Miscellaneous Exercise on Chapter 3 Solutions

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Miscellaneous Exercise on Chapter 3 Solutions

1. Let

{\text{A} = \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right],}

show that

{\left(a\text{I} + b\text{A}\right)^n = a^n\text{I} + na^{n-1}b\text{A},}

where I is the identity matrix of order 2 and

{n ∈ \textbf{N}}

We shall prove the result by using principle of mathematical induction

We have

{\textbf{P}(n) :}

{\text{A} = \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right],}

then

{\left(a\text{I} + b\text{A}\right)^n = a^n\text{I} + na^{n-1}b\text{A},}

{n ∈ \textbf{N}}

{\textbf{P}(1) : \text{A} = \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right],}

{\left(a\text{I} + b\text{A}\right)^1 = a\text{I} + b\text{A},}

{n ∈ \textbf{N}}

∴ The result is true for

{n = 1}

Let the result be true for

{n = k.}

{\textbf{P}(k) : \text{A} = \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right],}

then

{\left(a\text{I} + b\text{A}\right)^k = a^k\text{I} + na^{k-1}b\text{A},}

{n ∈ \textbf{N}}

Now, we prove that the result holds for

{n = k + 1}

Now,

{\left(a\text{I} + b\text{A}\right)^{k + 1}}

{= (a\text{I} + b\text{A}) × \left(a\text{I} + b\text{A}\right)^k}

{= (a\text{I} + b\text{A}) × \left(a^k\text{I} + ka^{k-1}b\text{A}\right)}

{= \left(a\text{I} × a^k\text{I}\right) + \left(a\text{I} × ka^{k-1}b\text{A}\right) + \left(b\text{A} × a^k\text{I}\right) + \left(b\text{A} × ka^{k-1}b\text{A}\right)}

{= \left(a^{k + 1}\text{I}\right) + \left(ka^kb\text{A}\right) + \left(a^kb\text{A}\right) + \left(ka^{k-1}b^2\text{A}^2\right)}

———–❶

Now,

{\text{A}^2}

{= \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right] \left[\begin{array}{cc} 0 & 1 \\[5pt] 0 & 0 \end{array}\right]}

{= \left[\begin{array}{cc} (0 × 0) + (1 × 0) & (0 × 1) + (1 × 0) \\[5pt] (0 × 0) + (0 × 0) & (0 × 1) + (0 × 0) \end{array}\right]}

{= \left[\begin{array}{cc} 0 & 0 \\[5pt] 0 & 0 \end{array}\right]}

= O

Substituting the value of

\text{A}^2

in ❶, we have

{\left(a\text{I} + b\text{A}\right)^{k + 1}}

{= \left(a^{k + 1}\text{I}\right) + \left((k + 1)a^kb\text{A}\right) + \text{O}}

{= a^{k + 1}\text{I} + (k + 1)a^kb\text{A}}

∴ The result is true for

{n = k + 1.}

Thus by principle of mathematical induction, we have

{\left(a\text{I} + b\text{A}\right)^n = a^n\text{I} + na^{n-1}b\text{A},}

holds for all natural numbers.

2. If

{\text{A} = \left[\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \end{array}\right],}

then prove that

{\text{A}^n = \left[\begin{array}{ccc} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{array}\right],}

{n ∈ \textbf{N}.}

We shall prove the result by using principle of mathematical induction.

We have

{\text{P}(n):}

{\text{A} = \left[\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \end{array}\right],}

then

{\text{A}^n = \left[\begin{array}{ccc} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{array}\right],}

{n ∈ \textbf{N}.}

{\text{P}(1): \text{A} = \left[\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \end{array}\right],}

{\text{A}^1 = \left[\begin{array}{ccc} 9 & 9 & 9 \\[5pt] 9 & 9 & 9 \\[5pt] 9 & 9 & 9 \end{array}\right]}

∴ The result is true for

{n = 1.}

Let the result be true for

n = k.

{\text{P}(k): \text{A} = \left[\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \end{array}\right],}

then

{\text{A}^k = \left[\begin{array}{ccc} 3^{k - 1} & 3^{k - 1} & 3^{k - 1} \\[5pt] 3^{k - 1} & 3^{k - 1} & 3^{k - 1} \\[5pt] 3^{k - 1} & 3^{k - 1} & 3^{k - 1} \end{array}\right]}

Now, we have to prove that the result holds for

{n = k + 1}

{\text{A}^{k + 1}}

{= \text{A}.\text{A}^k}

{= \left[\begin{array}{ccc} 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \\[5pt] 1 & 1 & 1 \end{array}\right] \left[\begin{array}{ccc} 3^{k - 1} & 3^{k - 1} & 3^{k - 1} \\[5pt] 3^{k - 1} & 3^{k - 1} & 3^{k - 1} \\[5pt] 3^{k + 1} & 3^{k + 1} & 3^{k + 1} \end{array}\right]}

{= \left[\begin{array}{ccc} 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} \\[5pt] 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} \\[5pt] 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{array}\right]}

{= \left[\begin{array}{ccc} \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) \\[5pt] \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) \\[5pt] \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) & \left(3 × 3^{k - 1}\right) \end{array}\right]}

{= \left[\begin{array}{ccc} 3^k & 3^k & 3^k \\[5pt] 3^k & 3^k & 3^k \\[5pt] 3^k & 3^k & 3^k \end{array}\right]}

{= \left[\begin{array}{ccc} 3^{(k + 1) -1} & 3^{(k + 1) -1} & 3^{(k + 1) -1} \\[5pt] 3^{(k + 1) -1} & 3^{(k + 1) -1} & 3^{(k + 1) -1} \\[5pt] 3^{(k + 1) -1} & 3^{(k + 1) -1} & 3^{(k + 1) -1} \end{array}\right]}

∴ The result is true for

{n = k + 1.}

Thus by principle of mathematical induction, we have

{\text{A}^n = \left[\begin{array}{ccc} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\[5pt] 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{array}\right],}

holds for all natural numbers.

3. If

{\text{A} = \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right],}

then prove that

{\text{A}^n = \left[\begin{array}{cc} 1 + 2n & -4n \\[5pt] n & 1 - 2n \end{array}\right],}

where

n

is any positive integer.

We shall prove the result by using principle of mathematical induction.

We have

{\text{P}(n):}

{\text{A} = \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right],}

then

{\text{A}^n = \left[\begin{array}{cc} 1 + 2n & -4n \\[5pt] n & 1 - 2n \end{array}\right],}

where

{n ∈ \text{Z}^+}

{\text{P}(1): \text{A} = \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right],}

then

{\text{A}^1 = \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right],}

where

{n ∈ \text{Z}^+}

∴ The result is true for

{n = 1.}

Let the result be true for

{n = k.}

{\text{P}(k): \text{A} = \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right],}

then

{\text{A}^n = \left[\begin{array}{cc} 1 + 2k & -4k \\[5pt] k & 1 - 2k \end{array}\right],}

Now, we prove that the result holds for

{n = k + 1}

Now

{\text{A}^{k + 1}}

{= \text{A}.\text{A}^k}

{= \left[\begin{array}{cc} 3 & -4 \\[5pt] 1 & -1 \end{array}\right] \left[\begin{array}{cc} 1 + 2k & -4k \\[5pt] k & 1 - 2k \end{array}\right]}

{= \left[\begin{array}{cc} 3 × (1 + 2k) + (-4 × k) & (3 × -4k) + (-4) × (1 - 2k) \\[5pt] 1 × (1 + 2k) + (-1) × k & 1 × (-4k) + (-1) × (1 - 2k) \end{array}\right]}

{= \left[\begin{array}{cc} 3 + 6k - 4k & -12k - 4 + 8k \\[5pt] 1 + 2k - k & -4k - 1 + 2k \end{array}\right]}

{= \left[\begin{array}{cc} 1 + + 2 + 2k & -4k - 4 \\[5pt] 1 + k & -2k -2 + 1 \end{array}\right]}

{= \left[\begin{array}{cc} 1 + 2(k + 1) & -4(k + ) \\[5pt] k + 1 & 1 - 2(k + 1) \end{array}\right]}

∴ The result is true for

{n = k + 1.}

Thus by principle of mathematical induction, we have

{\text{A}^n = \left[\begin{array}{cc} 1 + 2n & -4n \\[5pt] n & 1 - 2n \end{array}\right],}

holds for all positive integers.

4. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Given that the matrices A and B are symmetric.

⇒ A′ = A and B′ = B

Now,

(AB – BA)′

= (AB)′ – (BA)′

(∵ (A – B)′ = A′ – B′)

= B′A′ – ′B′

(∵ (AB)′ = B′A′)

= BA – AB

(∵ A′ = A and B′ = B)

= -(AB – BA)

So, as (AB – BA)′ = -(AB – BA), it is proved that (AB – BA) is a skew-symmetric matrix.

5. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

(i) To prove that B′AB is symmetric when A is symmetric.

When A is symmetric, it implies that A′ = A

So,

(B′AB)′

= [B′(AB)]′

= (AB)′(B′)′

(∵ (AB)′ = B′A′)

= B′A′B

(∵ (AB)′ = B′A′ and (A′)′ = A)

= B′AB

(∵ A′ = A)

∴ B′AB is symmetric.

(ii) To prove that B′AB is skew-symmetric when A is skew-symmetric.

When A is skew-symmetric, it implies that A′ = -A

So,

(B′AB)′

= [B′(AB)]′

= (AB)′(B′)′

(∵ (AB)′ = B′A′)

= B′A′B

(∵ (AB)′ = B′A′ and (A′)′ = A)

= B′(-A)B

(∵ A′ = -A)

= -B′AB

∴ B′AB is skew-symmetric.

Thus, it is proved that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

6. Find the values of

{x, y, z}

if the matrix

{\text{A} = \left[\begin{array}{cc} 0 & 2y & z \\[5pt] x & y & -z \\[5pt] x & -y & z \end{array}\right]}

satisfy the equation A′A = I.

Given that

{= \left[\begin{array}{cc} 0 & 2y & z \\[5pt] x & y & -z \\[5pt] x & -y & z \end{array}\right]}

⇒ A′

{= \left[\begin{array}{cc} 0 & x & x \\[5pt] 2y & y & -y \\[5pt] z & -z & z \end{array}\right]}

It is also given that the matrix A satisfies the equation

A′A = I.

{⇒ \left[\begin{array}{ccc} 0 & x & x \\[5pt] 2y & y & -y \\[5pt] z & -z & z \end{array}\right] \left[\begin{array}{ccc} 0 & 2y & z \\[5pt] x & y & -z \\[5pt] x & -y & z \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} (0 × 0) + (x × x) + (x × x) & (0 × 2y) + (x × y) + (x × -y) & (0 × z) + (x × -z) + (x × z) \\[5pt] (2y × 0) + (y × x) + (-y × x) & (2y × 2y) + (y × y) + (-y × -y) & (2y × z) + (y × -z) + (-y × z) \\[5pt] (z × 0) + (-z × x) + (z × x) & (z × 2y) + (-z × y) + (z × -y) & (z × z) + (-z × -z) + (z × z) \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} 0 + x^2 + x^2 & 0 + xy - xy & 0 - xz + xz \\[5pt] 0 + xy - xy & 4y^2 + y^2 + y^2 & 2yz - yz - yz \\[5pt] 0 - xz + xz & 2yz - yz - yz & z^2 + z^2 + z^2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} 2x^2 & 0 & 0 \\[5pt] 0 & 6y^2 & 0 \\[5pt] 0 & 0 & 3z^2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

{2x^2}

= 1

{⇒ x^2}

{= \dfrac12}

{\Rightarrow x}

{= \sqrt{\dfrac12}}

{\Rightarrow x}

{= \pm\dfrac{1}{\sqrt{2}}}

Similarly,

{6y^2}

= 1

{⇒ y^2}

{= \dfrac16}

{\Rightarrow y}

{= \sqrt{\dfrac16}}

{\Rightarrow y}

{= \pm\dfrac{1}{\sqrt{6}}}

Similarly,

{3z^2}

= 1

{⇒ z^2}

{= \dfrac13}

{\Rightarrow z}

{= \sqrt{\dfrac13}}

{\Rightarrow z}

{= \pm\dfrac{1}{\sqrt{3}}}

∴

{x = \dfrac{1}{\sqrt{2}},}

{y = \dfrac{1}{\sqrt{6}},}

and

{z = \dfrac{1}{\sqrt{3}}}

Note: If this needs to be solved for an MCQ or any other short problem, we can multiply only the

{1^{\text{st}}}

row and

{1^{\text{st}}}

column,

{2^{\text{nd}}}

row and

{2^{\text{nd}}}

column and

{3^{\text{rd}}}

row and

{3^{\text{rd}}}

columns and skip the rest as given below:

{⇒ \left[\begin{array}{ccc} 0 & x & x \\[5pt] 2y & y & -y \\[5pt] z & -z & z \end{array}\right] \left[\begin{array}{ccc} 0 & 2y & z \\[5pt] x & y & -z \\[5pt] x & -y & z \end{array}\right] = \left[\begin{array}{cc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} (0 × 0) + (x × x) + (x × x) & \dots & \dots \\[5pt] \dots & (2y × 2y) + (y × y) + (-y × -y) & \dots \\[5pt] \dots & \dots & (z × z) + (-z × -z) + (z × z) \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} 0 + x^2 + x^2 & \dots & \dots \\[5pt] \dots & 4y^2 + y^2 + y^2 & \dots \\[5pt] \dots & \dots & z^2 + z^2 + z^2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} 2x^2 & \dots & \dots \\[5pt] \dots & 6y^2 & \dots \\[5pt] \dots & \dots & 3z^2 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}

AND THE REST OF THE PROBLEM WILL BE SAME AS ABOVE.

7. For what values of

{x :}

{\left[\begin{array}{ccc} 1 & 2 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 0 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 0 & 2 \end{array}\right] \left[\begin{array}{c} 0 \\[5pt] 2 \\[5pt] x \end{array}\right] = \text{O}?}

Given that

{\left[\begin{array}{ccc} 1 & 2 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 0 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 0 & 2 \end{array}\right] \left[\begin{array}{c} 0 \\[5pt] 2 \\[5pt] x \end{array}\right] = \text{O}}

As the matrix multiplication is associative, let’s multiply the second and third matrices. So, from the given equation, it implies that

{⇒ \left[\begin{array}{ccc} 1 & 2 & 1 \end{array}\right] \left[\begin{array}{c} (1 × 0) + (2 × 2) + (0 × x) \\[5pt] (2 × 0) + (0 × 2) + (1 × x) \\[5pt] (1 × 0) + (0 × 2) + (2 × x) \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{ccc} 1 & 2 & 1 \end{array}\right] \left[\begin{array}{c} 4 \\[5pt] x \\[5pt] 2x \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{c} (1 × 4) + (2 × x) + (1 × 2x) \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{c} 4x + 4 \end{array}\right] = \left[\begin{array}{c} 0 \end{array}\right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

{4x + 4 = 0}

{\Rightarrow 4x = -4}

{⇒ x = -\dfrac44}

{\Rightarrow x = -1}

∴ The answer is

{x = -1}

8. If

{\text{A} = \left[\begin{array}{cc} 3 & 1 \\[5pt] -1 & 2 \end{array}\right],}

show that

{\text{A}^2 - 5\text{A} + 7I = O.}

Given that

{= \left[\begin{array}{cc} 3 & 1 \\[5pt] -1 & 2 \end{array}\right]}

{⇒ \text{A}^2}

{= \left[\begin{array}{cc} 3 & 1 \\[5pt] -1 & 2 \end{array}\right] \left[\begin{array}{cc} 3 & 1 \\[5pt] -1 & 2 \end{array}\right]}

{= \left[\begin{array}{cc} (3 × 3) + (1 × -1) & (3 × 1) + (1 × 2) \\[5pt] (-1 × 3) + (2 × -1) & (-1 × 1) + (2 × 2) \end{array}\right]}

{= \left[\begin{array}{cc} 9 - 1 & 3 + 2 \\[5pt] -3 - 2 & -1 + 4 \end{array}\right]}

{= \left[\begin{array}{cc} 8 & 5 \\[5pt] -5 & 3 \end{array}\right]}

Now,

{\text{A}^2 - 5\text{A} + 7\text{I}}

{= \left[\begin{array}{cc} 8 & 5 \\[5pt] -5 & 3 \end{array}\right] - 5\left[\begin{array}{cc} 3 & 1 \\[5pt] -1 & 2 \end{array}\right] + 7 \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}

{= \left[\begin{array}{cc} 8 - (5 × 3) + (7 × 1) & 5 - (5 × 1) + (7 × 0) \\[5pt] -5 - (5 × -1) + (7 × 0) & 3 - (5 × 2) + (7 × 1) \end{array}\right]}

{= \left[\begin{array}{cc} 8 - 15 + 7 & 5 - 5 + 0 \\[5pt] -5 + 5 + 0 & 3 - 10 + 7 \end{array}\right]}

{= \left[\begin{array}{cc} 0 & 0 \\[5pt] 0 & 0 \end{array}\right]}

= O

∴ It is showed that

{\text{A}^2 - 5\text{A} + 7\text{I} = O.}

9. Find

{x,}

{\left[\begin{array}{ccc} x & -5 & -1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3\end{array}\right] \left[\begin{array}{c} x \\[5pt] 4 \\[5pt] 1 \end{array}\right] = \text{O}}

Given that

{\left[\begin{array}{ccc} x & -5 & -1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3\end{array}\right] \left[\begin{array}{c} x \\[5pt] 4 \\[5pt] 1 \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{ccc} x & -5 & -1 \end{array}\right] \left[\begin{array}{c} (1 × x) + (0 × 4) + (2 × 1) \\[5pt] (0 × x) + (2 × 4) + (1 × 1) \\[5pt] (2 × x) + (0 × 4) + (3 × 1) \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{ccc} x & -5 & -1 \end{array}\right] \left[\begin{array}{c} x + 2 \\[5pt] 9 & 2x + 3 \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{c} x × (x + 2) + (-5 × 9) + (-1) × (2x + 3) \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{c} x^2 + 2x - 45 -2x - 3 \end{array}\right] = \text{O}}

{⇒ \left[\begin{array}{c} x^2 - 48 \end{array}\right] = \left[\begin{array}{c} 0 \end{array}\right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

{x^2 - 48}

= 0

{⇒ x^2}

= 48

{\Rightarrow x}

{= \sqrt{48}}

{= \pm4\sqrt{3}}

∴ The solution is

{x = \pm4\sqrt{3}}

10. A manufacturer produces three products

x, y, z

which he sells in two markets. Annual sales are indicated below:

Market

Products

10,000

2,000

18,000

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.

(b)

If the unit costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.

The annual sales of the products

{x,}

y

and

z

in the market I and market II can be represented in the matrix form as

{\text{V} = \left[\begin{array}{ccc} 10,000 & 2,000 & 18,000 \\[5pt] 6,000 & 20,000 & 8,000 \end{array}\right]}

(a) To find the total revenue in each market with the help of matrix algebra.

As the unit sales price of the products

{x,}

{y,}

and

{z,}

is given as ₹ 2.50, ₹ 1.50 and ₹ 1.00, it can be represented in the matrix form as

{\text{P} = \left[\begin{array}{c} 2.50 \\[5pt] 1.50 \\[5pt] 1.00 \end{array}\right]}

The annual sales revenue in market I and market II can be obtained by multiplying sales volume matrix V with the unit sales price matrix P.

Let the annual sales revenue matrix be R. Thus,

= VP

{= \left[\begin{array}{cc} 10,000 & 2,000 & 18,000 \\[5pt] 6,000 & 20,000 & 8,000 \end{array}\right] \left[\begin{array}{c} 2.50 & 1.50 & 1.00 \end{array}\right]}

{= \left[\begin{array}{c} (10,000 × 2.50) + (2,000 × 1.50) + (18,000 × 1.00) \\[5pt] (6,000 × 2.50) + (20,000 × 1.50) + (8,000 × 1.00) \end{array}\right]}

{= \left[\begin{array}{c} 25,000 + 3,000 + 18,000 \\[5pt] 15,000 + 30,000 + 8,000 \end{array}\right]}

{= \left[\begin{array}{c} 46,000 \\[5pt] 53,000 \end{array}\right]}

Thus, we have,

Total revenue in the market – I

= ₹ 46,000

Total revenue in the market – II

= ₹ 53,000

(b) To find the gross profit with the help of matrix algebra.

As the unit sales price of the products

{x,}

{y,}

and

{z,}

is given as ₹ 2.00, ₹ 1.00 and ₹ 0.50, it can be represented in the matrix form as

{\text{U} = \left[\begin{array}{c} 2.00 \\[5pt] 1.00 \\[5pt] 0.50 \end{array}\right]}

As the unit cost of the products

{x,}

{y,}

and

{z,}

is given as ₹ 2.00, ₹ 1.00 and 50 paise, it can be represented in the matrix form as

{\left[\begin{array}{c} 2.00 \\[5pt] 1.00 \\[5pt] 0.50 \end{array}\right]}

The annual cost of the products sold in market I and market II can be obtained by multiplying sales volume matrix V with the unit cost matrix U.

Let the annual cost matrix be C. Thus,

= VP

{= \left[\begin{array}{cc} 10,000 & 2,000 & 18,000 \\[5pt] 6,000 & 20,000 & 8,000 \end{array}\right] \left[\begin{array}{c} 2.00 & 1.00 & 0.50 \end{array}\right]}

{= \left[\begin{array}{c} (10,000 × 2.00) + (2,000 × 1.00) + (18,000 × 0.50) \\[5pt] (6,000 × 2.00) + (20,000 × 1.00) + (8,000 × 0.50) \end{array}\right]}

{= \left[\begin{array}{c} 20,000 + 2,000 + 9,000 \\[5pt] 12,000 + 20,000 + 4,000 \end{array}\right]}

{= \left[\begin{array}{c} 31,000 \\[5pt] 36,000 \end{array}\right]}

Now, the gross profit can be obtained by deducting the cost matric C from the revenue matrix R. Let the gross profit matrix be G. So, we have

= R – C

{= \left[\begin{array}{c} 46,000 \\[5pt] 53,000 \end{array}\right] \left[\begin{array}{c} 31,000 \\[5pt] 36,000 \end{array}\right]}

{= \left[\begin{array}{c} 46,000 - 31,000 \\[5pt] 53,000 - 36,000 \end{array}\right]}

{= \left[\begin{array}{c} 15,000 \\[5pt] 17,000 \end{array}\right]}

Thus, we have the gross profit through sales as

Gross profit in the market – I

= ₹ 15,000

Gross profit in the market – II

= ₹ 17,000

11. Find the matrix X so that

{\text{X} \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 4 & 5 & 6 \end{array}\right] = \left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

The condition that must be met for the multiplication of the matrices X and

{\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 4 & 5 & 6 \end{array}\right]}

to be possible is

No. of columns of X = No. of rows of

{\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 4 & 5 & 6 \end{array}\right]}

⇒ No. of columns of X = 2

Also, the resulting matrix

{\left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

will have the same number of rows as X. So,

No. of rows of X = No. of rows of

{\left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

⇒ No. of rows of X = 2

So, the order of the matrix X is 2 × 2

Now, let

{\text{X} = \left[\begin{array}{cc} x & y \\[5pt] z & w \end{array}\right]}

The given equation is

{\text{X} \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 4 & 5 & 6 \end{array}\right] = \left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

{⇒ \left[\begin{array}{cc} x & y \\[5pt] z & w \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 4 & 5 & 6 \end{array}\right] = \left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

{⇒ \left[\begin{array}{ccc} x + 4y & 2x + 5y & 3x + 6y\\[5pt] z + 4w & 2z + 5w & 3z + 6w \end{array}\right] = \left[\begin{array}{ccc} -7 & -8 & -9 \\[5pt] 2 & 4 & 6 \end{array}\right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

{x + 4y}

= -7

———-❶

{2x + 5y}

= -8

———-❷

{3x + 6y}

= -9

———-❸

{z + 4w}

= 2

———-❹

{2z + 5w}

= 4

———-❺

{3x + 6w}

= 6

———-❻

Now

❶ × 2 ➜

{2x + 8y}

= -14

———-❼

Also,

{2x + 5y}

= -8

———-❶

Subtracting ❶ from ❼, we get

{3y}

= -6

{\Rightarrow y}

= -2

Substituting

{y = -2}

in ❶, we get

{x + (4 \times -2)}

= -7

{\Rightarrow x - 8}

= -7

{\Rightarrow x}

= 8 – 7

{\Rightarrow x}

= 1

Similarly,

❹ × 2 ➜

{2z + 8w}

= 4

———-❽

Also,

{2z + 5w}

= 4

———-❺

Subtracting ❺ from ❽, we get

{3w}

= 0

{\Rightarrow w}

= 0

Substituting

{w = 0}

in ❺, we get

{z + (4 \times 0)}

= 2

{\Rightarrow z}

= 2

So, the matrix

{\text{X} = \left[\begin{array}{cc} x & y \\[5pt] z & w \end{array}\right] \left[\begin{array}{cc} 1 & -2 \\[5pt] 2 & 0 \end{array}\right]}

12. If A and B are square matrices of the same order such that AB = BA, then prove by induction that

{\text{AB}^n = \text{B}^n\text{A}.}

Further, prove that

{\text{(AB)}^n = \text{A}^n\text{B}^n}

for all

{n ∈ \textbf{N}.}

(i) To prove by induction that

{\text{AB}^n = \text{B}^n\text{A}.}

We shall prove the result by using principle of mathematical induction.

We have

{\text{P}(n): \text{AB = BA},}

then

{\text{AB}^n = \text{B}^n\text{A}.}

{\text{P}(1): \text{AB = BA},}

then

{\text{AB}^1}

{= \text{B}^1\text{A}}

= BA

= AB (∵ AB = BA)

∴ The result is true for

{n = 1.}

Let the result be true for

{n = k.}

So,

{\text{P}(k): \text{AB = BA},}

then

{\text{AB}^k = \text{B}^k\text{A}}

———-❶

Now, we prove that the result holds for

{n = k + 1}

Now,

{\text{AB}^{k + 1}}

{= \text{A.B}^k.\text{B}}

{= \text{AB}^k.\text{B}}

(∵ matrix multiplication is associative)

{= \left(\text{B}^k.\text{A}\right).\text{B}}

(from ❶)

{= \text{B}^k.\text{(AB)}}

(∵ matrix multiplication is associative)

{= \text{B}^k.\text{(BA)}}

(∵ AB = BA)

{= \left(\text{B}^k.\text{B}\right)\text{A}}

(∵ matrix multiplication is associative)

{= \text{B}^{k + 1}.\text{A}}

———-❷

∴ The result is true for

{n = k + 1.}

Thus, by principle of mathematical induction, we have

{\text{AB}^n = \text{B}^n\text{A},}

holds for all natural numbers.

(ii) To prove by induction that

{\text{(AB)}^n = \text{A}^n\text{B}^n}

We shall prove the result by using principle of mathematical induction.

We have,

{\text{P}(n): \text{AB = BA},}

then

{\text{(AB)}^n = \text{A}^n\text{B}^n}

{\text{P}(1): \text{AB = BA},}

{\text{(AB)}^1 = \text{AB}}

∴ The result is true for

{n = 1.}

Let the result be true for

{n = k.}

{\text{P}(k): \text{AB = BA},}

then

{\text{(AB)}^k = \text{A}^k\text{B}^k}

———-❸

Now, we prove that the result holds for

{n = k + 1}

{\left(\text{AB}\right)}^{k + 1}

{= \left(\text{AB}\right)^k(\text{AB})}

{= \text{A}^k\text{B}^k.(\text{AB})}

(from ❸)

{= \text{A}^k\text{B}^k.\text{BA}}

(∵ AB = BA)

{= \text{A}^k\left(\text{B}^k\text{B}\right)\text{A}}

(∵ matrix multiplication is associative)

{= \text{A}^k\text{B}^{k + 1}\text{A}}

{= \text{A}^k\left(\text{B}^{k + 1}\text{A}\right)}

(∵ matrix multiplication is associative)

{= \text{A}^k\left(\text{AB}^{k + 1}\right)}

(from ❷)

{= \left(\text{A}^k\text{A}\right)\text{B}^{k + 1}}

(∵ matrix multiplication is associative)

{= \text{A}^{k + 1}\text{B}^{k + 1}}

∴ The result is true for

{n = k + 1.}

Thus, by principle of mathematical induction, we have

{\text{(AB)}^n = \text{A}^n\text{B}^n,}

holds for all natural numbers.

Choose the correct answer in the following questions:

13. If

{\text{A} = \left[\begin{array}{cc} α & β \\[5pt] γ & -α \end{array}\right]}

is such that

{\text{A}^2 = \text{I},}

then

(A)

{1 + α^2 + βγ = 0}

(B)

{1 - α^2 + βγ = 0}

(C)

{1 - α^2 - βγ = 0}

✔

(D)

{1 + α^2 - βγ = 0}

Given that

{\text{A}^2 = I}

{⇒ \left[\begin{array}{cc} α & β \\[5pt] γ & -α \end{array}\right] \left[\begin{array}{cc} α & β \\[5pt] γ & -α \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}

{⇒ \left[\begin{array}{cc} α^2 + βγ & αβ - αβ \\[5pt] αγ - αγ & βγ + α^2 \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}

As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get

{α^2 + βγ = 1}

{⇒ 1 - α^2 - βγ = 0}

Hence option C is the correct answer

Choose the correct answer in the following questions:

14. If the matrix A is both symmetric and skew symmetric, then

(A) A is a diagonal matrix

(B) A is a zero matrix ✔

(D) None of these

By definition, if a matrix is a symmetric matrix, then

A′ = A ———–❶

Also, by definition, if a matrix is a skew symmetric matrix, then

A′ = -A ———–❷

Comparing ❶ and ❷, we have,

A = -A

⇒ 2A = O

⇒ A = O

Thus option B is the correct answer.

Choose the correct answer in the following questions:

15. If A is a square matrix such that

{\text{A}^2 = \text{A},}

then

{\left(\text{I + A}\right)^3 - 7\text{A}}

is equal to

(A) A

(B) I – A

(D) 3A

Given that

{\text{A}^2 = \text{A}}

———–❶

Multiplying with A on both sides, we have

{\text{A}^2.\text{A}}

= A.A

{⇒ \text{A}^3}

{= \text{A}^2}

{= \text{A}}

———–❷

Also, we know that

{\text{I}^2 = \text{I}}

and

{\text{I}^3 = \text{I}}

———-❸

Now, applying

{\left(a + b\right)^3 = a^3 + 3a^2b + 3ab^2 + b^3,}

we have,

{(\text{I + A})^3 - 7\text{A}}

{= \text{I}^3 + 3\text{I}^2\text{A} + 3\text{IA}^2 + \text{A}^3 - 7\text{A}}

Substituting the values from ❶, ❷ and ❸, we get

{(\text{I + A})^3 - 7\text{A}}

{= \text{I} + 3\text{IA} + 3\text{IA} + \text{A} - 7\text{A}}

{\left(∵ \text{A}^2 = A\right)}

{= \text{I} + 3\text{IA} + 3\text{IA} + \text{A} - 7\text{A}}

{(∵ \text{AI = IA = A})}

{= \text{I} + 3\text{A} + 3\text{A} + \text{A} - 7\text{A}}

= I

So, option C is the correct answer.