Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

E=[[1,4,2],[1,2,2]] and B=[[1,3],[2,3],[2,-2]] Let H=EB. Find H. H=[quad]

E=[142122] E=\left[\begin{array}{lll}1 & 4 & 2 \\ 1 & 2 & 2\end{array}\right] and B=[13232−2] \mathrm{B}=\left[\begin{array}{rr}1 & 3 \\ 2 & 3 \\ 2 & -2\end{array}\right] \newlineLet H=EB \mathrm{H}=\mathrm{EB} . Find H \mathrm{H} .\newlineH= \mathbf{H}=

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Unions and intersections of setsright chevron icon

Full solution

Q. E=[142122] E=\left[\begin{array}{lll}1 & 4 & 2 \\ 1 & 2 & 2\end{array}\right] and B=[13232−2] \mathrm{B}=\left[\begin{array}{rr}1 & 3 \\ 2 & 3 \\ 2 & -2\end{array}\right] \newlineLet H=EB \mathrm{H}=\mathrm{EB} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
  1. Understand matrix multiplication: Understand matrix multiplication.\newlineTo multiply two matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix. The resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.
  2. Verify matrix multiplication possibility: Verify if matrix multiplication is possible.\newlineMatrix EE has dimensions 2×32 \times 3 (22 rows and 33 columns), and matrix BB has dimensions 3×23 \times 2 (33 rows and 22 columns). Since the number of columns in EE is equal to the number of rows in BB, matrix multiplication is possible.
  3. Set up the multiplication: Set up the multiplication.\newlineTo find HH, we need to multiply each row of EE by each column of BB and sum the products. The resulting matrix HH will have dimensions 2×22 \times 2.
  4. Calculate first element of matrix H: Calculate the first element of matrix H. Multiply the first row of EE by the first column of BB: H[1,1]=(1×1)+(4×2)+(2×2)=1+8+4=13H[1,1] = (1 \times 1) + (4 \times 2) + (2 \times 2) = 1 + 8 + 4 = 13
  5. Calculate second element first row H: Calculate the second element of the first row of matrix HH. Multiply the first row of EE by the second column of BB: H[1,2]=(1×3)+(4×3)+(2×(−2))=3+12−4=11H[1,2] = (1 \times 3) + (4 \times 3) + (2 \times (-2)) = 3 + 12 - 4 = 11
  6. Calculate first element second row HH: Calculate the first element of the second row of matrix HH. Multiply the second row of EE by the first column of BB: H[2,1]=(1×1)+(2×2)+(2×2)=1+4+4=9H[2,1] = (1 \times 1) + (2 \times 2) + (2 \times 2) = 1 + 4 + 4 = 9
  7. Calculate second element second row HH: Calculate the second element of the second row of matrix HH. Multiply the second row of EE by the second column of BB: H[2,2]=(1×3)+(2×3)+(2×(−2))=3+6−4=5H[2,2] = (1 \times 3) + (2 \times 3) + (2 \times (-2)) = 3 + 6 - 4 = 5
  8. Combine results form matrix \newlineHH: Combine the results to form matrix \newlineHH.\newline\newlineH = \left[\begin{array}{cc}\(\newline\)\(13\) & \(11\) (\newline\)\(9\) & \(5\)\(\newline\)\end{array}\right]\newline

More problems from Unions and intersections of sets

Question
z=−24−12.7i z=-24-12.7 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−24 and Im⁡(z)=−12.7 \begin{array}{l} \operatorname{Re}(z)=-24 \text { and } \\ \operatorname{Im}(z)=-12.7 \end{array} \newline(B)\newlineRe⁡(z)=−24 and Im⁡(z)=−12.7i \begin{array}{l} \operatorname{Re}(z)=-24 \text { and } \\ \operatorname{Im}(z)=-12.7 i \end{array} \newline(C)\newlineRe⁡(z)=−12.7 and Im⁡(z)=−24 \begin{array}{l} \operatorname{Re}(z)=-12.7 \text { and } \\ \operatorname{Im}(z)=-24 \end{array} \newline(D)\newlineRe⁡(z)=−12.7i and Im⁡(z)=−24 \begin{array}{l} \operatorname{Re}(z)=-12.7 i \text { and } \\ \operatorname{Im}(z)=-24 \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
z=77−27.8i z=77-27.8 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=77 and Im⁡(z)=−27.8 \begin{array}{l} \operatorname{Re}(z)=77 \text { and } \\ \operatorname{Im}(z)=-27.8 \end{array} \newline(B)\newlineRe⁡(z)=−27.8i and Im⁡(z)=77 \begin{array}{l} \operatorname{Re}(z)=-27.8 i \text { and } \\ \operatorname{Im}(z)=77 \end{array} \newline(C)\newlineRe⁡(z)=−27.8 and Im⁡(z)=77 \begin{array}{l} \operatorname{Re}(z)=-27.8 \text { and } \\ \operatorname{Im}(z)=77 \end{array} \newline(D)\newlineRe⁡(z)=77 and Im⁡(z)=−27.8i \begin{array}{l} \operatorname{Re}(z)=77 \text { and } \\ \operatorname{Im}(z)=-27.8 i \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
z=−16i−92.3 z=-16 i-92.3 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−92.3 and Im⁡(z)=−16i \begin{array}{l} \operatorname{Re}(z)=-92.3 \text { and } \\ \operatorname{Im}(z)=-16 i \end{array} \newline(B)\newlineRe⁡(z)=−92.3 and Im⁡(z)=−16 \begin{array}{l} \operatorname{Re}(z)=-92.3 \text { and } \\ \operatorname{Im}(z)=-16 \end{array} \newline(C)\newlineRe⁡(z)=−16i and Im⁡(z)=−92.3 \begin{array}{l} \operatorname{Re}(z)=-16 i \text { and } \\ \operatorname{Im}(z)=-92.3 \end{array} \newline(D)\newlineRe⁡(z)=−16 and Im⁡(z)=−92.3 \begin{array}{l} \operatorname{Re}(z)=-16 \text { and } \\ \operatorname{Im}(z)=-92.3 \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
x2+8x+6=0 x^{2}+8 x+6=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A)\newlinex=−8−10 and x=−8+10 \begin{array}{l} x=-8-\sqrt{10} \text { and } \\ x=-8+\sqrt{10} \end{array} \newline(B)\newlinex=−8−7 and x=−8+7 \begin{array}{l} x=-8-\sqrt{7} \text { and } \\ x=-8+\sqrt{7} \end{array} \newline(C)\newlinex=−4−10 and x=−4+10 \begin{array}{l} x=-4-\sqrt{10} \text { and } \\ x=-4+\sqrt{10} \end{array} \newline(D)\newlinex=−4−7 and x=−4+7 \begin{array}{l} x=-4-\sqrt{7} \text { and } \\ x=-4+\sqrt{7} \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
If x2+6x−5=0 x^{2}+6 x-5=0 , what are the values of x x ?\newlineChoose 11 answer:\newline(A) −3−14 -3-\sqrt{14} and −3+14 -3+\sqrt{14} \newline(B) 3−14 3-\sqrt{14} and 3+14 3+\sqrt{14} \newline(C) −1-1 and −5-5\newline(D) 11 and 55
Get tutor helpright-arrow

Posted 3 months ago

Question
Consider the expression 4(8x+5) 4(8 x+5) .\newlineComplete 22 descriptions of the parts of the expression.\newline11. The entire expression is the product of □\square. \newline22. On its own, (8x+5) (8 x+5) is a sum with □\square.
Get tutor helpright-arrow

Posted 3 months ago

Question
Evaluate f−g+(−2) f-g+(-2) where f=−3.005 f=-3.005 and g=4.7 g=4.7 .
Get tutor helpright-arrow

Posted 3 months ago

Question
Evaluate −a+(−b) -a+(-b) where a=6.05 a=6.05 and b=3.611 b=3.611 .
Get tutor helpright-arrow

Posted 3 months ago

Question
E=[−1−2−2−1] E=\left[\begin{array}{ll}-1 & -2 \\ -2 & -1\end{array}\right] and A=[202−1]. A=\left[\begin{array}{rr} 2 & 0 \\ 2 & -1 \end{array}\right] . \newlineLet H=EA \mathrm{H}=\mathrm{EA} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
Get tutor helpright-arrow

Posted 3 months ago

Question
B=[5−2−15] \mathbf{B}=\left[\begin{array}{rr}5 & -2 \\ -1 & 5\end{array}\right] and A=[2002] \mathrm{A}=\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right] .\newlineLet H=BA \mathrm{H}=\mathrm{BA} . Find H \mathrm{H} .\newlineH= \mathbf{H}=
Get tutor helpright-arrow

Posted 3 months ago

View all (20)