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E=[[-1,-2],[-2,-1]] and A=[[2,0],[2,-1]] Let H=EA. Find H. H=[◻

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}=

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Full solution

Q. 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}=
  1. Define Matrices E and A: Define the matrices E and A. Matrix E is given by E=[−1−2 −2−1]E = \begin{bmatrix} -1 & -2 \ -2 & -1 \end{bmatrix} and matrix A is given by A=[20 2−1]A = \begin{bmatrix} 2 & 0 \ 2 & -1 \end{bmatrix}.
  2. Multiply Row 11 by Column 11: Multiply the first row of EE by the first column of AA. To find the element at the first row and first column of matrix HH, we calculate (−1×2)+(−2×2)=−2−4=−6(-1 \times 2) + (-2 \times 2) = -2 - 4 = -6.
  3. Multiply Row 11 by Column 22: Multiply the first row of EE by the second column of AA. To find the element at the first row and second column of matrix HH, we calculate (−1×0)+(−2×−1)=0+2=2(-1 \times 0) + (-2 \times -1) = 0 + 2 = 2.
  4. Multiply Row 22 by Column 11: Multiply the second row of EE by the first column of AA. To find the element at the second row and first column of matrix HH, we calculate (−2×2)+(−1×2)=−4−2=−6(-2 \times 2) + (-1 \times 2) = -4 - 2 = -6.
  5. Multiply Row 22 by Column 22: Multiply the second row of EE by the second column of AA. To find the element at the second row and second column of matrix HH, we calculate (−2×0)+(−1×−1)=0+1=1(-2 \times 0) + (-1 \times -1) = 0 + 1 = 1.

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