Ncert Solutions For Class 12 Maths Ex 3.4

Ncert Solutions For Class 12 Maths Chapter 3 Ex 3.4

Q-1: What will be the inverse of the matrix \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}\)M [ R2 → R2 – 2 R1 ]

⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)M [ R2 → \(\frac{ 1 }{ 5 } \) R2]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)M [ R2 → R1 + R2 ]

Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)

 

 

Q-2: What will be the inverse of the matrix \(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2 & -6 \\ 1 & -3 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}\)M [ C2 → C2 + 3 C1 ]

⟹ \(\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 3 \\ -1 & 1 \end{bmatrix}\)M [ C1 → C1 – C2]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 3 \\ \frac{ -1 }{ 2 } & 1 \end{bmatrix}\)M [ C1 → \(\frac{ 1 }{ 2 }\) C1]

Hence, M-1 = \(\begin{bmatrix} \frac{ 3 }{ 5 } & \frac{ 1 }{ 5 } \\ \frac{ -2 }{ 5 } & \frac{ 1 }{ 5 } \end{bmatrix}\)

 

 

Q-3: What will be the inverse of the matrix \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 } & 0 \\ 0 & 1 \end{bmatrix}\)M [ R1 → \(\frac{ 1 }{ 6 }\) R1 ]

⟹ \(\begin{bmatrix} 1 & \frac{ -1 }{ 2 } \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{ 1 }{ 6 } & 0 \\ \frac{ 1 }{ 3 } & 1 \end{bmatrix}\)M [ R2 → R2 +2R1]

Here, in the above matrix in LHS side, there is only zero in the second row.

Hence, M-1 does not exist.

 

 

Q-4: What will be the inverse of the matrix \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 + R2 ]

⟹ \(\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}\)M [ R2 → R2 + R1]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)M [ R2 → R1 + R2]

Hence, M-1 = \( \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}\)

 

 

Q-5: What will be the inverse of the matrix \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 0 & 0 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 1 & – \frac{ 1 }{ 2 } \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 – \(\frac{ 1 }{ 6 }\) R2 ]

Here, in the above matrix in LHS side, there is only zero in the second row.

Hence, M-1 does not exist.

 

 

Q-6: What will be the inverse of the matrix \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)

As we know, M = IM

⟹ \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)M

⟹ \(\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}\)M [ R1 → R1 – R2 ]

⟹ \(\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}\)M [ R2 → R2 – 2R1]

⟹ \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)M [ R1 → R1 – 3 R2]

Hence, M-1 = \( \begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)

 

 

Q-7: What will be the inverse of the matrix \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\), if any exists?

Solution:

Let, M = \(\begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\)

As we know, M = IM

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

⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M [ R1 → \(\frac{ 1 }{ 2 }\)R1 ]

⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & 3 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\)M [ R2 → R2 – 5 R1]

⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & \frac{ 1 }{ 2 } \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ \frac{ 5 }{ 2 } & -1 & 1 \end{bmatrix}\)M [ R3 → R3 – R2]

⟹ \(\begin{bmatrix} 1 & 0 & \frac{ -1 }{ 2 } \\ 0 & 1 & \frac{ 5 }{ 2 } \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} \frac{ 1 }{ 2 } & 0 & 0 \\ \frac{ -5 }{ 2 } & 1 & 0 \\ 5 & -2 & 2 \end{bmatrix}\)M [ R3 → 2R3]

⟹ \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)M [ R1 → R1 + \(\frac{ 1 }{ 2 }\) R3, and R2 → R2 – \(\frac{ 5 }{ 2 }\) R3]

Hence, M-1 = \(\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\)

 

 

Q-8: The matrices M and N can be inverse of each other if and only if

(a) MN = NM (b) MN = NM = 0

(c) MN = 0, NM = I (d) MN = NM = I

 

Solution:

As we know, if M is a square matrix of order a, and if there exists another square matrix N of the same order b, in such a manner that MN = NM = I, then N will be said to be the inverse of M. So, in this case, we can see that M is the inverse of N.

Therefore, matrices M and N will be inverses of each other only if MN = NM = I.