Q. E=[−1034] and B=[0552−22]Let H=EB. Find H.H=
Define Matrices E and B: Define the matrices E and B.Matrix E is given as:E=[−1amp;30amp;4]Matrix B is given as:B=[0amp;5amp;−25amp;2amp;2]
Multiply Matrices E and B: Multiply matrix E by matrix B to find matrix H.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.The element in the first row and first column of matrix H (H[1,1]) is calculated as:H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]H[1,1]=(−1)∗0+3∗5H[1,1]=0+15H[1,1]=15The element in the first row and second column of matrix H (H[1,2]) is calculated as:H[1,2]=E[1,1]∗B[1,2]+E[1,2]∗B[2,2]H[1,2]=(−1)∗5+3∗2H[1,2]=−5+6H[1,2]=1The 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:H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]1H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]2H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]3H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]4The 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:H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]6H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]7H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]8H[1,1]=E[1,1]∗B[1,1]+E[1,2]∗B[2,1]9The element in the second row and second column of matrix H (H[1,1]=(−1)∗0+3∗50) is calculated as:H[1,1]=(−1)∗0+3∗51H[1,1]=(−1)∗0+3∗52H[1,1]=(−1)∗0+3∗53H[1,1]=(−1)∗0+3∗54The element in the second row and third column of matrix H (H[1,1]=(−1)∗0+3∗55) is calculated as:H[1,1]=(−1)∗0+3∗56H[1,1]=(−1)∗0+3∗57H[1,1]=(−1)∗0+3∗58$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: \(H = \left[\begin{array}{ccc} 15 & 1 & 8 \ 20 & 8 & 8 \end{array}\right]\)
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