Construct a 3 times 4 matrix,whose elements a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In general,a $3 	imes 4$ matrix is given by $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \ a_{21} & a_{22} & a_{23} & a_{24} \ a_{31} &

Vedclass · Accepted Answer

$A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \ \frac{5}{2} & 2 & \frac{3}{2} & 1 \ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$

Vedclass · Answer

(A) In general,a $3 	imes 4$ matrix is given by $A=\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \ a_{21} & a_{22} & a_{23} & a_{24} \ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$.Given: $a_{ij}=\frac{1}{2}|-3i+j|$,where $i=1, 2, 3$ and $j=1, 2, 3, 4$.Calculating the elements:$a_{11} = \frac{1}{2}|-3(1)+1| = \frac{1}{2}|-2| = 1$$a_{12} = \frac{1}{2}|-3(1)+2| = \frac{1}{2}|-1| = \frac{1}{2}$$a_{13} = \frac{1}{2}|-3(1)+3| = \frac{1}{2}|0| = 0$$a_{14} = \frac{1}{2}|-3(1)+4| = \frac{1}{2}|1| = \frac{1}{2}$$a_{21} = \frac{1}{2}|-3(2)+1| = \frac{1}{2}|-5| = \frac{5}{2}$$a_{22} = \frac{1}{2}|-3(2)+2| = \frac{1}{2}|-4| = 2$$a_{23} = \frac{1}{2}|-3(2)+3| = \frac{1}{2}|-3| = \frac{3}{2}$$a_{24} = \frac{1}{2}|-3(2)+4| = \frac{1}{2}|-2| = 1$$a_{31} = \frac{1}{2}|-3(3)+1| = \frac{1}{2}|-8| = 4$$a_{32} = \frac{1}{2}|-3(3)+2| = \frac{1}{2}|-7| = \frac{7}{2}$$a_{33} = \frac{1}{2}|-3(3)+3| = \frac{1}{2}|-6| = 3$$a_{34} = \frac{1}{2}|-3(3)+4| = \frac{1}{2}|-5| = \frac{5}{2}$Thus,the matrix is $A=\begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \ \frac{5}{2} & 2 & \frac{3}{2} & 1 \ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}$.

Construct a $3 \times 4$ matrix,whose elements are given by $a_{i j}=\frac{1}{2}|-3 i+j|$.

Similar Questions

If $A$ is a $3 \times 4$ matrix and $B$ is a matrix such that $A^{\prime}B$ and $BA^{\prime}$ are both defined,then $B$ is of the type:

For matrix $A = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$, if $A^2 - 2I = KA$, then $K = \dots$

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $hA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$, then the values of $h, a, b$ are respectively

If $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$,then prove that $A^{n} = \begin{bmatrix} \cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta \end{bmatrix}$ for all $n \in N$.

Find the value of $x, y$ and $z$ from the following equation : $\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module