Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I.

Question 1.

Find the following products wherever possible.

Hint: (1 × 3) by (3 × 1) = 1 × 1

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

-1 & 4 & 2

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

5 \\

1 \\

3

\end{array}\right]\)

(ii) \(\left[\begin{array}{ccc}

2 & 1 & 4 \\

6 & -2 & 3

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

1 \\

2 \\

1

\end{array}\right]\)

(iii) \(\left[\begin{array}{cc}

3 & -2 \\

1 & 6

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

4 & -1 \\

2 & 5

\end{array}\right]\)

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

2 & 2 & 1 \\

1 & 0 & 2 \\

2 & 1 & 2

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

-2 & -3 & 4 \\

2 & 2 & -3 \\

1 & 2 & -2

\end{array}\right]\)

(v) \(\left[\begin{array}{ccc}

3 & 4 & 9 \\

0 & -1 & 5 \\

2 & 6 & 12

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

13 & -2 & 0 \\

0 & 4 & 1

\end{array}\right]\)

(vi) \(\left[\begin{array}{c}

1 \\

-2 \\

1

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

2 & 1 & 4 \\

6 & -2 & 3

\end{array}\right]\)

(vii) \(\left[\begin{array}{cc}

1 & -1 \\

-1 & 1

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

1 & 1 \\

1 & 1

\end{array}\right]\)

(viii) \(\left[\begin{array}{ccc}

0 & c & -b \\

-c & 0 & a \\

b & -a & 0

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

a^{2} & a b & a c \\

a b & b^{2} & b c \\

a c & b c & c^{2}

\end{array}\right]\)

Solution:

(v) \(\left[\begin{array}{ccc}

3 & 4 & 9 \\

0 & -1 & 5 \\

2 & 6 & 12

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

13 & -2 & 0 \\

0 & 4 & 1

\end{array}\right]\)

First matrix is a 3 × 3 matrix and second matrix is 2 × 3 matrix.

No. of columns in the first matrix ≠ No. of rows in the second matrix.

∴ Matrix product is not possible.

(vi) \(\left[\begin{array}{c}

1 \\

-2 \\

1

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

2 & 1 & 4 \\

6 & -2 & 3

\end{array}\right]\)

No. of columns in first matrix = 1

No. of rows in second matrix = 2

No. of columns in the first matrix ≠ No. of rows in the second matrix

Multiplication of matrices is not possible.

Question 2.

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

1 & -2 & 3 \\

-4 & 2 & 5

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

2 & 3 \\

4 & 5 \\

2 & 1

\end{array}\right]\), do AB and BA exist? If they exist, find them. Do A and B commute with respect to multiplication?

Solution:

AB ≠ BA

∴ A and B are not commutative with respect to the multiplication of matrices.

Question 3.

Find A^{2} where A = \(\left[\begin{array}{cc}

4 & 2 \\

-1 & 1

\end{array}\right]\)

Solution:

Question 4.

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

i & 0 \\

0 & i

\end{array}\right]\), find A^{2}.

Solution:

Question 5.

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

i & 0 \\

0 & -i

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

0 & -1 \\

1 & 0

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

0 & \mathbf{i} \\

\mathbf{i} & \mathbf{0}

\end{array}\right]\), and I is the unit matrix of order 2, then show that

(i) A^{2} = B^{2} = C^{2} = -I

(ii) AB = -BA = -C

Solution:

Question 6.

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

2 & 1 \\

1 & 3

\end{array}\right]\) and B = \(\left[\begin{array}{lll}

3 & 2 & 0 \\

1 & 0 & 4

\end{array}\right]\), find AB. Find BA if it exists.

Solution:

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

2 & 1 \\

1 & 3

\end{array}\right]\) and B = \(\left[\begin{array}{lll}

3 & 2 & 0 \\

1 & 0 & 4

\end{array}\right]\)

The order of AB is 2 × 3

BA does not exist since no. of columns in B ≠ No. of rows in A.

Question 7.

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

2 & 4 \\

-1 & k

\end{array}\right]\) and A^{2} = 0, then find the value of k.

Solution:

II.

Question 1.

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

3 & 0 & 0 \\

0 & 3 & 0 \\

0 & 0 & 3

\end{array}\right]\) then find A^{4}.

Solution:

Question 2.

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

1 & 1 & 3 \\

5 & 2 & 6 \\

-2 & -1 & -3

\end{array}\right]\) then find A^{3}.

Solution:

Question 3.

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

1 & -2 & 1 \\

0 & 1 & -1 \\

3 & -1 & 1

\end{array}\right]\), then find A^{3} – 3A^{2} – A – 3I, where I is unit matrix of order 3 × 3.

Solution:

Question 4.

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

1 & 0 \\

0 & 1

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

0 & 1 \\

0 & 0

\end{array}\right]\), show that (aI + bE)^{3} = a^{3}I + 3a^{2}bE, Where I is unit matrix of order 2.

Solution:

III.

Question 1.

If A = [a_{1}, a_{2}, a_{3},], then for any integer n ≥ 1 show that A^{n} = \(\left[\begin{array}{lll}

a_{1}, & a_{2}^{n}, & a_{3}^{n}

\end{array}\right]\)

Solution:

Given A = diag [a_{1}, a_{2}, a_{3},] = \(\left[\begin{array}{ccc}

a_{1} & 0 & 0 \\

0 & a_{2} & 0 \\

0 & 0 & a_{3}

\end{array}\right]\)

A^{n} = diag \(\left[\begin{array}{lll}

a_{1}^{n} & a_{2}^{n} & a_{3}^{n}

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

a_{1}^{n} & 0 & 0 \\

0 & a_{2}^{n} & 0 \\

0 & 0 & a_{3}^{n}

\end{array}\right]\)

This problem can be should by using Mathematical Induction

put n = 1

A^{1} = \(\left[\begin{array}{ccc}

a_{1} & 0 & 0 \\

0 & a_{2} & 0 \\

0 & 0 & a_{3}

\end{array}\right]\)

∴ The result is true for n = 1

Assume the result is true for n = k

A^{k} = \(\left[\begin{array}{ccc}

a_{1}^{k} & 0 & 0 \\

0 & a_{2}^{k} & 0 \\

0 & 0 & a_{3}^{k}

\end{array}\right]\)

Consider

∴ The result is true for n = k + 1

Hence by the Principle of Mathematical Induction, the statement is true ∀ n ∈ N

Question 2.

If θ – φ = \(\frac{\pi}{2}\), then show that \(\left[\begin{array}{cc}

\cos ^{2} \theta & \cos \theta \sin \theta \\

\cos \theta \sin \theta & \sin ^{2} \dot{\theta}

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

\cos ^{2} \phi & \cos \phi \sin \phi \\

\cos \phi \sin \phi & \sin ^{2} \phi

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

Solution:

Question 3.

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

3 & -4 \\

1 & -1

\end{array}\right]\) then show that A^{n} = \(\left[\begin{array}{cc}

1+2 n & -4 n \\

n & 1-2 n

\end{array}\right]\), for any integer n ≥ 1 by using Mathematical Induction.

Solution:

We shall prove the result by Mathematical Induction.

∴ The given result is true for n = k + 1

By Mathematical Induction, the given result is true for all positive integral values of n.

Question 4.

Give examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.

Solution:

Question 5.

A Trust fund has to invest ₹ 30,000 in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds if the trust fund must obtain an annual total interest of (a) ₹ 1800 (b) ₹ 2000

Solution:

Let the first bond be ‘x’ and the second bond be 30,000 – x respectively

The rate of interest is 0.05 and 0.07 respectively.

(a) \([x, 30,000-x]\left[\begin{array}{l}

0.05 \\

0.07

\end{array}\right] \quad=[1800]\)

[0.05x + 0.07(30,000 – x)] = 1800

\(\frac{5}{100} x+\frac{7}{100}(30,000-x)=1800\)

5x + 21,0000 – 7x = 1,80,000

-2x = 1,80,000 – 2,10,000 = -30,000

x = 15,000

∴ First bond = 15,000

Second bond = 30,000 – 15,000 = 15,000

(b) \(\left[\begin{array}{ll}

x & 30,000-x

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

0.05 \\

0.07

\end{array}\right]=[2000]\)

[0.05x + 0.07(30,000 – x)] = [2000]

\(\frac{5 x}{100} \times \frac{7}{100}(30,000-x)=2000\)

5x + 2,10,000 – 7x = 2,00,000

-2x = 2,00,000 – 2,10,000

-2x = -10,000

x = 5,000

∴ First bond = 5000

Second bond = 30,000 – 5000 = 25,000