Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

{:[E=[[-1,3],[0,4]]" and "],[B=[[0,5,-2],[5,2,2]]]:}
Let
H=EB. Find
H.

H=[]

$\begin{array}{l} \mathrm{E}=\left[\begin{array}{rr} -1 & 3 \\ 0 & 4 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{llr} 0 & 5 & -2 \\ 5 & 2 & 2 \end{array}\right] \end{array}$ $\newline$ Let $\mathrm{H}=\mathrm{EB}$ . Find $\mathrm{H}$ . $\newline$ $\mathbf{H}=$

View step-by-step help

View step-by-step help

Multiply two decimals: where does the decimal point go?

Full solution

Q.

\begin{array}{l} \mathrm{E}=\left[\begin{array}{rr} -1 & 3 \\ 0 & 4 \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{llr} 0 & 5 & -2 \\ 5 & 2 & 2 \end{array}\right] \end{array}

\newline

Let

\mathrm{H}=\mathrm{EB}

. Find

\mathrm{H}

.

\newline

\mathbf{H}=

Define Matrices $E$ and $B$ : Define the matrices $E$ and $B$ . $\newline$ Matrix $E$ is given as: $\newline$ $E = \left[\begin{array}{cc}-1 & 3\ 0 & 4\end{array}\right]$ $\newline$ Matrix $B$ is given as: $\newline$ $B = \left[\begin{array}{ccc}0 & 5 & -2\ 5 & 2 & 2\end{array}\right]$
Multiply Matrices E and B: Multiply matrix E by matrix B to find matrix H. $\newline$ To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of the second matrix. The product matrix H will have the same number of rows as the first matrix E and the same number of columns as the second matrix B. $\newline$ The element in the first row and first column of matrix H ( $H[1,1]$ ) is calculated as: $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $\newline$ $H[1,1] = 0 + 15$ $\newline$ $H[1,1] = 15$ $\newline$ The element in the first row and second column of matrix H ( $H[1,2]$ ) is calculated as: $\newline$ $H[1,2] = E[1,1]*B[1,2] + E[1,2]*B[2,2]$ $\newline$ $H[1,2] = (-1)*5 + 3*2$ $\newline$ $H[1,2] = -5 + 6$ $\newline$ $H[1,2] = 1$ $\newline$ The element in the first row and third column of matrix H ( $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $0$ ) is calculated as: $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $1$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $2$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $3$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $4$ $\newline$ The element in the second row and first column of matrix H ( $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $5$ ) is calculated as: $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $6$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $7$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $8$ $\newline$ $H[1,1] = E[1,1]*B[1,1] + E[1,2]*B[2,1]$ $9$ $\newline$ The element in the second row and second column of matrix H ( $H[1,1] = (-1)*0 + 3*5$ $0$ ) is calculated as: $\newline$ $H[1,1] = (-1)*0 + 3*5$ $1$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $2$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $3$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $4$ $\newline$ The element in the second row and third column of matrix H ( $H[1,1] = (-1)*0 + 3*5$ $5$ ) is calculated as: $\newline$ $H[1,1] = (-1)*0 + 3*5$ $6$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $7$ $\newline$ $H[1,1] = (-1)*0 + 3*5$ $8$ $\newline$ $H[\(2\),\(3\)] = \(8\)
Combine Elements to Form Matrix H: Combine the calculated elements to form matrix H. Matrix H, which is the product of matrix E and matrix B, is: \(H = \left[\begin{array}{ccc} 15 & 1 & 8 \ 20 & 8 & 8 \end{array}\right]\)

More problems from Multiply two decimals: where does the decimal point go?

Question

Evaluate the following expression.

\newline

9+\frac{8^{3}}{2}=

Posted 4 months ago

Question

\newline

\frac{14-4}{2}=

Posted 4 months ago

Question

\begin{array}{l}g(r)=25-3 r \\ g(4)=\end{array}

Posted 4 months ago

Question

\begin{array}{l}f(x)=14-0.5 x \\ f(30)=\square\end{array}

Posted 4 months ago

Question

z

\newline

\begin{array}{l} \frac{4}{z}=\frac{12}{5} \\ z= \end{array}

Posted 3 months ago

Question

n

\newline

\begin{array}{l} \frac{n}{8}=\frac{11}{5} \\ n= \end{array}

Posted 3 months ago

Question

r

\newline

\begin{array}{l} \frac{4}{r}=\frac{5}{7} \\ r=\square \end{array}

Posted 3 months ago

Question

r

\newline

\begin{array}{l} \frac{r}{5}=\frac{4}{7} \\ r=\square \end{array}

Posted 3 months ago

Question

Reduce to simplest form.

\newline

\frac{5}{8}-\left(-\frac{7}{2}\right)=

Posted 3 months ago

Question

Reduce to simplest form.

\newline

-\frac{9}{12}-\left(-\frac{7}{4}\right)=

Posted 3 months ago