2nd PUC Basic Maths Question Bank Chapter 1 Matrices and Determinants Ex 1.1

Students can Download Basic Maths Exercise 1.1 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 1.1

Part – A

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 One Mark Questions and Answers

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

Question 2.
If A = \(\left[ \begin{matrix} 1 & -3 & 5 \\ 6 & 2 & 4 \end{matrix} \right]\) find 5A’.
Answer:

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

Question 4.
If  \(\left[ \begin{matrix} x+y & 3 \\ 2 & -y+x \end{matrix} \right] +\left[ \begin{matrix} 2 & 3 \\ 4 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 6 \\ 6 & 0 \end{matrix} \right]\) . Find x and y.
Answer:

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

⇒ x² + 2x = -1
x² + 2x + 1 = 0
(x + 1)² = 0 ⇒ x + 1 = 0
⇒ x = -1

Question 6.
If A = \(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right]\) . Find A – A’
Answer:

Question 7.
If A = \(\left[ \begin{matrix} 2 & -x \\ x & -7 \end{matrix} \right]\) . Find A + A’.
Answer:

Question 8.
If A = \(\left[ \begin{matrix} 1 & 2 \\ 3 & -1 \end{matrix} \right]\) B = \(\left[ \begin{matrix} 2 & 5 \\ -3 & -1 \end{matrix} \right]\) C = \(\left[ \begin{matrix} 1 & -1 \\ 4 & -3 \end{matrix} \right]\) Find 2A – 3B – C
Answer:

Question 9.
Find the matrix A. if 2A + B = \(\left[ \begin{matrix} 2 & 0 \\ 1 & -3 \end{matrix} \right]\) , where u = \(\left[ \begin{matrix} 1 & -1 \\ 3 & 0 \end{matrix} \right]\)
Answer:

Question 10.
If A = \(\left[ \begin{matrix} 3 & -1 \\ 4 & 5 \end{matrix} \right]\) , Find X such that A – 2x = \(\left[ \begin{matrix} 1 & 4 \\ 2 & -3 \end{matrix} \right]\) .
Answer:

Question 11.
If A + B + C = 0 where A = \(\left[ \begin{matrix} 3 & -2 \\ 2 & 0 \\ 1 & 4 \end{matrix} \right]\) , B = \(\left[ \begin{matrix} 2 & -1 \\ -4 & 2 \\ -3 & 3 \end{matrix} \right]\) . Find C
Answer:

Question 12.
If A = B = \(\left[ \begin{matrix} 3 & 1 \\ -3 & 4 \end{matrix} \right]\) verify (A – B)’ = A’ – B’
Answer:

Part – B

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 Two Marks Questions and Answers

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

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

Question 3.
If A = \(\left[ \begin{matrix} 3 & 2 \\ 1 & 4 \end{matrix} \right]\) B = \(\left[ \begin{matrix} 1 & -1 \\ -2 & 3 \end{matrix} \right]\) and C = \(\left[ \begin{matrix} -3 & 4 \\ 2 & -1 \end{matrix} \right]\) . Find 34A – 2B – 4C
Answer:

Question 4.
If A = \(\left[ \begin{matrix} 1 & -3 \\ -4 & -1 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 3 & 4 \\ -5 & 1 \end{matrix} \right]\) and 0 is a null matrix of order 2 × 2. Find the matrix C such that
(i) 2C = A + B
(ii) A + C = 0
(iii) B + 5c = A
(iv) 3A + 5B + 2C = 0
Answer:

Question 5.

Answer:

Question 6.
Find x and y given that \(\left[ \begin{matrix} -9 \\ 2 \end{matrix} \right] \quad -\quad \left[ \begin{matrix} 5 \\ -1 \end{matrix} \right] =\quad \left[ \begin{matrix} x \\ y \end{matrix} \right]\)
Answer:

Question 7.
If A = \(\left[ \begin{matrix} 3 & 1 & 4 \\ 5 & 6 & 3x+1 \end{matrix} \right]\) and B = \(\left[ \begin{matrix} 3 & 5 \\ 1 & 6 \\ 4 & 3 \end{matrix} \right]\) .
Find x given that A = B’
Answer:

Question 8.
Solve for x and y \(x\left[ \begin{matrix} 2 \\ 1 \end{matrix} \right] +y\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] +\left[ \begin{matrix} 4 \\ 6 \end{matrix} \right] =\left[ \begin{matrix} 12 \\ 17 \end{matrix} \right]\) .
Answer:

⇒ 2x + 3y + 4 = 12
x + 5y + 6 = 7
⇒ 2x + 3y = 8 …. (1) solving these equations
x + 5y = 11 ….. (2) × 2

and x = 11 – 5y = 11 – 10 = 1
∴ x = 1, y = 2

Question 9.
Find x and y given that \(\left[ \begin{matrix} x+y & 3 \\ -1 & x-y \end{matrix} \right] =\left[ \begin{matrix} 4 & 3 \\ -1 & 8 \end{matrix} \right]\)
Answer:

Question 10.
If \(\left[ \begin{matrix} 2 & 3 \\ 7 & 5 \end{matrix} \right] +\left[ \begin{matrix} 2 & x-2 \\ y-1 & 5 \end{matrix} \right] =\left[ \begin{matrix} 4 & 1 \\ 7 & 10 \end{matrix} \right]\) Find x and y-1
Answer:

⇒x + 1 = 1 ⇒ x = 0
y + 6 = 7 ⇒ y = 1
∴ x = 0, y = 1

Part – C

2nd PUC Basic Maths Matrices and Determinants Ex 1.1 Three Marks Questions and Answers

Question 1.
Find A and B if
(a) 2A + B = \(\left[ \begin{matrix} 1 & -1 \\ 0 & 1 \end{matrix} \right]\) and A – 3B = \(\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right]\) .
(b) 2A + B = \(\left[ \begin{matrix} 2 & 3 & 1 \\ 1 & 4 & 0 \end{matrix} \right]\) , 3A + B = \(\left[ \begin{matrix} 4 & 6 & 1 \\ 2 & 3 & 5 \end{matrix} \right]\)
(c) 2A – 3B = \(\left[ \begin{matrix} 2 & -4 \\ -12 & 1 \end{matrix} \right]\) , A + 5B = \(\left[ \begin{matrix} 1 & 24 \\ 33 & 7 \end{matrix} \right]\)
(d) 2A + B = \(\left[ \begin{matrix} 3 & -1 \\ -2 & 5 \end{matrix} \right]\) and A – 2B = \(\left[ \begin{matrix} 4 & 2 \\ -1 & 5 \end{matrix} \right]\) .
Answer:

Question 2.
If A = \(\left[ \begin{matrix} 3 & 8 & 1 \\ 2 & -6 & 3 \\ 7 & 4 & -5 \end{matrix} \right]\) , B = \(\left[ \begin{matrix} 4 & 0 & 2 \\ 6 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right]\) . verify 3(A + B) = 3A + 3B.
Answer:

Question 3.
Find x, y, a, b if = \(\left[ \begin{matrix} 3x+4y & 2 & x-2y \\ a+b & 2a-b & -1 \end{matrix} \right] =\left[ \begin{matrix} 2 & 2 & 4 \\ 5 & -5 & -1 \end{matrix} \right]\)
Answer:

Question 4.
Find x if \(\left[ \begin{matrix} { x }^{ 3 } & 1 \\ 3 & 3 \end{matrix} \right] +\left[ \begin{matrix} { -2x }^{ 2 } & 3 \\ 1 & 4 \end{matrix} \right] =\left[ \begin{matrix} -x & 4 \\ 4 & 7 \end{matrix} \right]\)
Answer:

⇒ x³ – 2x² = x; x³ – 2x² + x = 0 ; x(x² – 2x + 1) = 0
x = 0 or (x – 1)² = 0 ⇒ x = 0 or x = 1

Question 5.
If A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -2 & 1 & 0 \end{matrix} \right]\) , B = \(\left[ \begin{matrix} -1 & 0 & 4 \\ 5 & 1 & -3 \end{matrix} \right]\) C = \(\left[ \begin{matrix} 4 & -1 & 1 \\ 6 & 2 & -1 \end{matrix} \right]\) . Find the matrix X such that 2A – 3B – x = C
Answer:

Question 6.
Find a,b,c given that \(4\left[ \begin{matrix} a & 2 & 3 \\ 4 & 5 & -6 \\ 7 & -8 & 9 \end{matrix} \right] -5\left[ \begin{matrix} 3 & 2 & 1 \\ b & 6 & -4 \\ 9 & -8 & 7 \end{matrix} \right] =\left[ \begin{matrix} 13 & -2 & 7 \\ 6 & -5 & -4 \\ -17 & 2c & 1 \end{matrix} \right]\)
Answer: