Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.6 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.6

Question 1.

Evaluate:

i. \(\left[\begin{array}{l}

3 \\

2 \\

1

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

{[2} & -4 & 3

\end{array}\right]\)

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

2 & -1 & 3

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

4 \\

3 \\

1

\end{array}\right]\)

Solution:

i. \(\begin{aligned}

\left[\begin{array}{l}

3 \\

2 \\

1

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

2 & -4 & 3

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

3(2) & 3(-4) & 3(3) \\

2(2) & 2(-4) & 2(3) \\

1(2) & 1(-4) & 1(3)

\end{array}\right] \\

&=\left[\begin{array}{ccc}

6 & -12 & 9 \\

4 & -8 & 6 \\

2 & -4 & 3

\end{array}\right]

\end{aligned}\)

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

2 & -1 & 3

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

4 \\

3 \\

1

\end{array}\right]\)

= [2(4)-1(3)+ 3(1)]

= [8 – 3 + 3] = [8]

Question 2.

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

1 & -3 \\

4 & 2

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

4 & 1 \\

3 & -2

\end{array}\right]\), = show that AB ≠ BA.

Solution:

From (i) and (ii), we get

AB ≠ BA

Question 3.

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

-1 & 1 & 1 \\

2 & 3 & 0 \\

1 & -3 & 1

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

2 & 1 & 4 \\

3 & 0 & 2 \\

1 & 2 & 1

\end{array}\right]\) state whether AB = BA? Justify your answer.

Solution:

From (i) and (ii), we get

AB ≠ BA

Question 4.

Show that AB = BA, where

i. A = \(\left[\begin{array}{rrr}

-2 & 3 & -1 \\

-1 & 2 & -1 \\

-6 & 9 & -4

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

1 & 3 & -1 \\

2 & 2 & -1 \\

3 & 0 & -1

\end{array}\right]\)

ii. A = \(\left[\begin{array}{cc}

\cos \theta & \sin \theta \\

-\sin \theta & \cos \theta

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

\cos \theta & -\sin \theta \\

\sin \theta & \cos \theta

\end{array}\right]\)

Solution:

From (i) and (ii), we get

AB = BA

From (i) and (ii), we get

AB = BA

[Note: The question has been modified.]

Question 5.

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

4 & 8 \\

-2 & -4

\end{array}\right]\), prove that A^{2} = 0.

Solution:

A^{2} = A.A

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

4 & 8 \\

-2 & -4

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

4 & 8 \\

-2 & -4

\end{array}\right]\)

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

16-16 & 32-32 \\

-8+8 & -16+16

\end{array}\right] \)

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

0 & 0 \\

0 & 0

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

Question 6.

Verify A(BC) = (AB)C in each of the following cases:

i. A = \(=\left[\begin{array}{cc}

4 & -2 \\

2 & 3

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

-1 & 1 \\

3 & -2

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

4 & 1 \\

2 & -1

\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}

1 & -1 & 3 \\

2 & 3 & 2

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

1 & 0 \\

-2 & 3 \\

4 & 3

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

1 & 2 \\

-2 & 0 \\

4 & -3

\end{array}\right]\)

Solution:

From (i) and (ii), we get

A(BC) = (AB)C.

Question 7.

Verify that A(B + C) = AB + AC in each of the following matrices:

i. A = \(\left[\begin{array}{cc}

4 & -2 \\

2 & 3

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

-1 & 1 \\

3 & -2

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

4 & 1 \\

2 & -1

\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}

1 & -1 & 3 \\

2 & 3 & 2

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

1 & 0 \\

-2 & 3 \\

4 & 3

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

1 & 2 \\

-2 & 0 \\

4 & -3

\end{array}\right]\)

Solution:

From (i) and (ii), we get

A(B + C) = AB + AC.

[Note: The question has been modified.]

Question 8.

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

1 & -2 \\

5 & 6

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

3 & -1 \\

3 & 7

\end{array}\right]\), find AB – 2I, where I is unit matrix of order 2.

Solution:

Question 9.

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

4 & 3 & 2 \\

-1 & 2 & 0

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

1 & 2 \\

-1 & 0 \\

1 & -2

\end{array}\right]\), show that matrix AB is non singular.

Solution:

im

∴ AB is non-singular matrix.

Question 10.

If A = \(\), find the product (A + I)(A – I).

Solution:

[Note : Answer given in the textbook is \(\left[\begin{array}{ccc}

9 & 6 & 4 \\

15 & 32 & -2 \\

35 & -7 & 29

\end{array}\right]\)

However, as per our calculation it is \(\left[\begin{array}{ccc}

10 & 10 & 4 \\

25 & 39 & 2 \\

35 & 7 & 22

\end{array}\right]\). ]

Question 11.

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

\alpha & 0 \\

1 & 1

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

1 & 0 \\

2 & 1

\end{array}\right]\), find α, if A^{2} = B.

Solution:

A^{2} = B

∴ By equality of matrices, we get

α^{2} = 1 and α + 1 = 2

∴ α = ± 1 and α = 1

∴ α = 1

Question 12.

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

1 & 2 & 2 \\

2 & 1 & 2 \\

2 & 2 & 1

\end{array}\right]\), show that A^{2} – 4A is scalar matrix.

Solution:

A^{2} – 4A = A.A – 4A

Question 13.

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

1 & 0 \\

-1 & 7

\end{array}\right]\), find k so that A^{2} – 8A – kI = O, where I is a unit matrix and O is a null matrix of order 2.

Solution:

A^{2} – 8A – kI = O

∴ A.A – 8A – kI = O

∴ by equality of matrices, we get

1 – 8 – k = 0

∴ k = -7

Question 14.

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

8 & 4 \\

10 & 5

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

5 & -4 \\

10 & -8

\end{array}\right]\), show that (A+B)^{2} = A^{2} + AB + B^{2}.

Solution:

We have to prove that (A + B)^{2} = A^{2} + AB + B^{2},

i.e., to prove A^{2} + AB + BA + B^{2} = A^{2} + AB + B^{2},

i.e., to prove BA = 0.

BA = \(\left[\begin{array}{cc}

5 & -4 \\

10 & -8

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

8 & 4 \\

10 & 5

\end{array}\right]\)

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

40-40 & 20-20 \\

80-80 & 40-40

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

0 & 0 \\

0 & 0

\end{array}\right]\)

Question 15.

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

3 & 1 \\

-1 & 2

\end{array}\right]\), prove that A^{2} – 5A + 7I = 0, where I is unit matrix of order 2.

Solution:

A^{2} – 5A + 7I = 0 = A.A – 5A + 7I = 0

Question 16.

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

3 & 4 \\

-4 & 3

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

2 & 1 \\

-1 & 2

\end{array}\right]\), show that (A + B)(A – B) = A^{2} – B^{2}.

Solution:

We have to prove that (A + B)(A – B) = A^{2} – B^{2},

i.e., to prove A^{2} – AB + BA – B^{2} = A^{2} – B^{2},

i.e., to prove – AB + BA = 0,

i.e., to prove AB – BA.

From (i) and (ii), we get AB = BA

Question 17.

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

1 & 2 \\

-1 & -2

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

2 & a \\

-1 & b

\end{array}\right]\) and (A + B)^{2} = A^{2} + B^{2}, find the values of a and b.

Solution:

Given, (A + B)^{2} = A^{2} + B^{2}

∴ A^{2} + AB + BA + B^{2} = A^{2} + B^{2}

∴ AB + BA = 0

∴ AB = -BA

∴ by equality of matrices, we get

– 2 + a = 0 and 1 + b = 0

a = 2 and b = -1

[Note: The question has been modified.]

Question 18.

Find matrix X such that AX = B,

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

1 & -2 \\

-2 & 1

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

-3 \\

-1

\end{array}\right]\)

Solution:

Let X = \(\left[\begin{array}{c}

a \\

b

\end{array}\right]\)

But AX = B

∴ \(\left[\begin{array}{cc}

1 & -2 \\

-2 & 1

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

\mathrm{a} \\

\mathrm{b}

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

-3 \\

-1

\end{array}\right]\)

∴ \(\left[\begin{array}{c}

a-2 b \\

-2 a+b

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

-3 \\

-1

\end{array}\right]\)

By equality of matrices, we get

a – 2b = -3 …(i)

-2a + b = -l …(ii)

By (i) x 2 + (ii), we get

-3b =-7

∴ b = \(\frac{7}{3}\)

Substituting b = \(\frac{7}{3}\) in (i), we get

a – 2 (\(\frac{7}{3}\)) = -3

∴ a = -3 + \(\frac{14}{3}=\frac{5}{3}\)

∴ X = \(\left[\begin{array}{l}

\frac{5}{3} \\

\frac{7}{3}

\end{array}\right]\)

Question 19.

Find k, if A = \(\left[\begin{array}{ll}

3 & -2 \\

4 & -2

\end{array}\right]\) and A^{2} = KA – 2I

Solution:

A^{2} = kA – 2I

∴ AA + 2I = kA

∴ \(\left[\begin{array}{ll}

3 & -2 \\

4 & -2

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

3 k & -2 k \\

4 k & -2 k

\end{array}\right]\)

∴ By equality of matrices, we get

3k = 3

∴ k = 1

Question 20.

Find x, if \(\left[\begin{array}{lll}

1 & x & 1

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

1 & 2 & 3 \\

4 & 5 & 6 \\

3 & 2 & 5

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

1 \\

-2 \\

3

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

Solution:

∴ [6 + 12x + 14] =[0]

∴ By equality of matrices, we get

∴ 12x + 20 = 0

∴ 12x =-20

∴ x = \(\frac{-5}{3}\)

Question 21.

Find x and y, if \(\left\{4\left[\begin{array}{ccc}

2 & -1 & 3 \\

1 & 0 & 2

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

3 & -3 & 4 \\

2 & 1 & 1

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

2 \\

-1 \\

1

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

x \\

y

\end{array}\right]\)

Solution:

∴ By equality of matrices, we get

x = 19 andy = 12

Question 22.

Find x, y, z if

\(\left\{3\left[\begin{array}{ll}

2 & 0 \\

0 & 2 \\

2 & 2

\end{array}\right]-4\left[\begin{array}{cc}

1 & 1 \\

-1 & 2 \\

3 & 1

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

1 \\

2

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

x-3 \\

y-1 \\

2 z

\end{array}\right]\)

Solution:

∴ By equality of matrices, we get

x – 3 = -6,y – 1 = 0, 2z = -2

∴ x = – 3, y = 1, z = – 1

Question 23.

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

\cos \alpha & \sin \alpha \\

-\sin \alpha & \cos \alpha

\end{array}\right]\) show that A^{2} = \(=\left[\begin{array}{cc}

\cos 2 \alpha & \sin 2 \alpha \\

-\sin 2 \alpha & \cos 2 \alpha

\end{array}\right]\)

Solution:

Question 24.

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

1 & 2 \\

3 & 5

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

0 & 4 \\

2 & -1

\end{array}\right]\)

show that AB ≠ BA, but |AB| = |A| . |B|.

Solution:

AB = \(\left[\begin{array}{ll}

1 & 2 \\

3 & 5

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

0 & 4 \\

2 & -1

\end{array}\right]\)

Now, |AB| = \(\left|\begin{array}{cc}

4 & 2 \\

10 & 7

\end{array}\right|\) = 28 – 20 = 8

|A| = \(\left|\begin{array}{ll}

1 & 2 \\

3 & 5

\end{array}\right|\) = 5 – 6 = -1

|B| = \(\left|\begin{array}{cc}

0 & 4 \\

2 & -1

\end{array}\right|\) = 0 – 8 = -8

∴ |A| . |B| = (-1).(-8) = 8 = |AB|

∴ AB ≠ BA, but |AB| = |A|.|B|

Question 25.

Jay and Ram are two friends in a class. Jay wanted to buy 4 pens and 8 notebooks, Ram wanted to buy 5 pens and 12 notebooks. Both of them went to a shop. The price of a pen and a notebook which they have selected was 6 and ₹ 10. Using matrix multiplication, find the amount required from each one of them.

Solution:

Let A be the matrix of pens and notebooks and B be the matrix of prices of one pen and one notebook.

The total amount required for each one of them is obtained by matrix AB.

∴ Jay needs ₹ 104 and Ram needs ₹ 150.