If$ A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 2 \\ 3 & -2 \end{bmatrix}$, whether AB is singular or non-singular matrix?

1.1kviews

written 5.2 years ago by

teamques10 ★ 68k

modified 2.6 years ago by

pedsangini276 • 4.8k

engineering mathematics

ADD COMMENT EDIT

1 Answer

84views

written 5.2 years ago by

teamques10 ★ 68k

• modified 5.2 years ago

Solution:

$AB = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} % \begin{bmatrix} 1 & 2 \\ 3 & -2 \end{bmatrix}$

$AB = \quad \begin{bmatrix} 2+3 & 4-2 \\ 0+9 & 0-6 \end{bmatrix}$

$AB = \quad \begin{bmatrix} 5 & 2 \\ 9 & -6 \end{bmatrix}$

$ |AB| = \Bigg| \begin{array}{ll}{2} & {1} \\ {0} & {3}\end{array} \Bigg|$ = -30-18 = -48

$|AB| \ne 0$

AB is non singular matrix

ADD COMMENT EDIT

Please log in to add an answer.