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## NCERT Solutions for Class 12 Maths Chapter 3 Matrices Miscellaneous Exercise

Question 1.

Let A = \(\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]\) show that (aI + bA)ⁿ = aⁿI + na^{n-1} bA, where I is the identity matrix of order 2 and n ∈ N .

Solution:

Let P(n) : (aI + bA)ⁿ = aⁿI + na^{n-1} bA

P(1) : (aI + bA)^{1 }= a^{1}I + 1a°bA

⇒ P(1) : aI + bA = aI + bA

∴ P(1) is true

Let us assume that P(k) is true.

i.e., P(k) : (aI + bA)^{k} = a^{k}I + ka^{k-1}bA

P(k+ 1) : (aI + bA)^{k+1}

= (aI + bA)^{k}.(aI + bA)

= (a^{k}I + ka^{k-1}A) (aI + bA)

= a^{k}.a.I.l + a^{k}.b. I.A + ka^{k-1}.b.a. AI + ka^{k-1}b.bA.A

= a^{k+1}I + a^{k}bA + ka^{k}bA + ka^{k-1}b²A²

= a^{k+1}I + (k + 1)a^{k}bA + ka^{k-1}b²A² … (1)

A² = \(\left[\begin{array}{ll}

0 & 1 \\

0 & 0

\end{array}\right]\left[\begin{array}{ll}

0 & 1 \\

0 & 0

\end{array}\right]\) = \(\left[\begin{array}{ll}

0 & 0 \\

0 & 0

\end{array}\right]\) = 0

∴ (1) ⇒ P(k + 1) = a^{k+1} I + (k + 1)a^{k}bA + kak^{k-1}b² x 0

= a^{k+1} I + (k + 1)a^{(k+1)-1}Ab

which is true whenever P(L) is true.

Hence by the Principle of Mathematical Induction, the result is true for all n ∈ N .

Question 2.

If A = \(\left[\begin{array}{lll}

1 & 1 & 1 \\

1 & 1 & 1 \\

1 & 1 & 1

\end{array}\right]\), prove that A^{n} = \(\left[\begin{array}{ccc}

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{array}\right]\), n ∈ N

Solution:

which is true

whenever P(k) is true.

Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 3.

If A = \(\left[\begin{array}{ll}

3 & -4 \\

1 & -1

\end{array}\right]\), then prove that Aⁿ = \(\left[\begin{array}{cc}

1+2 n & -4 n \\

n & 1-2 n

\end{array}\right]\), where n is any positive integer.

Solution:

∴ P(k + 1) is true, whenever P(k) is true.

Hence by the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Question 4.

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

Solution:

A and B are symmetric matrices

∴ A’ = A and B’ = B

(i) (AB – BA)’ = (AB)’ – (BA)’

= B’A’ – A’B’ since (AB)’ = B’A’

= BA – AB since A’ = A and

B’ = B

= – (AB – BA)

∴ AB – BA is a skew-symmetric matrix.

(ii) Let AB = BA

Taking transpose on both sides,

(AB)’ = (BA)’

i.e.,(AB)’ = A’B’ since (BA)’ = A’B’

i.e., (AB)’ = AB, since A’ A and B’ = B

∴ AB ¡s a symmetric matrix.

Let AB be a symmetric matrix.

Then (AB)’ = AB

⇒ B’A’ = AB

⇒ BA = AB since B’ = B and A’ = A

∴ AB = BA

Question 5.

Show that the matrix B’AB is symmetric or skew symmetric according as A is sym¬metric or skew symmetric.

Solution:

Let A be a symmetric matrix.

Then A’ = A

∴ (B’AB)’ = (B’(AB))’

= (AB)’(B’)’

= B’A’B

= B’AB since A’ = A

∴ B’AB is a symmetric matrix

Let A be a skew-symmetric matrix.

Then A’ = – A

∴ (B’AB)’ (B’(AB))’

= (AB)’(B’)’

= B’A’B

= B’(-A)B since A’= – A

= – B’AB

∴ B’AB is a skew-symmetric matrix.

Question 6.

Find the values of x, y, z if the matrix A = \(\left[\begin{array}{ccc}

0 & 2 y & z \\

x & y & -z \\

x & -y & z

\end{array}\right]\) satisfied the equation A’A = I.

Solution:

Question 7.

For what values of x if

\(\left[\begin{array}{lll}

1 & 2 & 1

\end{array}\right]\left[\begin{array}{lll}

1 & 2 & 0 \\

2 & 0 & 1 \\

1 & 0 & 2

\end{array}\right]\left[\begin{array}{l}

0 \\

2 \\

x

\end{array}\right]\) = 0

Solution:

Question 8.

If A = \(\left[\begin{array}{cc}

3 & 1 \\

-1 & 2

\end{array}\right]\), show that A² – 5A + 7I = 0.

Solution:

Question 9.

Find x, if

\(\left[\begin{array}{lll}

x & -5 & -1

\end{array}\right]\left[\begin{array}{lll}

1 & 0 & 2 \\

0 & 2 & 1 \\

2 & 0 & 3

\end{array}\right]\left[\begin{array}{l}

x \\

4 \\

1

\end{array}\right]\) = 0

Solution:

Question 10.

A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(i) 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.

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

Solution:

(i) Let

represents annual sale of products in two markets and the column matrix

B = \(\left[\begin{array}{l|l}

2.50 \\

1.50 \\

1.00

\end{array}\right] \begin{aligned}

&x \\

&y \\

&z

\end{aligned}\) represents the unit sale price of the commodities x, y, z.

∴ The revenue collected by each market is given by AB.

The revenue in market I = ₹ 46,000 and

the revenue in market II = ₹ 53,000

Hence gross revenue in two markets

= ₹ 46,000 + 53,000 = ₹ 99,000

(ii) Let the column matrix C = \(\left[\begin{array}{c}

2.00 \\

1.00 \\

0.50

\end{array}\right]\)

Repre-sent the unit cost price of the commodities x, y and z.

∴ The cost price of the articles in the two markets is AC.

The cost price of articles in market I = ₹ 31,000 and the cost price of articles in market II = ₹ 36,000.

Hence the gross cost price

= ₹ 31,000 + 36,000 = ₹ 67,000

Gross profit = Gross revenue – Gross cost price

= ₹ 99,000 – 67, 000 = ₹ 32, 000.

Question 11.

Find the matrix X so that

X\(\left[\begin{array}{lll}

1 & 2 & 3 \\

4 & 5 & 6

\end{array}\right]\) = \(\left[\begin{array}{ccc}

-7 & -8 & -9 \\

2 & 4 & 6

\end{array}\right]\)

Solution:

X\(\left[\begin{array}{lll}

1 & 2 & 3 \\

4 & 5 & 6

\end{array}\right]\) = \(\left[\begin{array}{ccc}

-7 & -8 & -9 \\

2 & 4 & 6

\end{array}\right]\)

It is clear that the order of X is 2 x 2

since the product exists. Let X = \(\left[\begin{array}{ll}

a & b \\

c & d

\end{array}\right]\).

Equating the corresponding elements

∴ a + 4b = – 7 … (1)

2a + 56 = – 8 … (2)

3a + 66 = – 9 … (3)

c + 4d = 2 … (4)

2c + 5d = 4 … (5)

3c + 6d = 6 … (6)

(1) + (2) ⇒ 3a + 96 = – 15

(3) ⇒ 3a + 6b = – 9

Subtracting we get 3b = – 6

⇒ 6 = – 2

∴ (1) ⇒ a = -7 + 8 = 1

(4) + (5) ⇒ 3c + 9d = 6

(6) ⇒ 3c + 6d = 6

Subtracting 3d = 0 ⇒ d = 0

(4) ⇒ c = 2

∴ X = \(\left[\begin{array}{ll}

a & b \\

c & d

\end{array}\right]\) = \(\left[\begin{array}{cc}

1 & -2 \\

2 & 0

\end{array}\right]\)

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:

Let P(n) : ABⁿ = BⁿA be the given statement.

P(1): AB’ = B’A which is true.

Assume that P(&) is true, i.e., AB^{k} = B^{k}A

AB^{k+1} = (AB)B^{k} = (BA)B^{k}

= B(AB^{k})

= B(B^{k}A) since P(k) is true

= B^{k+1}A

Hence by the Principle of Mathematical In-duction, P(n) is true for all values of n ∈ N .

Let P(n) : (AB)ⁿ = AⁿBⁿ

P(1) : (AB)’ = A’B’ ⇒ AB = AB which is true.

Assume that P(k) is true i.e., (AB)^{k} = A^{k}B^{k}

(AB)^{k+1} = (AB)^{k}.(AB) = A^{k}.B^{k}(AB)

= A^{k}(B^{k}A)B

= A^{k}(AB^{k})B (since B^{k} A = AB^{k})

= (A^{k}.A)(B^{k}B)

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

∴ By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Choose the correct answer in the following questions.

Question 13.

If A = \(\left[\begin{array}{cc}

\alpha & \beta \\

\gamma & -\alpha

\end{array}\right]\) 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:

c. 1 – α² – βγ = 0

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:

b. A is a zero matrix

For symmetric matrix a_{ij} = a_{ji}

For skew symmetric matrix a_{ij} = – a_{ji}

These conditions are satisfied only if a_{ij} = 0 for all i and j.

Hence the matrix is a zero matrix.

Question 15.

If A is square matrix such that A² = A, then (I + A)³ – 7A is equal to

a. A

b. I – A

c. I

d. 3A

Solution:

c. I

(I + A)³ – 7A = I³ + 3I²A + 3IA² + A³ – 7A

= I + 3A + 3A² + A²A – 7A

= I + 3A + 3A + A.A – 7A

= I + 6A + A – 7A

= I + 7A – 7A = I