RBSE Maths Class 12 Chapter 3: Matrix Important Questions and Solutions

RBSE Class 12 Maths Chapter 3 – Matrix Important questions and solutions are provided here. All the important questions and solutions of matrices of Class 12 available here contain stepwise solutions. The RBSE Class 12 important questions and solutions available at BYJU’S will guide the students in clearing the exams with flying colours.

Chapter 3 of RBSE Class 12 has two exercises which cover various concepts of matrices along with the required solved examples. This chapter contains questions in different patterns such as finding the order of a given matrix, finding the given matrix when a particular equation is provided to get the elements of that matrix. Also, students will learn simple arithmetic operations on matrices, namely addition, subtraction and multiplication.

RBSE Maths Chapter 3: Exercise 3.1 Textbook Important Questions and Solutions

Question 1: If \(\begin{bmatrix} k+4 & -1\\ 3 & k-6 \end{bmatrix} = \begin{bmatrix} a & -1\\ 3 & -4 \end{bmatrix}\), then find the value of a.

Solution:

Given,

\(\begin{bmatrix} k+4 & -1\\ 3 & k-6 \end{bmatrix} = \begin{bmatrix} a & -1\\ 3 & -4 \end{bmatrix}\)

By equating the corresponding elements,

k – 6 = -4

k = 6 – 4

k = 2

And

k + 4 = a

2 + 4 = a

⇒ a = 6

Question 2: Find a matrix A = [aij] of order 2 x 2 whose elements are defined by aij = 2i – 3j.

Solution:

Given,

aij = 2i – 3j

a11 = 2(1) – 3(1) = 2 – 3 = -1

a12 = 2(1) – 3(2) = 2 – 6 = -4

a21 = 2(2) – 3(1) = 4 – 3 = 1

a22 = 2(2) – 3(2) = 4 – 6 = -2

Therefore, the required matrix \(=\begin{bmatrix} -1 & -4\\ 1 & -2 \end{bmatrix}\)

Question 3: Find the values of a and b, if

RBSE class 12 maths chapter 3 imp que 3

Solution:

Given,

RBSE class 12 maths chapter 3 imp que 3 sol

By comparing the corresponding elements,

a + b = 6

b = 6 – a ….(i)

ab = 8 ….(ii)

From (i) and (ii),

a(6 – a)= 8

6a – a2 – 8=0

⇒ a2 – 6a + 8 = 0

⇒ a2 – 2a – 4a + 8 = 0

⇒ a(a – 2) – 4(a – 2) = 0

⇒ (a – 2)(a – 4) = 0

⇒ a – 2 = 0, a – 4 = 0

⇒ a = 2, a = 4

If a = 2, b = 6 – 2 = 4

If a = 4, b = 6 – 4 = 2

Therefore, the values of a and b are 2, 4 and 4, 2, respectively.

Question 4: For what values of a, b and c, matrices A and B are equal matrices. Where:

RBSE class 12 maths chapter 3 imp que 4

Solution:

Given,

RBSE class 12 maths chapter 3 imp que 4 sol

By equating the corresponding terms,

a – 2 = b

⇒ a – b = 2 …(i)

3 = c

12c = 6b

⇒ b = 12c/6 = 2c = 2 × 3 = 6

From (i),

a – 6 = 2

a = 2 + 6 = 8

Therefore, a = 8, b = 6 and c = 3.

RBSE Maths Chapter 3: Exercise 3.2 Textbook Important Questions and Solutions

Question 5: If \(A=\begin{bmatrix} -3& 2& 1\\ 1 & -4 & 7 \end{bmatrix}\) and \(B=\begin{bmatrix} 3& 5& -2\\ -1 & 4 & -2 \end{bmatrix}\), then find A + B and A – B.

Solution:

Given,

\(A=\begin{bmatrix} -3& 2& 1\\ 1 & -4 & 7 \end{bmatrix}\) and \(B=\begin{bmatrix} 3& 5& -2\\ -1 & 4 & -2 \end{bmatrix}\)

RBSE class 12 maths chapter 3 imp que 5 sol

Question 6: If \(A = \begin{bmatrix} 1 & 3\\ 2 & 1\\ 3 & -1 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 1\\ 1 & 2\\ -1 & 0 \end{bmatrix}\), then find matrix C where A + 2B + C = O and O is a zero matrix.

Solution:

Given,

\(A = \begin{bmatrix} 1 & 3\\ 2 & 1\\ 3 & -1 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & 1\\ 1 & 2\\ -1 & 0 \end{bmatrix}\)

Now,

A + 2B + C = O

RBSE class 12 maths chapter 3 imp que 6 sol

Question 7: If \(A=\begin{bmatrix} 0 & 1 & 2 & 3\\ 3 & 2 & 1 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 0 & 3\\ 1& 2\\ 2 & 1\\ 3& 0 \end{bmatrix}\), then show that AB ≠ BA.

Solution:

Given,

RBSE class 12 maths chapter 3 imp que 7 sol

Therefore, AB ≠ BA

Question 8: Prove that:

RBSE class 12 maths chapter 3 imp que 8

Solution:

LHS = \(\begin{bmatrix} x & y & z \end{bmatrix}\begin{bmatrix} a & h & g\\ h & b & f\\ g & f & c \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}\) \(=\begin{bmatrix} ax+hy+gz & hx+by+fz & gx+fy+cz \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}\)

= (ax + hy + gz)x + (hx + by + fz)v + (gr + fy + cz)z

= ax2 + hxy + gzx + hxy + by2 + fyz + gzx + fyz + cz2

= ax2 + by2 + cz2 + 2hxy + 2fyz + 2gzx

= RHS

Hence Proved.

Question 9: If \(A=\begin{bmatrix} 1 & 0\\ -1 &7 \end{bmatrix}\) and \(I=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}\), then find the value of K where A2 = 8A + KI.

Solution:

Given,

\(A=\begin{bmatrix} 1 & 0\\ -1 &7 \end{bmatrix}\) and \(I=\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}\)

Consider, A2 = 8A + KI

RBSE class 12 maths chapter 3 imp que 9 sol

Now, by equating the corresponding elements,

1 = 8 + K

⇒ K = 1 – 8 = -7

Therefore, the value of K = -7.

Question 10: If \(A=\begin{pmatrix} cos\alpha &sin\alpha \\ -sin\alpha &cos\alpha \end{pmatrix}\), then prove that \(A^n=\begin{pmatrix} cos\ n\alpha &sin\ n\alpha \\ -sin\ n\alpha &cos\ n\alpha \end{pmatrix}\), where n is a positive integer.

Solution:

Given,

\(A=\begin{pmatrix} cos\alpha &sin\alpha \\ -sin\alpha &cos\alpha \end{pmatrix}\)

A2 = A.A

RBSE class 12 maths chapter 3 imp que 10.1 sol

A3 = A2.A

RBSE class 12 maths chapter 3 imp que 10.2 sol

Similarly,

\(A^n=\begin{pmatrix} cos\ n\alpha &sin\ n\alpha \\ -sin\ n\alpha &cos\ n\alpha \end{pmatrix}\)

Hence proved.

RBSE Maths Chapter 3: Additional Important Questions and Solutions

Question 1: If \(A=\begin{bmatrix} -i & 0\\ 0& i \end{bmatrix}\) and \(B=\begin{bmatrix} 0& i\\ i& 0 \end{bmatrix}\) where i =√(-1), then find BA.

Solution:

Given,

\(A=\begin{bmatrix} -i & 0\\ 0& i \end{bmatrix}\) and \(B=\begin{bmatrix} 0& i\\ i& 0 \end{bmatrix}\)

RBSE class 12 maths chapter 3 important Q1 sol

Question 2: The order of a matrix A is 3 x 4 and B is a matrix such that ATB and ABT both are defined, then find the order of B.

Solution:

Given,

Order of the matrix A = 3 x 4

∴ Order of the matrix AT = 4 x 3

Also, given that: ATB and ABT are defined.

Hence, the order of B is 3 x 4.

Question 3: If \(A=\begin{bmatrix} -2 & -1 & 1\\ -1 & 7 & 4\\ 1 & -x & -3 \end{bmatrix}\) is a symmetric matrix, then find the value of x.

Solution:

Given,

\(A=\begin{bmatrix} -2 & -1 & 1\\ -1 & 7 & 4\\ 1 & -x & -3 \end{bmatrix}\)

A is a symmetric matrix.

Thus, aij = aji

Now,

a32 = a23

⇒ -x = 4

⇒ x = -4

Question 4: If \(A=\begin{bmatrix} 2 & 3 & -4\\ -1 & 2 & 3 \end{bmatrix}\) and \(B=\begin{bmatrix} -1 & 2\\ 3 & 4\\ -5 & -6 \end{bmatrix}\), then find A + BT.

Solution:

Given,

\(A=\begin{bmatrix} 2 & 3 & -4\\ -1 & 2 & 3 \end{bmatrix}\) and \(B=\begin{bmatrix} -1 & 2\\ 3 & 4\\ -5 & -6 \end{bmatrix}\)

RBSE class 12 maths chapter 3 important Q4 sol

Question 5: If \(A=\begin{bmatrix} 2 & 3\\ -1 & 4 \end{bmatrix}\) and \(B=\begin{bmatrix} 1 & -1\\ 2 & 5 \end{bmatrix}\), then prove that: (AB)T = BTAT

Solution:

Given,

\(A=\begin{bmatrix} 2 & 3\\ -1 & 4 \end{bmatrix}\) and \(B=\begin{bmatrix} 1 & -1\\ 2 & 5 \end{bmatrix}\)

RBSE class 12 maths chapter 3 important Q5.1 sol

Now multiply the transpose of B and A,

RBSE class 12 maths chapter 3 important Q5.2 sol

Therefore, (AB)T = BTAT

Hence proved.

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