# 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) :} If {\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],} so {\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.} So
{\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):} If {\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],} so {\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. So
{\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):} If {\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.} So
{\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)
= BAB
(∵ (AB) = BA)
= 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 BAB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
(i) To prove that BAB is symmetric when A is symmetric.
When A is symmetric, it implies that A = A
So,
(BAB)
= [B(AB)]
= (AB)(B)
(∵ (AB) = BA)
= BAB
(∵ (AB) = BA and (A) = A)
= BAB
(∵ A = A)
BAB is symmetric.
(ii) To prove that BAB is skew-symmetric when A is skew-symmetric.
When A is skew-symmetric, it implies that A = -A
So,
(BAB)
= [B(AB)]
= (AB)(B)
(∵ (AB) = BA)
= BAB
(∵ (AB) = BA and (A) = A)
= B(-A)B
(∵ A = -A)
= -BAB
BAB is skew-symmetric.
Thus, it is proved that the matrix BAB 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 AA = I.
Given that
A
{= \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
AA = 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}
{⇒ x}
{= \sqrt{\dfrac12}}
{⇒ x}
{= \pm\dfrac{1}{\sqrt{2}}}
Similarly,
{6y^2}
= 1
{⇒ y^2}
{= \dfrac16}
{⇒ y}
{= \sqrt{\dfrac16}}
{⇒ y}
{= \pm\dfrac{1}{\sqrt{6}}}
Similarly,
{3z^2}
= 1
{⇒ z^2}
{= \dfrac13}
{⇒ z}
{= \sqrt{\dfrac13}}
{⇒ 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}
{⇒ 4x = -4}
{⇒ x = -\dfrac44}
{⇒ 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
A
{= \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,} if {\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
{⇒ 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
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.
(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,
R
= 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,
R
= 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
G
= 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
{⇒ y}
= -2
Substituting {y = -2} in ❶, we get
{x + (4 × -2)}
= -7
{⇒ x - 8}
= -7
{⇒ x}
= 8 – 7
{⇒ x}
= 1
Similarly,
❹ × 2 ➜
{2z + 8w}
= 4
———-❽
Also,
{2z + 5w}
= 4
———-❺
Subtracting ❺ from ❽, we get
{3w}
= 0
{⇒ w}
= 0
Substituting {w = 0} in ❺, we get
{z + (4 × 0)}
= 2
{⇒ 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},} so {\text{(AB)}^1 = \text{AB}}
∴ 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{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 ✔
(C) A is a square 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
(C) I ✔
(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.