Question 1. 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 2. 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 3. 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 4: 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 5: 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 6: 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 7: 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 8. 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]
Choose the correct answer in the following questions:
Question 9: 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 10. 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 11. 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.
Deleted Questions
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.
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.
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.
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.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...