Chapter 3 of the Class 12 NCERT Mathematics textbook, titled “Matrices,” covers fundamental concepts related to matrices, including operations such as addition, multiplication, and finding determinants and inverses. The Miscellaneous Exercise in this chapter provides a range of problems that integrate these concepts, helping students apply their understanding of matrices in various contexts and deepen their problem-solving skills.
NCERT Solutions for Class 12 – Mathematics – Chapter 3 Matrices – Miscellaneous Exercise
This section provides detailed solutions for the Miscellaneous Exercise from Chapter 3 of the Class 12 NCERT Mathematics textbook. The exercise includes problems on matrix operations, determinants, and inverses, designed to challenge students and reinforce their comprehension of matrix theory. The solutions offer step-by-step explanations to ensure a thorough understanding of each concept.
Question 1: Let [Tex]A =\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} [/Tex], show that (aI + bA)n = an I + nan – 1 bA, where I is the identity matrix of order 2 and n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
(aI + bA)n = (aI + bA)1 = (aI + bA)
anI + nan – 1 bA = aI + 1a1 – 1 bA = (aI + bA)
It is true for P(1)
Step 2: Now take n=k
(aI + bA)k = akI + kak – 1 bA …………………(1)
Step 3: Let’s check whether, its true for n = k+1
(aI + bA)k+1 = (aI + bA)k (aI + bA)
= (akI + kak – 1 bA) (aI + bA)
= ak+1I×I + kak bAI + ak bAI + kak-1 b2AA
AA = [Tex]\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} = 0 [/Tex]
= ak+1I×I + kak bAI + ak bAI + 0
= ak+1I + (k+1)ak+1-1 bA
= P(k+1)
Hence, P(n) is true.
Question 2: If [Tex]A =\begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix} [/Tex], prove that [Tex]A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1}\\ 3^{n-1} & 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix} ,n\in N[/Tex]
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
[Tex]A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1}\\ 3^{1-1} & 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1} & 3^{1-1} \end{bmatrix}=\begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix}[/Tex]
It is true for P(1)
Step 2: Now take n=k
[Tex]A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1} \end{bmatrix}[/Tex]
Step 3: Let’s check whether, its true for n = k+1
[Tex]A^{k+1}=A^kA=\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1}\\ 3^{k-1} & 3^{k-1} &3^{k-1} \end{bmatrix} \begin{bmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{bmatrix}\\ A^{k+1} = \begin{bmatrix} 3^k & 3^k &3^k\\ 3^k & 3^k &3^k\\ 3^k & 3^k &3^k \end{bmatrix}\\ A^{k+1} = \begin{bmatrix} 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1}\\ 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1}\\ 3^{k+1-1} & 3^{k+1-1} &3^{k+1-1} \end{bmatrix}[/Tex]
= P(k+1)
Hence, P(n) is true.
Question 3: If [Tex]A =\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}[/Tex], prove that [Tex]A^n =\begin{bmatrix} 1+2n & -4n\\ n & 1-2n \end{bmatrix} [/Tex] ,where n is any positive integer.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
[Tex]A^1 = A =\begin{bmatrix} 1+2(1) & -4(1)\\ n & 1-2(1) \end{bmatrix}=\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix} [/Tex]
It is true for P(1)
Step 2: Now take n=k
[Tex]A^k =\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}[/Tex]
Step 3: Let’s check whether, its true for n = k+1
[Tex]A^{k+1} = A^kA =\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}\begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}\\ = \begin{bmatrix} 3(1+2k)+1(-4k) & -4(1+2k)+(-1)(-4k)\\ 3k+1(1-2k) & (-4)(k)+(-1)(1-2k) \end{bmatrix}\\ = \begin{bmatrix} 3+6k-4k & -4-8k+4k\\ 3k+1-2k & -4k-1+2k \end{bmatrix}\\ = \begin{bmatrix} 3+2k & -4-4k\\ k+1 & -2k-1 \end{bmatrix}\\ = \begin{bmatrix} 1+2(k+1) & -4(k+1)\\ k+1 & 1-2(k+1) \end{bmatrix}\\[/Tex]
= P(k+1)
Hence, P(n) is true.
Question 4. If A and B are symmetric matrices, prove that AB – BA is a skew-symmetric matrix.
Solution:
As, it is mentioned that A and B are symmetric matrices,
A’ = A and B’ = B
(AB – BA)’ = (AB)’ – (BA)’ (using, (A-B)’ = A’ – B’)
= B’A’ – A’B’ (using, (AB)’ = B’A’)
= BA – AB
(AB – BA)’ = – (AB – BA)
Hence, AB – BA is a skew symmetric matrix
Question 5. Show that the matrix B′AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric.
Solution:
Let’s take A as symmetric matrix
A’ = A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B’A B
As, here (B′AB)’ = B’A B. It is a symmetric matrix.
Let’s take A as skew matrix
A’ = -A
Then,
(B′AB)’ = {B'(AB)}’
= (AB)’ (B’)’ (using, (AB)’ = B’A’)
= B’A’ (B) (using, (AB)’ = B’A’ and (B’)’ = B)
= B'(-A) B
= – B’A B
As, here (B′AB)’ = -B’A B. It is a skew matrix.
Hence, we can conclude that B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Question 6. Find the values of x, y, z if the matrix [Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix} [/Tex] satisfy the equation A′A = I
Solution:
[Tex]A =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix} [/Tex]
[Tex]A’ =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}^T=\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} [/Tex]
A’A = [Tex]I =\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} [/Tex]
[Tex]\begin{bmatrix} 0 & 2y &z\\ x & y &-z\\ x & -y &z \end{bmatrix}\begin{bmatrix} 0 & x &x\\ 2y & y &-y\\ z & -z &z \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}[/Tex]
[Tex]\begin{bmatrix} 0+x^2+x^2 & 0+xy-xy &0-xz+xz\\ 0+xy-xy & 4y^2+y^2+y^2 &2yz-yz-yz\\ 0-zx+zx & 2yz-yz-yz &z^2+z^2+z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}[/Tex]
[Tex]\begin{bmatrix} 2x^2 & 0 &0\\ 0 & 6y^2 &0\\ 0 & 0 &3z^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 &0\\ 0 & 1 &0\\ 0 & 0 &1 \end{bmatrix}[/Tex]
By evaluating the values, we have
2x2 = 1
x = ± [Tex]\frac{1}{\sqrt{2}}[/Tex]
6y2 = 1
y = ± [Tex]\frac{1}{\sqrt{6}}[/Tex]
3z2 = 1
z = ± [Tex]\frac{1}{\sqrt{3}}[/Tex]
Question 7: For what values of x : [Tex]\begin{bmatrix} 1 & 2 &1 \end{bmatrix} \begin{bmatrix} 1 & 2 &0\\ 2 & 0 &1\\ 1 & 0 &2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]
Solution:
[Tex]\begin{bmatrix} 1 & 2 &1 \end{bmatrix} \begin{bmatrix} 1 & 2 &0\\ 2 & 0 &1\\ 1 & 0 &2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]
[Tex]\begin{bmatrix} 1+4+1 & 2+0+0 &0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]
[Tex]\begin{bmatrix} 6 & 2 &4 \end{bmatrix} \begin{bmatrix} 0\\ 2\\ x \end{bmatrix} = 0[/Tex]
[Tex]\begin{bmatrix} 6(0) + 2(2) +4(x) \end{bmatrix}= 0\\ \begin{bmatrix} 0 + 4 +4x \end{bmatrix}= 0\\ 4(x+1) = 0\\ x+1 = 0\\ x = -1[/Tex]
Question 8: If [Tex]A =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} [/Tex], show that A2 – 5A + 7I = 0.
Solution:
[Tex]A^2 = AA =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\\ = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}\\ = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}[/Tex]
[Tex]5A =5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} =\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} [/Tex]
[Tex]7I =7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}[/Tex]
A2 – 5A + 7I = [Tex]\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}+ \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\\ =\begin{bmatrix} 8-15+7 & 3-3+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}\\ = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}[/Tex]
Hence proved!
Question 9: Find x, if [Tex]\begin{bmatrix} x & -5 &-1 \end{bmatrix} \begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1\\ 2 & 0 &3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0[/Tex]
Solution:
[Tex]\begin{bmatrix} x & -5 &-1 \end{bmatrix} \begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1\\ 2 & 0 &3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x+0-2 & 0-10+0 &2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x-2 & -10 &2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0\\ \begin{bmatrix} x(x-2) + (-10)(4) +1(2x-8) \end{bmatrix}= 0\\ \begin{bmatrix} x^2-2x -40+2x-8 \end{bmatrix}= 0\\ \begin{bmatrix} x^2-48 \end{bmatrix}= 0\\ x^2 = 48\\ x = \pm \sqrt{48}\\ x = \pm 4\sqrt{3}[/Tex]
Question 10: A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:
| Market | Products | ||
| I | 10,000 | 2,000 | 18,000 |
| II | 6,000 | 20,000 | 8,000 |
(a) If unit sale prices of x, y and z are ₹ 2.50, ₹ 1.50 and ₹ 1.00, respectively, find the total revenue in each market with the help of matrix algebra.
Solution:
Total revenue in market I and II can be arranged from given data as follows:
[Tex]\begin{bmatrix} 10,000 & 2,000 &18,000\\6,000 & 20,000 &8,000 \end{bmatrix} \begin{bmatrix} 2.5 \\ 1.5 \\1 \end{bmatrix}[/Tex]
After multiplication, we get
[Tex]\begin{bmatrix} 25,000 + 3,000 +18,000\\15,000 + 30,000 +8,000 \end{bmatrix}=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}[/Tex]
Hence, the total revenue in Market I and market II are ₹ 46,000 and ₹ 53,000 respectively.
(b) If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Solution:
Total cost prices of all the products in market I and market II can be arranged from given data as follows:
[Tex]\begin{bmatrix} 10,000 & 2,000 &18,000\\6,000 & 20,000 &8,000 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\0.5 \end{bmatrix}[/Tex]
After multiplication, we get
[Tex]\begin{bmatrix} 20,000 + 2,000 +9,000\\12,000 + 20,000 +4,000 \end{bmatrix}=\begin{bmatrix} 31,000\\36,000 \end{bmatrix}[/Tex]
As, Profit earned = Total revenue – Cost price
Profit earned [Tex]=\begin{bmatrix} 46,000\\53,000 \end{bmatrix}-\begin{bmatrix} 31,000\\36,000 \end{bmatrix}[/Tex]
Profit earned = [Tex]=\begin{bmatrix} 15,000\\17,000 \end{bmatrix}[/Tex]
Hence, profit earned in Market I and market II are ₹ 15,000 and ₹ 17,000 respectively. Which is equal to ₹ 32,000
Question 11. Find the matrix X so that [Tex]X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}[/Tex]
Solution:
[Tex]X\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}[/Tex]
Here, the RHS is a 2×3 matrix and LHS is 2×3. So, X will be 2×2 matrix.
Let’s take X as,
[Tex]X= \begin{bmatrix} p & q \\ r & s \end{bmatrix}[/Tex]
Now solving the matrix, we have
[Tex]\begin{bmatrix} p & q\\ r & s \end{bmatrix}\begin{bmatrix} 1 & 0 &2\\ 0 & 2 &1 \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}\\ \begin{bmatrix} p+4q & 2p+5qb &3p+6q\\ r+4s & 2r+5s &3r+6s \end{bmatrix}= \begin{bmatrix} -7 & -8 &-9\\ 2 & 4 &6 \end{bmatrix}\\[/Tex]
Equating each of them, we get
p+4q = -7 ………..(1)
2p+5q = -8 ………….(2)
3p + 6q = -9
r + 4s = 2 …………(3)
2r + 5s = 4 ……………(5)
3r + 6s = 6
Solving (1) and (2), we get
p = 1 and q = -2
Solving (3) and (4), we get
r = 2 and s = 0
Hence, matrix X is
[Tex]X= \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}[/Tex]
Question 12: If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N.
Solution:
Using mathematical induction,
Step 1: Let’s check for n=1
ABn = AB1 = AB
BnA = B1A = BA
It is true for P(1)
Step 2: Now take n=k
ABk = BkA
Step 3: Let’s check whether, its true for n = k+1
AB(k+1) = ABkB
= BkAB
= Bk+1 A
= P(k+1)
Hence, P(n) is true.
Now, for (AB)n = AnBn
Using mathematical induction,
Step 1: Let’s check for n=1
(AB)1 = AB
B1A1 = BA
It is true for P(1)
Step 2: Now take n=k
(AB)k = AkBk
Step 3: Let’s check whether, its true for n = k+1
(AB)(k+1) = (AB)k(AB)
= AkBk AB
= Ak+1 Bk+1
= (AB)k+1
= P(k+1)
Hence, P(n) is true.
Choose the correct answer in the following questions:
Question 13: If [Tex]A= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} [/Tex] is such that A² = I, then
(A) 1 + α² + βγ = 0
(B) 1 – α² + βγ = 0
(C) 1 – α² – βγ = 0
(D) 1 + α² – βγ = 0
Solution:
[Tex]A^2 = AA= \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\\ = \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix} [/Tex]
As, A2 = I
[Tex]\begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \beta\gamma+\alpha^2 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}[/Tex]
α² + βγ = 1
1 – α² – βγ = 0
Hence, Option (C) is correct.
Question 14. If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Solution:
If the matrix A is both symmetric and skew symmetric, then
A = A’
and A = -A
Only zero matrix satisfies both the conditions.
Hence, Option (B) is correct.
Question 15. If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
Solution:
(I + A)³ – 7 A = I3 + A3 + 3A^2 + 3AI^2 – 7A
= I3 + A3 + 3A2 + 3A – 7A
= I + A3 + 3A2 – 4A
As, A2 = A
A3 = A2A = AA = A
So, I + A3 + 3A2 – 4A = I + A + 3A – 4A = I
Hence, Option (C) is correct.
Summary
Chapter 3 of the Class 12 NCERT Mathematics textbook, “Matrices,” explores various matrix operations and concepts. The Miscellaneous Exercise provides diverse problems on matrix addition, multiplication, determinants, and inverses. It challenges students to apply these concepts and reinforces their understanding of matrix theory through detailed problem-solving.
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FAQs on Matrices
What are the basic operations you can perform on matrices?
The basic operations on matrices include addition, subtraction, and multiplication. You can also find the determinant and inverse of a matrix, provided the matrix is square and has a non-zero determinant.
Why are matrices important in mathematics and applied fields?
Matrices are crucial in mathematics for solving systems of linear equations, performing transformations, and representing data. In applied fields such as computer graphics, engineering, and economics, matrices are used for modeling and solving complex problems efficiently.
How do you perform matrix multiplication?
Matrix multiplication involves taking the dot product of rows from the first matrix with columns from the second matrix. The result is a new matrix where each element is the sum of the products of corresponding elements from the rows and columns.
