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:

B=[[5,-2],[-1,5]] and A=[[2,0],[0,2]] Let H=BA. Find H. H=[quad]

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}=

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. 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}=
  1. Understand matrix multiplication: Understand matrix multiplication. To multiply two matrices, we take the dot product of the rows of the first matrix with the columns of 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. Write down given matrices: Write down the given matrices. Matrix BB is given as B=[5−2 −15]B=\left[\begin{array}{cc}5 & -2 \ -1 & 5\end{array}\right] and matrix AA is given as A=[20 02]A=\left[\begin{array}{cc}2 & 0 \ 0 & 2\end{array}\right].
  3. Multiply first row: Multiply the first row of BB by the first column of AA. The first row of BB is [5,−2][5, -2] and the first column of AA is [2,0][2, 0]. The dot product is (5×2)+(−2×0)=10+0=10(5 \times 2) + (-2 \times 0) = 10 + 0 = 10.
  4. Multiply second row: Multiply the first row of BB by the second column of AA. The first row of BB is [5,−2][5, -2] and the second column of AA is [0,2][0, 2]. The dot product is (5×0)+(−2×2)=0−4=−4(5\times0) + (-2\times2) = 0 - 4 = -4.
  5. Combine results: Multiply the second row of BB by the first column of AA. The second row of BB is [−1,5][-1, 5] and the first column of AA is [2,0][2, 0]. The dot product is (−1×2)+(5×0)=−2+0=−2(-1 \times 2) + (5 \times 0) = -2 + 0 = -2.
  6. Combine results: Multiply the second row of BB by the first column of AA. The second row of BB is [−1,5][-1, 5] and the first column of AA is [2,0][2, 0]. The dot product is (−1×2)+(5×0)=−2+0=−2(-1\times2) + (5\times0) = -2 + 0 = -2.Multiply the second row of BB by the second column of AA. The second row of BB is [−1,5][-1, 5] and the second column of AA is AA22. The dot product is AA33.
  7. Combine results: Multiply the second row of BB by the first column of AA. The second row of BB is [−1,5][-1, 5] and the first column of AA is [2,0][2, 0]. The dot product is (−1×2)+(5×0)=−2+0=−2(-1\times2) + (5\times0) = -2 + 0 = -2. Multiply the second row of BB by the second column of AA. The second row of BB is [−1,5][-1, 5] and the second column of AA is AA22. The dot product is AA33. Combine the results to form the matrix AA44. The results from steps 33 to 66 give us the matrix AA44: AA66

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 4 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 4 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 4 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 4 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 4 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 −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=[0−12] \mathrm{B}=\left[\begin{array}{r}0 \\ -1 \\ 2\end{array}\right] and A=[12] \mathrm{A}=\left[\begin{array}{ll}1 & 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

Question
e2xe^{2x}
Get tutor helpright-arrow

Posted 9 hours ago

View all (19)