 3 & 1 \\ 2 & 4 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} -1 \\ 3 \end{matrix} \right]\) . Find AB
Answers:

Question 2.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 0 & -3 \end{matrix} \right]\) . Find A²
Answer:

Question 3.
If A = \(\left[ \begin{matrix} 3 & -1 \\ 2 & -4 \end{matrix} \right]\) . Find A²
Answer:

Question 4.
If A = \(\left[ \begin{matrix} 3 & 1 & 2 \\ 0 & -1 & 2 \\ -4 & 1 & -3 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]\) . Find AB
Answer:

Question 5.
If \(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 0 \end{matrix} \right]\) . Find x and y.
Answer:

Question 6.
If \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & -1 \\ 2 & 0 & -1 \\ 0 & 1 & -2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right]\) . Find x,y,z.
Answer:

Question 7.
If \(\left[ \begin{matrix} 4 \\ 1 \\ 3 \end{matrix} \right] \left[ \begin{matrix} x & y & z \end{matrix} \right] =\left[ \begin{matrix} 4 & 8 & 4 \\ 1 & 2 & 1 \\ 3 & 6 & 2 \end{matrix} \right]\) . Find x,y,z.
Answer:

Question 8.
If A = \(\left[ \begin{matrix} 1 \\ 4 \\ 2 \end{matrix} \right]\) anf B = \(\left[ \begin{matrix} 1 & 3 & 4 \end{matrix} \right]\) . Find AB.
Answer:

Question 9.
Find A.B, if A = \(\left[ \begin{matrix} 1 & 3 & 2 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 4 \\ 2 \\ 1 \end{matrix} \right]\) .
Answer:

Question 10.
Find x, if \(\left[ \begin{matrix} 2 & \times & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 4 \\ 2 \end{matrix} \right] =\left[ 3 \right] \) .
Answer:

[2 + 4x + 4] = [3]
⇒ 4x = 3 – 6 = -3 ; x = \(\frac { -3 }{ 4 }\)

Part – B

2nd PUC Basic Maths Matrices and Determinants Two Marks Questions and Answers

Question 1.
If A = \(\left[ \begin{matrix} 2 & -2 \\ 3 & -1 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 3 & -1 \\ -2 & 0 \end{matrix} \right]\) . Find AB and BA
Answer:

Question 2.
If A = \(\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 1 & 4 & 2 \end{matrix} \right]\) .Find AB and BA.
Answer:

Question 3.
If A = \(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & 0 \end{matrix} \right]\) Find AA’ and A’A
Answer:

Question 4.
If A = \(\left[ \begin{matrix} 3 & -1 & 2 \\ 1 & 2 & 2 \\ 2 & 0 & 5 \end{matrix} \right]\) , Find A2
Answer:

Question 5.
If A = \(\left[ \begin{matrix} 2 & 3 \\ -1 & 4 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 1 & -1 \\ 2 & 4 \end{matrix} \right]\) Find AB’ and A’B
Answer:

Question 6.
If A = \(\left[ \begin{matrix} 2 & -1 \\ 1 & 4 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} -3 & 2 \\ -1 & 4 \end{matrix} \right]\) . show that (AB)’ = mB’A’
Answer:

Question 7.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 1 & 4 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 4 & -3 \\ 2 & 1 \end{matrix} \right]\) and C = \(\left[ \begin{matrix} 1 & 0 \\ -2 & 4 \end{matrix} \right]\) .
verify that (i) A(BC) = (AB)A (ii) A(B+C) = AB + AC
Answer:

Question 8.
If A = \(\left[ \begin{matrix} 1 & 3 \\ 1 & 0 \end{matrix} \right]\) . prove that A² – A – 31 = 0
Answer:

Question 9.
If A = \(\left[ \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right]\) . Show that A² – 4A + 31 = 0
Answer:

Question 10.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right]\) Show that A² – 5A = 21.
Answer:

Question 11.
Solve for x, y:
(i) \(\left[ \begin{matrix} 2 & -1 \\ 3 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 2 \end{matrix} \right]\)
(ii) \(\left[ \begin{matrix} 1 & 3 \\ -2 & 4 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 0 \end{matrix} \right]\)
(iii) \(\left[ \begin{matrix} x & y \end{matrix} \right] \left[ \begin{matrix} 2 & -1 \\ 3 & 1 \end{matrix} \right] =\left[ \begin{matrix} -1 & 4 \end{matrix} \right]\) .
Answer:

Question 12.
If x = \(\left[ \begin{matrix} 2 & 1 & 3 \\ 0 & 1 & 4 \end{matrix} \right]\) , y = \(\left[ \begin{matrix} 1 \\ 0 \\ 3 \end{matrix} \right]\) and z = [2 1]. verify that x(yz) = (xy)z.
Answer:

From 1 & 2 we get (xy)z = x(yz).

Part – C

2nd PUC Basic Maths Matrices and Determinants Three Marks Questions and Answers

Question 1.
If A = \(\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 2 & 1 & 4 \\ 0 & 1 & 3 \end{matrix} \right]\) . and C = \(\left[ \begin{matrix} 1 & -3 & -1 \\ -2 & 4 & 5 \\ 1 & 3 & -2 \end{matrix} \right]\) . verify (AB)C = A(BC).
Answer:

Question 2.
If A = \(\left[ \begin{matrix} 1 & 2 & -3 \\ 1 & -4 & 1 \\ 0 & 5 & 3 \end{matrix} \right]\) , B = \(\left[ \begin{matrix} 4 & -2 & -3 \\ 2 & -4 & -1 \\ 0 & 1 & 3 \end{matrix} \right]\) , C = \(\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 1 \\ -1 & -3 & 4 \end{matrix} \right]\) . verify (A + B)C = AC + BC.
Answer:


From 1 & 2 we get (A + B)C = AC + BC

Question 3.
If A = \(\left[ \begin{matrix} i & 0 \\ 0 & -1 \end{matrix} \right]\) B = \(\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right]\) and C = \(\left[ \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right]\) where i2 = -1, show that
(i) A2 = B2 = C2 = -1
(ii) AB = -BA = -C
Answer:

Question 4.
If A = \(\left[ \begin{matrix} -1 & 1 & 0 \\ 3 & -3 & 3 \\ 5 & -5 & 5 \end{matrix} \right]\) , B = \(\left[ \begin{matrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{matrix} \right]\) . Show that A² . B² = A².
Answer:

Hence A²B² = A² is proved.

Question 5.
If f(x) = x² – 5x + 7, find f(A) where A = \(\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \end{matrix} \right]\) .
Answer:

Question 6.
If A = \(\left[ \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right]\) B = \(\left[ \begin{matrix} 2 & 2 & -4 \\ -4 & 2 & 4 \\ 2 & -1 & 5 \end{matrix} \right]\) Show that BA = 61
Answer:

Hence BA = 61 is proved

Question 7.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix} \right]\) and A + 2B = A², find B
Answer:
Given A + 2B = A²
2B = A² – A

Question 8.
If \(\left[ \begin{matrix} 1 & 2 \\ 4 & 7 \end{matrix} \right] \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]\) . Find a, b, c, d.
Answer:

∴ a = -7, b = 2, c = 4, d = -1.

Question 9.
If A = \(\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right]\) and I = \(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]\) . Show that A² – (a + d)A – (bc – ad) I is a null matrix.
Answer:

Question 10.
If A = \(\left[ \begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right]\) . Prove that A² – 4A – 51 is a null matrix of order 3 × 3.
Answer:

∴ A² – 4A – 51 is a null matrix.