Ncert Solutions For Class 12 Maths Ex 3.2

Ncert Solutions For Class 12 Maths Chapter 3 Ex 3.2

Q-1: A = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\), B = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\), C = \(\begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

Find the value of each of the following from the matrices given:

(a) A + B (b) A – B (c) 3A – C (d) AB (e) BA

 

Solution:

(a) A + B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3 + 2 & 5 + 4 \\ 4 – 3 & 3 + 6 \end{bmatrix} \)

= \( \begin{bmatrix} 5 & 9 \\ 1 & 9 \end{bmatrix}\)

 

(b) A – B = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3 – 2 & 5 – 4 \\ 4 + 3 & 3 – 6 \end{bmatrix} \)

= \( \begin{bmatrix} 1 & 1 \\ 7 & -3 \end{bmatrix}\)

 

(c) 3A – C = \(3\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 3 \times 3 & 5 \times 3 \\ 4 \times 3 & 3 \times 3 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 9 & 15 \\ 12 & 9 \end{bmatrix} – \begin{bmatrix} -3 & 6 \\ 4 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 9 + 3 & 15 – 6 \\ 12 – 4 & 9 – 5 \end{bmatrix}\)

= \(\begin{bmatrix} 12 & 9 \\ 8 & 4 \end{bmatrix}\)

(d) Since, the matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then

Multiplication of the two matrices A and B will be defined as:

AB = \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)\(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 2 ) + 5( -3 ) & 3( 4 ) + 5( 6 ) \\ 4( 2 ) + 3( -3 ) & 4( 4 ) + 3( 6 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 – 15 & 12 + 30 \\ 8 – 9 & 16 + 18 \end{bmatrix}\)

= \(\begin{bmatrix} -9 & 42 \\ -1 & 34 \end{bmatrix}\)

 

(e) Since, the matrix A and B, both are a square matrix as the number of rows and the number of columns is the same in both of the matrices, then

Multiplication of the two matrices B and A will be defined as:

BA = \(\begin{bmatrix} 2 & 4 \\ -3 & 6 \end{bmatrix}\) \(\begin{bmatrix} 3 & 5 \\ 4 & 3 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 3 ) + 3( 4 ) & 2( 5 ) + 4( 3 ) \\ -3( 3 ) + 6( 4 ) & -3( 5 ) + 6( 3 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 + 12 & 10 + 12 \\ – 9 + 24 & -15 + 18 \end{bmatrix}\)

= \(\begin{bmatrix} 18 & 22 \\ 15 & 3 \end{bmatrix}\)

 

Q-2: Solve the following:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)

(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)

(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)

(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)

 

Solution:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix} + \begin{bmatrix} x & y \\ y & x \end{bmatrix}\)

= \(\begin{bmatrix} x + x & y + y \\ -y + y & x + x \end{bmatrix}\)

= \(\begin{bmatrix} 2x & 2y \\ 0 & 2x \end{bmatrix}\)

 

(ii) \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & y^{ 2 } + z^{ 2 } \\ x^{ 2 } + z^{ 2 } & x^{ 2 } + y^{ 2 } \end{bmatrix} + \begin{bmatrix} 2xy & 2yz \\ -2xz & -2xy \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } + 2xy & y^{ 2 } + z^{ 2 } + 2yz \\ x^{ 2 } + z^{ 2 } – 2xz & x^{ 2 } + y^{ 2 } – 2ab \end{bmatrix}\)

= \(\begin{bmatrix} \left ( x + y \right )^{ 2 } & \left ( y + z \right )^{ 2 } \\ \left ( x – y \right )^{ 2 } & \left ( x – y \right )^{ 2 } \end{bmatrix}\)

 

(iii) \(\begin{bmatrix} -2 & 5 & -7 \\ 9 & 6 & 17 \\ 3 & 9 & 6 \end{bmatrix} + \begin{bmatrix} 13 & 8 & 7\\ 9 & 1 & 6 \\ 4 & 3 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} -2 + 13 & 5 + 8 & -7 + 7 \\ 9 + 9 & 6 + 1 & 17 + 6 \\ 3 + 4 & 9 + 3 & 6 + 5 \end{bmatrix}\)

= \(\begin{bmatrix} 11 & 13 & 0 \\ 18 & 6 & 23 \\ 7 & 12 & 11 \end{bmatrix}\)

 

(iv) \(\begin{bmatrix} cos^{ 2 }a & sin^{ 2 }a \\ sin^{ 2 }a & cos^{ 2 }a \end{bmatrix} + \begin{bmatrix} sin^{ 2 }a & cos^{ 2 }a \\ cos^{ 2 }a & sin^{ 2 }a \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }a + sin^{ 2 }a & sin^{ 2 }a + cos^{ 2 }a \\ sin^{ 2 }a + cos^{ 2 }a & cos^{ 2 }a + sin^{ 2 }a \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \;\;\;\;\;\;\;\;\left ( as sin^{ 2 }a + cos^{ 2 }a = 1 \right )\)

Q-3: Compute the following:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)

(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)

(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)

(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)

(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)

(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)

 

Solution:

(i) \(\begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & -y \\ y & a \end{bmatrix}\)

= \(\begin{bmatrix} x( x )+ y( y ) & x( -y ) + y( x ) \\ -y( x ) + x( y ) & -y( -y ) + x( x ) \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & -xy + xy \\ -yx + xy & y^{ 2 } + x^{ 2 } \end{bmatrix}\)

= \(\begin{bmatrix} x^{ 2 } + y^{ 2 } & 0 \\ 0 & y^{ 2 } + x^{ 2 } \end{bmatrix}\)

 

(ii) \(\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 1 ) & 2( 2 ) & 2( 3 )\\ 3( 1 ) & 3( 2 ) & 3( 3 )\\ 4( 1 ) & 4( 2 ) & 4( 3 ) \end{bmatrix}\)

= \(\begin{bmatrix} 2 & 4 & 6 \\ 3 & 6 & 9 \\ 4 & 8 & 12 \end{bmatrix}\)

 

(iii) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 2( 2 ) – 3( 3 ) & 2( 3 ) – 3( 4 ) & 2( 4 ) – 3( 2 )\\ 3( 2 ) + 4( 3 ) & 3( 3 ) + 4( 4 ) & 3( 4 ) + 4( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 4 – 9 & 6 – 12 & 8 – 6 \\ 6 + 12 & 9 + 16 & 12 + 8 \end{bmatrix}\)

= \(\begin{bmatrix} -5 & -6 & -2 \\ 18 & 25 & 20 \end{bmatrix}\)

 

(iv) \(\begin{bmatrix} 3 & 4 & 5 \\ 4 & 5 & 6 \\ 5 & 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & -4 & 6 \\ 0 & 3 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 2 ) + 4( 0 ) + 5( 3 ) & 3( -4 ) + 4( 3 ) + 5( 0 ) & 3( 6 ) + 4( 4 ) + 5( 5 ) \\ 4( 2 ) + 5( 0 ) + 6( 3 ) & 4( -4 ) + 5( 3 ) + 6( 0 ) & 4( 6 ) + 5( 4 ) + 6( 5 ) \\ 5( 2 ) + 6( 0 ) + 7( 3 ) & 5( -4 ) + 6( 3 ) + 7( 0 ) & 5( 6 ) + 6( 4 ) + 7( 5 ) \end{bmatrix}\)

= \(\begin{bmatrix} 6 + 0 + 15 & -12 + 12 + 0 & 18 + 16 + 25 \\ 8 + 0 + 18 & -16 + 15 + 0 & 24 + 20 + 30 \\ 10 + 0 + 21 & -20 + 18 + 0 & 30 + 24 + 35 \end{bmatrix}\)

= \(\begin{bmatrix} 21 & 0 & 59 \\ 26 & -1 & 74 \\ 31 & -2 & 89 \end{bmatrix}\)

(v) \(\begin{bmatrix} 3 & 2 \\ 4 & 3 \\ -2 & 2 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 3 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 3( 1 ) + 2( -1 ) & 3( 0 ) + 2( 3 ) & 3( 1 ) + 2( 2 ) \\ 4( 1 ) + 3( -1 ) & 4( 0 ) + 3( 3 ) & 4( 1 ) + 3( 2 ) \\ -2( 1 ) + 2( -1 ) & -2( 0 ) + 2( 3 ) & -2( 1 ) + 2( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 3 – 2 & 0 + 6 & 3 + 4 \\ 4 – 3 & 0 + 9 & 4 + 6 \\ -2 – 2 & 0 + 6 & -2 + 4 \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 6 & 7 \\ 1 & 9 & 10 \\ -4 & 6 & 2 \end{bmatrix}\)

 

(vi) \(\begin{bmatrix} 4 & -2 & 4 \\ -2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 2 & 1 \\ 4 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 4( 3 ) – 2( 2 ) + 4( 4 ) & 4( -4 ) – 2( 1 ) + 4( 2 )\\ -2( 3 ) + 0( 2 ) + 3( 4 ) & -2( -4 ) + 0( 1 ) + 4( 2 ) \end{bmatrix}\)

= \(\begin{bmatrix} 12 – 4 + 16 & -16 – 2 + 8 \\ -6 + 0 + 12 & 8 + 0 + 6 \end{bmatrix}\)

= \(\begin{bmatrix} 24 & -10 \\ 6 & 14 \end{bmatrix}\)

 

Q-4: If, A = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\), B = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) and C = \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\), then

Find ( A + B ) and ( B – C ). Also, check whether A + ( B – C ) = ( A + B ) – C

 

Solution:

A + B = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 2 + 4 & 3 – 2 & -4 + 3 \\ 5 + 5 & 0 + 3 & 3 + 6 \\ 2 + 3 & -2 + 0 & 2 + 4 \end{bmatrix}\)

= \(\begin{bmatrix} 6 & 1 & -1 \\ 10 & 3 & 9 \\ 5 & -2 & 6 \end{bmatrix}\)

 

B – C = \(\begin{bmatrix} 4 & -2 & 3 \\ 5 & 3 & 6 \\ 3 & 1 & 4 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 4 – 5 & -2 – 2 & 3 – 3 \\ 5 – 1 & 3 – 4 & 6 – 3 \\ 3 – 2 & 1 – ( -3 ) & 4 – 4 \end{bmatrix}\)

= \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)

 

A + ( B – C ) = \(\begin{bmatrix} 2 & 3 & -4 \\ 5 & 0 & 3 \\ 2 & -2 & 2 \end{bmatrix}\) + \(\begin{bmatrix} -1 & -4 & 0 \\ 4 & -1 & 3 \\ 1 & 4 & 0 \end{bmatrix}\)

= \(\begin{bmatrix} 2 – 1 & 3 – 4 & -4 + 0 \\ 5 + 4 & 0 – 1 & 3 + 3 \\ 2 + 1 & -2 + 4 & 2 + 0 \end{bmatrix}\)

= \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 2 & 2 \end{bmatrix}\)

 

( A + B ) – C = \(\begin{bmatrix} 6 & 1 & -1 \\ 10 & 3 & 9 \\ 5 & -2 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 2 & 3 \\ 1 & 4 & 3 \\ 2 & -3 & 4 \end{bmatrix}\)

= \(\begin{bmatrix} 6 – 5 & 1 – 2 & -1 – 3 \\ 10 – 1 & 3 – 4 & 9 – 3 \\ 5 – 2 & -2 + 3 & 6 – 4 \end{bmatrix}\)

= \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)

Therefore, A + ( B – C ) = ( A + B ) – C = \(\begin{bmatrix} 1 & -1 & -4 \\ 9 & -1 & 6 \\ 3 & 1 & 2 \end{bmatrix}\)

Hence, proved.

 

 

Q-5: If, A = \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\), then

Find the value of 3A – 5B.

Solution:

3A – 5B = 3 \(\begin{bmatrix} \frac{ 2 }{ 3 } & 2 & \frac{ 5 }{ 3 } \\ \\ \frac{ 1 }{ 3 } & \frac{ 2 }{ 3 } & \frac{ 4 }{ 3 } \\ \\ \frac{ 7 }{ 3 } & 3 & \frac{ 2 }{ 3 } \end{bmatrix}\) – 5 \(\begin{bmatrix} \frac{ 2 }{ 5 } & \frac{ 3 }{ 5 } & 2 \\ \\ \frac{ 1 }{ 5 } & \frac{ 2 }{ 5 } & \frac{ 4 }{ 5 } \\ \\ \frac{ 7 }{ 5 } & \frac{ 6 }{ 5 } & \frac{ 2 }{ 5 } \end{bmatrix}\)

= \(\begin{bmatrix} 2 & 6 & 5 \\ \\ 1 & 2 & 4 \\ \\ 7 & 9 & 2 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 3 & 10 \\ \\ 1 & 2 & 4 \\ \\ 7 & 6 & 2 \end{bmatrix}\)

= \(\begin{bmatrix} 0 & 3 & -5 \\ \\ 0 & 0 & 0 \\ \\ 0 & 3 & 0 \end{bmatrix}\)

 

Q-6: Find the value of:

\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\).

Solution:

\(cos\Theta \begin{bmatrix} cos\Theta & sin\Theta \\ -sin\Theta & cos\Theta \end{bmatrix}\) + \(sin\Theta \begin{bmatrix} sin\Theta & -cos\Theta \\ cos\Theta & sin\Theta \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }\Theta & cos\Theta sin\Theta \\ -sin\Theta cos\Theta & cos^{ 2 }\Theta \end{bmatrix}\) + \(\begin{bmatrix} sin^{ 2 }\Theta & -sin\Theta cos\Theta \\ sin\Theta cos\Theta & sin^{ 2 }\Theta \end{bmatrix}\)

= \(\begin{bmatrix} cos^{ 2 }\Theta + sin^{ 2 }\Theta & cos\Theta sin\Theta – cos\Theta sin\Theta \\ -sin\Theta cos\Theta + cos\Theta sin\Theta & cos^{ 2 }\Theta + sin^{ 2 }\Theta \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

 

 

Q-7: What will be the value of A and B, if

(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) and A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) and 3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)

 

Solution:

(i) A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) …………………..(a)

A – B = \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\) …………………..(b)

By adding equation (a) and (b), we will get:

2A = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) + \(\begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}\)

2A = \(\begin{bmatrix} 8 + 4 & 0 + 0 \\ 0 + 4 & 6 + 4 \end{bmatrix}\)

2A = \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)

Thus,

A = \(\frac{ 1 }{ 2 } \) \(\begin{bmatrix} 12 & 0 \\ 4 & 10 \end{bmatrix}\)

A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)

A + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)

\(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) + B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\)

B = \(\begin{bmatrix} 8 & 0 \\ 4 & 6 \end{bmatrix}\) – \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\)

B = \(\begin{bmatrix} 8 – 6 & 0 – 0 \\ 4 – 2 & 6 – 5 \end{bmatrix}\)

B = \(\begin{bmatrix} 2 & 0 \\ 2 & 1 \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} 6 & 0 \\ 2 & 5 \end{bmatrix}\) and, B = \(\begin{bmatrix} 2 & 0 \\ 2 & 1 \end{bmatrix}\)

 

(ii) 2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) …………..(a)

3A + 2B = \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\) ……………(b)

By multiplying equation (a) by 2, we will get:

2( 2A + 3B ) = 2 \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)

4A + 6B = \(\begin{bmatrix} 6 & 8 \\ 10 & 0 \end{bmatrix}\) ……………(c)

By multiplying equation (b) by 3, we will get:

3( 3A + 2B ) = 3 \(\begin{bmatrix} 3 & -3 \\ -2 & 6 \end{bmatrix}\)

9A + 6B = \(\begin{bmatrix} 9 & -9 \\ -6 & 12 \end{bmatrix}\) …………….(d)

By subtracting equation (c) and (d), we will get:

( 4A + 6B ) – ( 9A + 6B ) = \(\begin{bmatrix} 6 – 9 & 8 + 9 \\ 10 + 6 & 0 – 12 \end{bmatrix}\)

-5A = \(\begin{bmatrix} -3 & 17 \\ 16 & -12 \end{bmatrix}\)

A = – \(\frac{ 1 }{ 5 }\) \(\begin{bmatrix} -3 & 17 \\ 16 & -12 \end{bmatrix}\)

A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\)

Now,

2A + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)

2 \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\) + 3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\)

3B = \(\begin{bmatrix} 3 & 4 \\ 5 & 0 \end{bmatrix}\) – \(\begin{bmatrix} \frac{ 6 }{ 5 } & \frac{ -34 }{ 5 } \\ \frac{ -32 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)

3B = \(\begin{bmatrix} 3 – \frac{ 6 }{ 5 } & 4 – \frac{ -34 }{ 5 } \\ 5 – \frac{ -32 }{ 5 } & 0 – \frac{ 24 }{ 5 } \end{bmatrix}\)

3B = \(\begin{bmatrix} \frac{ 9 }{ 5 } & \frac{ 54 }{ 5 } \\ \frac{ 57 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)

B = \(\frac{ 1 }{ 3 }\) \(\begin{bmatrix} \frac{ 9 }{ 5 } & \frac{ 54 }{ 5 } \\ \frac{ 57 }{ 5 } & \frac{ 24 }{ 5 } \end{bmatrix}\)

B = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 18 }{ 5 } \\ \frac{ 19 }{ 5 } & \frac{ 8 }{ 5 } \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} \frac{ 3 }{ 5 }& \frac{ -17 }{ 5 } \\ \frac{ -16 }{ 5 } & \frac{ 12 }{ 5 } \end{bmatrix}\) and B = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 18 }{ 5 } \\ \frac{ 19 }{ 5 } & \frac{ 8 }{ 5 } \end{bmatrix}\)

 

Q-8: What will be the value of A if,

B = \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\) and 2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)

Solution:

2A + B = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\)

2A = \(\begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\)

2A = \(\begin{bmatrix} 2 – 4 & 1 – 3 \\ -4 – 2 & 3 – 5 \end{bmatrix}\)

2A = \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)

A = \(\frac{ 1 }{ 2 }\) \(\begin{bmatrix} -2 & -2 \\ -6 & -2 \end{bmatrix}\)

A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\)

Hence, A = \(\begin{bmatrix} -1 & -1 \\ -3 & -1 \end{bmatrix}\).

 

 

Q-9: Calculate the value of a and b, if

\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\).

Solution:

\(2 \begin{bmatrix} 2 & 4 \\ 1 & a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \( \begin{bmatrix} 4 & 8 \\ 2 & 2a \end{bmatrix} + \begin{bmatrix} b & 1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} 4 + b & 8 + 1 \\ 2 + 2 & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

⟹ \(\begin{bmatrix} 4 + b & 9 \\ 4 & 2a + 2 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 2 & 9 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

4 + b = 6

b = 2

2a + 2 = 9

2a = 7

a = \(\frac{ 7 }{ 2 }\)

Hence, a = \(\frac{ 7 }{ 2 }\) and b = 2.

 

Q-10: Calculate the value of a, b, c and d, if

\(2 \begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} = 3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\).

Solution:

\(2\begin{bmatrix} a & b \\ c & d \end{bmatrix} + 3 \begin{bmatrix} 2 & -2 \\ 0 & 3 \end{bmatrix} = 3 \begin{bmatrix} 4 & 6 \\ 5 & 7 \end{bmatrix}\)

\(\begin{bmatrix} 2a & 2b \\ 2c & 2d \end{bmatrix} + \begin{bmatrix} 6 & -6 \\ 0 & 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)

\(\begin{bmatrix} 2a + 6 & 2b – 6 \\ 2c + 0 & 2d + 9 \end{bmatrix} = \begin{bmatrix} 12 & 18 \\ 15 & 21 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

2a + 6 = 12

⟹ 2a = 6

a = 3

2b – 6 = 18

⟹ 2b = 24

b = 12

2c + 0 = 15

⟹ 2c = 15

c = \(\frac{ 15 }{ 2 }\)

2d + 9 = 21

⟹ 2d = 12

d = 6

Hence, a = 3, b = 12, c = \(\frac{ 15 }{ 2 }\) and d = 6.

 

 

Q-11: What will be the value of x and y in the matrices given below:

\(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)?

Solution:

\(x \begin{bmatrix} 3 \\ 4 \end{bmatrix} + y \begin{bmatrix} -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)

⟹ \(\\\begin{bmatrix} 3x \\ 4x \end{bmatrix} + \begin{bmatrix} -2y \\ 2y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\\\)

⟹ \(\\\begin{bmatrix} 3x – 2y \\ 4x + 2y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

3x – 2y = 10………….. (i)

4x + 2y = 5 ……………(ii)

Adding equation (i) and (ii), we will get:

7x = 15

x = \(\frac{ 15 }{ 7 }\)

Putting the value of x in equation (i), we will get:

3 × \(\frac{ 15 }{ 7 }\) – 2y = 10

⟹ \(\frac{ 45 }{ 7 }\) – 2y = 10

⟹ 2y = \(\frac{ 45 }{ 7 }\) – 10

⟹ 2y = \(\frac{ 45 – 70 }{ 7 }\)

⟹ 2y = \(\frac{ -25 }{ 7 }\)

⟹ y = \(\frac{ -25 }{ 14 }\)

Hence, x = \(\frac{ 15 }{ 7 }\) and, y = \(\frac{ -25 }{ 14 }\)

 

 

Q-12: What will be the value of a, b, c and d in the following matrix

\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)?

 

Solution:

\(3\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a & 7 \\ -2 & 2d \end{bmatrix} + \begin{bmatrix} 5 & a + b \\ c + d & 4 \end{bmatrix}\)

⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5 & 7 + a + b \\ -2 + c + d & 2d + 4 \end{bmatrix}\\\\\)

⟹ \( \begin{bmatrix} 3a & 3b \\ 3c & 3d \end{bmatrix} = \begin{bmatrix} a + 5 & a + b + 7 \\ c + d – 2 & 2d + 4 \end{bmatrix}\)

By comparing the corresponding elements of the above two matrices, we will get:

3a = a + 5

2a = 5

a = \(\frac{ 5 }{ 2 }\)

3b = a + b + 7

⟹ 3b = \(\frac{ 5 }{ 2 }\) + b + 7

⟹ 2b = \(\frac{ 5 }{ 2 }\) + 7

⟹ 2b = \(\frac{ 5 + 14 }{ 2 }\)

⟹ 2b = \(\frac{ 19 }{ 2 }\)

b = \(\frac{ 19 }{ 4 }\)

3d = 2d + 4

d = 4

3c = c + d – 2

⟹ 2c = 4 – 2

⟹ 2c = 2

c = 1

Hence, a = \(\frac{ 5 }{ 2 }\), b = \(\frac{ 19 }{ 4 }\), c = 1 and d = 4.

 

 

Q-13:If F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), prove that F(a) F(b) = F( a + b ).

Solution:

F(a) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\), F(b) = \(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

F(a + b) = \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

F(a) F(b) = \(\begin{bmatrix} cos a & -sin a & 0 \\ sin a & cos a & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\begin{bmatrix} cos b & -sin b & 0 \\ sin b & cos b & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

= \(\begin{bmatrix} \left ( cos a \;cos b – sin a \;sin b + 0 \right ) & \left ( -cos a\; sin b – sin a \;cos b + 0 \right ) & 0 \\ \left ( sin a \;cos b + cos a \;sin b + 0 \right ) & -sin a \;sin b + cos a\; cos b + 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

= \(\begin{bmatrix} cos \left ( a + b \right ) & -sin \left ( a + b \right ) & 0 \\ sin \left ( a + b \right ) & cos \left ( a + b \right ) & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = F(a + b)

Hence, proved.

 

Q-14: Prove that:

(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\neq \begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\).

(ii) \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\neq \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

Solution:

(i) \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\)

= \(\\\begin{bmatrix} 6( 3 ) – 2( 4 ) & 6( 2 ) – 2( 5 ) \\ 7( 3 ) – 8( 4 ) & 7( 2 ) – 8( 5 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 18 – 8 & 12 – 10 \\ 21 – 24 & 14 – 40 \end{bmatrix}\)

= \(\\\begin{bmatrix} 10 & 2 \\ -3 & -26 \end{bmatrix}\)

\(\\\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\)

= \(\\\begin{bmatrix} 3( 6 ) + 2( 7 ) & 3( -2 ) + 2( 8 ) \\ 4( 6 ) + 5( 7 ) & 4( -2 ) + 5( 8 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 18 + 14 & -6 + 16 \\ 24 + 35 & -8 + 40 \end{bmatrix}\)

= \(\\\begin{bmatrix} 32 & 10 \\ 59 & 32 \end{bmatrix}\)

Therefore, \(\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}\begin{bmatrix} 6 & -2 \\ 7 & 8 \end{bmatrix}\)

Hence, proved.

 

(ii) \(\\\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\)

= \(\\\begin{bmatrix} 2( -2 ) + 3( 0 ) + 4( 3 ) & 2( 1 ) + 3( -2 ) + 4( 4 ) & 2( 0 ) + 3( 1 ) + 4( 5 ) \\ 0( -2 ) + 1( 0 ) + 0( 3 ) & 0( 1 ) + 1( -2 ) + 0( 4 ) & 0( 0 ) + 1( 1 ) + 0( 5 ) \\ 1( -2 ) + 1( 0 ) + 0( 3 ) & 1( 1 ) + 1( -2 ) + 0( 4 ) & 1( 0 ) + 1( 1 ) + 0( 5 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} -4 + 0 + 12 & 2 – 6 + 16 & 0 + 3 + 20 \\ 0 + 0 + 0 & 0 – 2 + 0 & 0 + 1 + 0 \\ -2 + 0 + 0 & 1 – 2 + 0 & 0 + 1 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 8 & 12 & 23 \\ 0 & – 2 & 1 \\ -2 & -1 & 1 \end{bmatrix}\)

\(\\\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} -2( 2 ) + 1( 0 ) + 0( 1 ) & -2( 3 ) + 1( 1 ) + 0( 1 ) & -2( 4 ) + 1( 0 ) + 0( 0 ) \\ 0( 2 ) – 2( 0 ) + 1( 1 ) & 0( 3 ) – 2( 1 ) + 1( 1 ) & 0( 4 ) – 2( 0 ) + 1( 0 ) \\ 3( 2 ) + 4( 0 ) + 5( 1 ) & 3( 3 ) + 4( 1 ) + 5( 1 ) & 3( 4 ) + 4( 0 ) + 5( 0 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} -4 + 0 + 0 & -6 + 1 + 0 & -8 + 0 + 0 \\ 0 + 0 + 1 & 0 – 2 + 1 & 0 – 0 + 0 \\ 6 + 0 + 5 & 9 + 4 + 5 & 12 + 0 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} -4 & -5 & -8 \\ 1 & – 1 & 0 \\ 11 & 18 & 12 \end{bmatrix}\)

Therefore, \(\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\) ≠ \(\begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 3 & 4 & 5 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)

Hence, proved.

 

Q-15: What will be the value of A2 – 5A + 6I, if A = \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)?

Solution:

A2 = A × A

A2 = \(\\\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 3( 3 ) + 3( 3 ) + 1( 1 ) & 3( 0 ) + 3( 2 ) + 1( -1 ) & 3( 1 ) + 3( 4 ) + 1( 0 ) \\ 3( 3 ) + 2( 3 ) + 4( 1 ) & 3( 0 ) + 2( 2 ) + 4( -1 ) & 3( 1 ) + 2( 4 ) + 4( 0 ) \\ 1( 3 ) – 1( 3 ) + 0( 1 ) & 1( 0 ) – 1( 2 ) + 0( -1 ) & 1( 1 ) – 1( 4 ) + 0( 0 ) \end{bmatrix}\)

= \(\\\begin{bmatrix} 9 + 9 + 1 & 0 + 6 – 1 & 3 + 12 + 0 \\ 9 + 6 + 4 & 0 + 4 – 4 & 3 + 8 + 0 \\ 3 – 3 + 0 & 0 – 2 + 0 & 1 – 4 + 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 19 & 5 & 15 \\ 19 & 0 & 11 \\ 0 & – 2 & -3 \end{bmatrix}\)

 

5A = 5\(\begin{bmatrix} 3 & 0 & 1 \\ 3 & 2 & 4 \\ 1 & -1 & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 15 & 0 & 5 \\ 15 & 10 & 20 \\ 5 & -5 & 0 \end{bmatrix}\)

 

6I = 6 \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

= \(\\\begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\)

A2 – 5A + 6I = \(\begin{bmatrix} 19 & 5 & 15 \\ 19 & 0 & 11 \\ 0 & – 2 & -3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 0 & 5 \\ 15 & 10 & 20 \\ 5 & -5 & 0 \end{bmatrix}\) + \(\begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}\)

= \(\\\begin{bmatrix} 19 + 15 + 6 & 5 + 0 + 0 & 15 + 5 + 0 \\ 19 + 15 + 0 & 0 + 10 + 6 & 11 + 20 + 0 \\ 0 + 5 + 0 & -2 -5 + 0 & -3 + 0 + 6 \end{bmatrix}\)

= \(\\\begin{bmatrix} 40 & 5 & 20 \\ 34 & 16 & 31 \\ 5 & -7 & 3 \end{bmatrix}\)

Hence,

A2 – 5A + 6I = \(\begin{bmatrix} 40 & 5 & 20 \\ 34 & 16 & 31 \\ 5 & -7 & 3 \end{bmatrix}\)

 

 

Q-16: If A = \(\begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix}\) and the identity matrix of order 2 be represented by I. Prove that:

I + A = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

Solution:

LHS = I + A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) + \(\begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\) ……… (i)

 

RHS = ( I – A ) \(\begin{bmatrix} cos \beta & -sin \beta \\ sin \beta & cos \beta \end{bmatrix}\)

= \(\\\left ( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} – \begin{bmatrix} 0 & -tan\frac{ \beta }{ 2 } \\ tan\frac{ \beta }{ 2 } & 0 \end{bmatrix} \right ) \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & tan\frac{ \beta }{ 2 } \\ -tan\frac{ \beta }{ 2 } & 1 \end{bmatrix} \;\; \begin{bmatrix} cos \beta & – sin\beta \\ sin\beta & cos\beta \end{bmatrix}\)

= \(\\\begin{bmatrix} cos \beta + sin \beta tan\frac{ \beta }{ 2 } & -sin \beta + cos \beta tan\frac{ \beta }{ 2 } \\ -cos \beta tan\frac{ \beta }{ 2 } + sin \beta tan\frac{ \beta }{ 2 } & sin \beta + cos \beta \end{bmatrix}\) ………..(ii)

= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + \left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } \\ -\left ( 2cos^{ 2 }\frac{ \beta }{ 2 } – 1 \right )tan\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } & 2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 }tan\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } + 2sin^{ 2 }\frac{ \beta }{ 2 } & -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } – tan\frac{ \beta }{ 2 } \\ -2sin\frac{ \beta }{ 2 }cos\frac{ \beta }{ 2 } + 2sin\frac{ \beta }{ 2 }cos\frac{ \beta } + tan\frac{ \beta }{ 2 } & 2sin^{ 2 }\frac{ \beta }{ 2 } + 1 – 2sin^{ 2 }\frac{ \beta }{ 2 } \end{bmatrix}\)

= \(\\\begin{bmatrix} 1 & -tan\frac{ \alpha }{ 2 } \\ tan\frac{ \alpha }{ 2 } & 1 \end{bmatrix}\)

= LHS

Hence, proved.

 

 

Q-17: There is the fund of Rs 30,000 from a trust which should be invested in any of the two different types of bonds. The first bond will pay 5% interest every year, and the second bond will pay 7% interest every year. Using the principle of matrix multiplication, determine how Rs 30,000 will be divided between the two different types of bonds, if the trust fund will obtain an annual total interest of:

(a) Rs 2,000 (b) Rs 1,800

 

Solution:

(a) Let, in the first bond, Rs. q is invested.

The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).

As per the data given in the question, the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.

Hence, in order to have an annual interest of Rs. 2000, we have

S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)

⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 2000\)

⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 2000\)

⟹ 5q + 7( 30000 – q ) = 200000

⟹ 5q + 210000 – 7q = 200000

⟹ 210000 – 200000 = 7q – 5q

⟹ 2q = 10000

q = 5000

In order to have an annual interest of Rs. 2000, the trusty providing fund must invest Rs. 5000 in the first bond and remaining Rs. 25000 in the second bond.

 

(b) Let, in the first bond, Rs. q is invested.

The sum of the total money thus invested in the second bond is Rs. ( 30000 – q ).

As per the data given in the question, the first bond needs to pay 5% interest every year and the second bond needs to pay 7% interest every year.

Hence, in order to have an annual interest of Rs. 1800, we have

S.I for 1 year = \(\frac{ Principle \times Rate \times 1 }{ 100 }\)

⟹ \(\begin{bmatrix} q & \left ( 30000 – q \right ) \end{bmatrix}\begin{bmatrix} \frac{ 5 }{ 100 } \\ \frac{ 7 }{ 100 } \end{bmatrix} = 1800\)

⟹ \(\frac{ 5q }{ 100 } + \frac{ 7 \left ( 30000 – q \right ) }{ 100 } = 1800\)

⟹ 5q + 7( 30000 – q ) = 180000

⟹ 5q + 210000 – 7q = 180000

⟹ 210000 – 180000 = 7q – 5q

⟹ 2q = 30000

q = 15000

In order to have an annual interest of Rs. 1800, the trusty providing fund must invest Rs. 15000 in the first bond and remaining Rs. 15000 in the second bond.

Q-18: A particular school has the bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books. Their selling prices are Rs 60, Rs 40 and Rs 80 each, respectively. What will be the total amount of the bookshop which will be received by selling all of the books using the matrix algebra?

Solution:

The bookshop of 8 dozen of the physics books, 10 dozen of the economics books, 10 dozen of the chemistry books.

The selling prices for the physics, chemistry and the economics book are Rs. 60, Rs. 40 and Rs. 80, respectively.

The total amount of the bookshop received by selling all of the books by using the matrix algebra will be represented in the form given below:

\(12\begin{bmatrix} 8 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 40 \\ 80 \end{bmatrix}\)

= 12 [8 × 60 + 10 × 40 + 10 × 80]

= 12 [480 + 400 + 800]

= 12 [1680]

= 20160

Hence, the bookshop will get Rs. 20160 by the selling of all of the books.

 

 

Q-19: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. Then, the restriction on m, n and o such that, TQ + SQ is defined as:

  1. n = 3, o = m
  2. n is arbitrary, o = 2
  3. o is arbitrary, n = 3
  4. n = 2, o = 3

Solution:

Matrices T and Q are having the orders o × n and 3 × n, respectively.

Thus, hence formed matrix TQ will be defined if n = 3. Consequently, TQ matrix will be have the order o × n. Matrices S and Q are having the orders of m × 3 and 3 × n, respectively.

As the number of columns in the matrix S is equal to the number of rows in the matrix Q, so, the matrix SQ hence is well-defined and will be of the order m × n. Matrices PY and WY will be added at the situation, when their orders will be the same.

Since, the matrix TQ has the order o × n and the matrix SQ has the order m × n. Hence, we will have o = m, surely.

Therefore, n = 3 and o = m are the restrictions on m, n and o, such that TQ + SQ will be defined.

 

 

Q-20: Let us consider P, Q, R, S and T be the matrices with order 2 × m, 3 × n, 2 × o, m× 3 and o× n, respectively. If m = o, then find the order for the matrix 7P – 5R.

(a). o × 2

(b). 2 × m

(c). m × 3

(d). o × n

Solution:

The matrix P has the order 2 × m.

Thus, the matrix 7P will also have the same order. The matrix R is of the order 2 × o, i.e., 2 × m [As m = o]

Thus, the matrix 5R will also have the same order.

Now, both of the matrices 7P and 5R are of the order 2 × m.

Thus, the matrix 7P – 5R is well-defined and will have the order 2 × m.