Class 12Mathematics3 and 4 .Determinants and MatricesSpecial types of matrices, Transpose of matrices and Trace of matrices
MediumFor the matrices $A$ and $B$,verify that $(AB)^{\prime} = B^{\prime} A^{\prime}$ where $A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$.
(A) First,we calculate the product $AB$:
$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 \times 1 & 0 \times 5 & 0 \times 7 \\ 1 \times 1 & 1 \times 5 & 1 \times 7 \\ 2 \times 1 & 2 \times 5 & 2 \times 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$
Now,we find the transpose $(AB)^{\prime}$ by interchanging rows and columns:
$(AB)^{\prime} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Next,we find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$
Now,we calculate the product $B^{\prime} A^{\prime}$:
$B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime} A^{\prime}$,the property is verified.
$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 \times 1 & 0 \times 5 & 0 \times 7 \\ 1 \times 1 & 1 \times 5 & 1 \times 7 \\ 2 \times 1 & 2 \times 5 & 2 \times 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$
Now,we find the transpose $(AB)^{\prime}$ by interchanging rows and columns:
$(AB)^{\prime} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Next,we find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$
Now,we calculate the product $B^{\prime} A^{\prime}$:
$B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime} A^{\prime}$,the property is verified.
Explore More
Similar Questions
If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then $A' = $ . . . . . . .
If $A$ is a skew-symmetric matrix,then (given $n \in N$):
$1$. $A^{2n}$ is a skew-symmetric matrix.
$2$. $A^{2n+1}$ is a skew-symmetric matrix.
$1$. $A^{2n}$ is a skew-symmetric matrix.
$2$. $A^{2n+1}$ is a skew-symmetric matrix.
If $A$ and $B$ are square matrices of the same order and $B$ is a skew-symmetric matrix,then $A^{\prime} B A$ is
MediumKCET 2020
View SolutionIf $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$,verify that $(A')' = A$.
Easy
View SolutionIf $A$ is a symmetric matrix and $B$ is a skew-symmetric matrix such that $A + B = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}$,then $AB$ is equal to
DifficultJEE MAIN 2019
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo