If A=left[begin{array}{cccc}2 & 1 & 3 & -1 1 | vedclass

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

Question

(D) Given the equation $2A + 3B - 5C = 0$.Rearranging for $C$,we get $5C = 2A + 3B$.Substitute the matrices $A$ and $B$:$5C = 2\left[\begin{array}{ccc

Vedclass · Accepted Answer

$\left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \ 1 & -7/5 & 2 & 3/5\end{array}ight]$

Vedclass · Answer

(D) Given the equation $2A + 3B - 5C = 0$.Rearranging for $C$,we get $5C = 2A + 3B$.Substitute the matrices $A$ and $B$:$5C = 2\left[\begin{array}{cccc}2 & 1 & 3 & -1 \ 1 & -2 & 2 & -3\end{array}ight] + 3\left[\begin{array}{cccc}2 & 1 & 0 & 3 \ 1 & -1 & 2 & 3\end{array}ight]$$5C = \left[\begin{array}{cccc}4 & 2 & 6 & -2 \ 2 & -4 & 4 & -6\end{array}ight] + \left[\begin{array}{cccc}6 & 3 & 0 & 9 \ 3 & -3 & 6 & 9\end{array}ight]$$5C = \left[\begin{array}{cccc}4+6 & 2+3 & 6+0 & -2+9 \ 2+3 & -4-3 & 4+6 & -6+9\end{array}ight]$$5C = \left[\begin{array}{cccc}10 & 5 & 6 & 7 \ 5 & -7 & 10 & 3\end{array}ight]$Dividing by $5$,we get $C = \left[\begin{array}{cccc}10/5 & 5/5 & 6/5 & 7/5 \ 5/5 & -7/5 & 10/5 & 3/5\end{array}ight] = \left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \ 1 & -7/5 & 2 & 3/5\end{array}ight]$.

If $A=\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right]$,$B=\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$ and $2A+3B-5C=0$,then $C=$

Similar Questions

If $A$ and $B$ are square matrices of the same order,then which of the following properties holds true?

If $A = \begin{bmatrix} 1 & 3 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 2 & 3 \\ -1 & 1 & 2 \end{bmatrix}$,then $AB =$

If $A = \begin{bmatrix} -1 & 0 \\ 0 & 2 \end{bmatrix}$,then $A^3 - A^2$ is equal to

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 $n \in N$.

The symmetric part of the matrix $A = \begin{bmatrix} 1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7 \end{bmatrix}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module