Question 1. Matrices A and B will be inverse of each other only if:
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0, BA = I
(D) AB = BA = I
Solution:
According to the definition of inverse of square matrix,
Option (D) is correct
i.e. AB=BA=I
Deleted Question
Using elementary transformations, find the inverse of each of the matrices, if it exists.
Solution:
Solution:
Let A=
[Tex]\begin{array}{l} \left(R_{1} \rightarrow R_{1}-R_{2}\right) \\ \left(R_{2} \rightarrow R_{2}-R_{1}\right) \end{array}[/Tex]
Solution:
Solution:
Solution:
Solution:
Using elementary transformation, find the inverse of each of the matrices, if it exists.
Solution:
Solution:
Solution:
Solution:
Solution:
}
Solution:
Here, both the elements of R2 of L.H.S. are 0.
Therefore, A-1 does not exist.
Solution:
Solution:
Here, both the elements of R2 of L.H.S. are 0.
Therefore, A-1 does not exist.
Solution:
Let A=
W.K.T. , A=IA
[Tex]\begin{array}{l} {\left[\mathrm{R}{1} \rightarrow \mathrm{R}{1}-\mathrm{R}_{3}\right]}\\\\ {\left[\mathrm{R}{1} \rightarrow(-1) \mathrm{R}{1}\right]} \end{array}[/Tex]
Therefore, A-1 =
Solution:
Let A=
W.K.T. , A=IA
[Tex]\begin{aligned} &\left[\begin{array}{lll} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 25 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -15 & 1 & 9 \end{array}\right] \text { A }\left[R_{1} \rightarrow R_{1}-3 R_{2}\right. \text { and } R_{3} \rightarrow R_{3}-9 R_{2}\\ &\left[\begin{array}{ccc} 1 & 0 & 10 \\ 0 & 1 & -4 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} -5 & 0 & 3 \\ 2 & 0 & -1 \\ -3 / 5 & 1 / 25 & 9 / 25 \end{array}\right] \text { A }\left[R_{1} \rightarrow \frac{1}{25} R_{3}\right]\\ &\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -\frac{2}{5} & -\frac{3}{5} \\ -\frac{2}{5} & \frac{4}{25} & \frac{11}{25} \\ -\frac{3}{5} & \frac{1}{25} & \frac{9}{25} \end{array}\right] A\left[R_{1} \rightarrow R_{1}-10 R_{1} \text { and } R_{2} \rightarrow R_{2}+4 R_{3}\right]\\ &Therefore,A^{-1}=\left[\begin{array}{ccc} 1 & -2 / 25 & -3 / 25 \\ -2 / 25 & 4 / 25 & 11 / 25 \\ -3 / 25 & 1 / 25 & 9 / 25 \end{array}\right]\\ \end{aligned}[/Tex]
Solution:
Let A=
W.K.T. , A=IA