ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS
More Exercises
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices Ex 8.1
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices Ex 8.2
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices Ex 8.3
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS
- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices Chapter Test
Choose the correct answer from the given four options (1 to 14) :
Question 1.
 If A = [aij]2×2 where aij = i + j, then A is equal to
 (a) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \)
 (b) \(\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} \)
 (d) \(\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \)
 Solution:
 A = [aij]2×2 where aij = i + j, then A is equal to
 \(\begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} \) (b)
Question 2.
 If \(\begin{bmatrix} x+3 & 4 \\ y-4 & x+y \end{bmatrix}=\begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \) then the values of x and y are
 (a) x = 2, y = 7
 (b) x = 7, y = 2
 (c) x = 3, y = 6
 (d) x = – 2, y = 7
 Solution:
 \(\begin{bmatrix} x+3 & 4 \\ y-4 & x+y \end{bmatrix}=\begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix} \)
 Comparing we get
 x + 3 = 5
 ⇒ x = 5 – 3 = 2
 and y – 4 = 3
 ⇒ y = 3 + 4 = 7
 x = 2, y = 7 (a)
Question 3.
 If \(\begin{bmatrix} x+2y & -y \\ 3x & 7 \end{bmatrix}=\begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \) then the values of x and y are
 (a) x = 2, y = 3
 (b) x = 2, y = – 3
 (c) x = – 2, y = 3
 (d) x = 3, y = 2
 Solution:
 \(\begin{bmatrix} x+2y & -y \\ 3x & 7 \end{bmatrix}=\begin{bmatrix} -4 & 3 \\ 6 & 4 \end{bmatrix} \)
 Comparing, we get
 3x = 6
 ⇒ \(x= \frac { 6 }{ 3 } \) = 2
 ⇒ -y = 3
 ⇒ y = – 3
 x = 2, y = -3 (b)
Question 4.
 If \(\begin{bmatrix} x-2y & 5 \\ 3 & y \end{bmatrix}=\begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \) then the value of x is
 (a) – 2
 (b) 0
 (c) 1
 (d) 2
 Solution:
 \(\begin{bmatrix} x-2y & 5 \\ 3 & y \end{bmatrix}=\begin{bmatrix} 6 & 5 \\ 3 & -2 \end{bmatrix} \)
 Comparing, we get
 y = -2
 and x – 2y = 6
 ⇒ x – 2 x (-2) = 6
 ⇒ x + 4 = 6
 ⇒ x = 6 – 4 = 2 (d)
Question 5.
 If \(\begin{bmatrix} x+2y & 3y \\ 4x & 2 \end{bmatrix}=\begin{bmatrix} 0 & -3 \\ 8 & 2 \end{bmatrix} \) then the value of x – y is
 (a) – 3
 (b) 1
 (c) 3
 (d) 5
 Solution:
 \(\begin{bmatrix} x+2y & 3y \\ 4x & 2 \end{bmatrix}=\begin{bmatrix} 0 & -3 \\ 8 & 2 \end{bmatrix} \)
 Comparing, we get
 3y = -3
 ⇒ \(y= \frac { -3 }{ 3 } \) = -1
 4x = 8
 ⇒ \(x= \frac { 8 }{ 4 } \) = 2
 x – y = 2 – (-1) = 2 + 1 = 3 (c)
Question 6.
 If \(x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 6 \end{matrix} \right] \) then the values of x and y are
 (a) x = 2, y = 6
 (b) x = 2, y = – 6
 (c) x = 3, y = – 4
 (d) x = 3, y = – 6
 Solution:
 Given
 \(x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 6 \end{matrix} \right] \)
 
Question 7.
 If B = \(\begin{bmatrix} -1 & 5 \\ 0 & 3 \end{bmatrix} \) and A – 2B = \(\begin{bmatrix} 0 & 4 \\ -7 & 5 \end{bmatrix} \)
 then the matrix A is equal to
 (a) \(\begin{bmatrix} 2 & 14 \\ -7 & 11 \end{bmatrix} \)
 (b) \(\begin{bmatrix} -2 & 14 \\ 7 & 11 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 2 & -14 \\ 7 & 11 \end{bmatrix} \)
 (d) \(\begin{bmatrix} -2 & 14 \\ -7 & 11 \end{bmatrix} \)
 Solution:
 Given
 B = \(\begin{bmatrix} -1 & 5 \\ 0 & 3 \end{bmatrix} \) and
 A – 2B = \(\begin{bmatrix} 0 & 4 \\ -7 & 5 \end{bmatrix} \)
 
Question 8.
 If A + B = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) and A – 2B = \(\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} \)
 then A is equal to
 (a) \(\frac { 1 }{ 3 } \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \)
 (b) \(\frac { 1 }{ 3 } \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \)
 (d) \(\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \)
 Solution:
 A + B = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) and
 A – 2B = \(\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix} \)
 
Question 9.
 A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) then A² =
 (a) \(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
 (b) \(\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
 (d) \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
 Solution:
 Given
 A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
 
Question 10.
 If A = \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \) , then A² =
 (a) \(\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \)
 (b) \(\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
 (d) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
 Solution:
 Given
 A = \(\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \)
 
Question 11.
 If A = \(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \) , then A² =
 (a) A
 (b) O
 (c) I
 (d) 2A
 Solution:
 Given
 A = \(\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \)
 
Question 12.
 If A = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \) , then A² =
 (a) \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
 (b) \(\begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} \)
 (d) none of these
 Solution:
 Given
 A = \(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \)
 
Question 13.
 If A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) , then A² =
 (a) \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \)
 (b) \(\begin{bmatrix} 8 & -5 \\ 5 & 3 \end{bmatrix} \)
 (c) \(\begin{bmatrix} 8 & -5 \\ -5 & -3 \end{bmatrix} \)
 (d) \(\begin{bmatrix} 8 & -5 \\ -5 & 3 \end{bmatrix} \)
 Solution:
 A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
 A² = A x A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \)
 
Question 14.
 If A = \(\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \) , then A² = pA, then the value of p is
 (a) 2
 (b) 4
 (c) – 2
 (d) – 4
 Solution:
 A = \(\begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} \)
 and A² = pA
 
Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 8 Matrices MCQS are helpful to complete your math homework.
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