Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 1.

Evaluate:

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

3 \\

2 \\

1

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

2 & -4 & 3

\end{array}\right]\)

Solution:

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

3 \\

2 \\

1

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

2 & -4 & 3

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

6 & -12 & 9 \\

4 & -8 & 6 \\

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:

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

2 & -1 & 3

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

4 \\

3 \\

1

\end{array}\right]\) = [8 – 3 + 3] = [8]

Question 2.

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 (1) and (2), AB ≠ BA.

Question 3.

Show that AB = BA, where A = \(\left[\begin{array}{lll}

-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]\)

Solution:

From (1) and (2), AB = BA.

Question 4.

Verify A(BC) = (AB)C, if A = \(\left[\begin{array}{lll}

1 & 0 & 1 \\

2 & 3 & 0 \\

0 & 4 & 5

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

2 & -2 \\

-1 & 1 \\

0 & 3

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

3 & 2 & -1 \\

2 & 0 & -2

\end{array}\right]\)

Solution:

From (1) and (2), A(BC) = (AB)C.

Question 5.

Verify that A(B + C) = AB + AC, if 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]\)

Solution:

From (1) and (2), A(B + C) = AB + AC.

Question 6.

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:

Hence, AB is a non-singular matrix.

Question 7.

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

1 & 2 & 0 \\

5 & 4 & 2 \\

0 & 7 & -3

\end{array}\right]\), find the product (A + I)(A – I).

Solution:

Question 8.

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

1 & 2 & 2 \\

2 & 1 & 2 \\

2 & 2 & 1

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

Solution:

which is a scalar matrix.

Question 9.

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 2 × 2 unit matrix and O is null matrix of order 2.

Solution:

By equality of matrices,

-k – 7 = 0

∴ k = -7.

Question 10.

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

3 & 1 \\

-1 & 2

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

Solution:

Question 11.

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 if(A + B)^{2} = A^{2} + B^{2}, find values of a and b.

Solution:

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

∴ (A + B)(A + B) = A^{2} + B^{2}

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

∴ AB + BA = 0

∴ AB = -BA

By the equality of matrices, we get

0 = a – 2 ……..(1)

0 = 1 + b ……..(2)

a + 2b = 2a – 4 ……..(3)

-a – 2b = 2 + 2b ……..(4)

From equations (1) and (2), we get

a = 2 and b = -1

The values of a and b satisfy equations (3) and (4) also.

Hence, a = 2 and b = -1.

Question 12.

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

3 & -2 \\

4 & -2

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

Solution:

By equality of matrices,

1 = 3k – 2 ……..(1)

-2 = -2k ……..(2)

4 = 4k ……..(3)

-4 = -2k – 2 ……..(4)

From (2), k = 1.

k = 1 also satisfies equation (1), (3) and (4).

Hence, k = 1.

Question 13.

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}{c}

x \\

y

\end{array}\right]\)

Solution:

By equality of matrices,

x = 19 and y = 12.

Question 14.

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,

-6 = x – 3, 0 = y – 1 and -2 = 2z

∴ x = -3, y = 1 and z = -1.

Question 15.

Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks. Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.

Solution:

The given data can be written in matrix form as:

Number of Pens and Notebooks

For finding the amount each one of them requires to buy the pens and notebook, we require the multiplication of the two matrices A and B.

Hence, Jay requires ₹ 104 and Ram requires ₹ 150 to buy the pens and notebooks.