Compute the indicated products left[begin{arr | vedclass

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

Question

(A) To compute the product of the two matrices,we multiply the rows of the first matrix by the columns of the second matrix:$\left[\begin{array}{lll}2

Vedclass · Accepted Answer

$\left[\begin{array}{ccc}14 & 0 & 42 \ 18 & -1 & 56 \ 22 & -2 & 70\end{array}ight]$

Vedclass · Answer

(A) To compute the product of the two matrices,we multiply the rows of the first matrix by the columns of the second matrix:$\left[\begin{array}{lll}2 & 3 & 4 \ 3 & 4 & 5 \ 4 & 5 & 6\end{array}ight]\left[\begin{array}{ccc}1 & -3 & 5 \ 0 & 2 & 4 \ 3 & 0 & 5\end{array}ight]$$= \left[\begin{array}{lll}2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)\end{array}ight]$$= \left[\begin{array}{lll}2+0+12 & -6+6+0 & 10+12+20 \ 3+0+15 & -9+8+0 & 15+16+25 \ 4+0+18 & -12+10+0 & 20+20+30\end{array}ight]$$= \left[\begin{array}{ccc}14 & 0 & 42 \ 18 & -1 & 56 \ 22 & -2 & 70\end{array}ight]$

Compute the indicated products $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

Similar Questions

If $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$ and $A^2 + xA + yI_2 = O_2$, where $I_2$ and $O_2$ are the identity matrix and null matrix of order $2$ respectively, then:

If $A = \text{diag}(2, -1, 3)$ and $B = \text{diag}(-1, 3, 2)$,then $A^2B = $

If $\begin{bmatrix} 2 & -3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} a & c \\ b & d \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 2 & -5 \end{bmatrix}$,then $(a, b, c, d) = $

If $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,then $(BB^TA)^5$ is equal to

If $A$ and $B$ are square matrices of the same order such that $AB = BA$,then prove by induction that $AB^{n} = B^{n}A$. Further,prove that $(AB)^{n} = A^{n}B^{n}$ for all $n \in N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module