B=[[2,3],[1,3]]" and "A=[[-1,3],[-1,0]] Let H=BA. Find H. H=[◻

Write Matrices B and A: First, let's write down the matrices B and A to visualize them better. Matrix B is given as B = [[2, 3], [1, 3]] and matrix A is given as A = [[-1, 3], [-1, 0]]. To find the product of two matrices, we multiply the rows of the first matrix by the columns of the second matrix.Calculate Element H[1,1]: Now, let's calculate the elements of the first row of the matrix H. The element at the first row and first column of H (H[1,1]) is calculated by multiplying the first row of B by the first column of A and summing the results: (2 * -1) + (3 * -1).Calculate Element H[1,2]: Performing the calculation for H[1,1]: (2 * -1) + (3 * -1) = -2 - 3 = -5. So, the first element of matrix H is -5.Calculate Element H[2,1: Next, we calculate the element at the first row and second column of H (H[1,2]) by multiplying the first row of B by the second column of A and summing the results: (2 * 3) + (3 * 0).Calculate Element H[2,2]: Performing the calculation for H[1,2]: (2 * 3) + (3 * 0) = 6 + 0 = 6. So, the second element of the first row of matrix H is 6.Now, let's calculate the elements of the second row of the matrix H. The element at the second row and first column of H (H[2,1]) is calculated by multiplying the second row of B by the first column of A and summing the results: (1 * -1) + (3 * -1).Performing the calculation for H[2,1]: (1 * -1) + (3 * -1) = -1 - 3 = -4. So, the first element of the second row of matrix H is -4.Finally, we calculate the element at the second row and second column of H (H[2,2]) by multiplying the second row of B by the second column of A and summing the results: (1 * 3) + (3 * 0).Performing the calculation for H[2,2]: (1 * 3) + (3 * 0) = 3 + 0 = 3. So, the second element of the second row of matrix H is 3.

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B=[[2,3],[1,3]]" and "A=[[-1,3],[-1,0]]
Let
H=BA. Find
H.

H=[◻

$B=\left[\begin{array}{ll} 2 & 3 \\ 1 & 3 \end{array}\right] \text { and } A=\left[\begin{array}{ll} -1 & 3 \\ -1 & 0 \end{array}\right]$ $\newline$ Let $\mathrm{H}=\mathrm{BA}$ . Find $\mathrm{H}$ . $\newline$ $\mathbf{H}=$

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Algebra 1

Compare linear, exponential, and quadratic growth

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B=\left[\begin{array}{ll} 2 & 3 \\ 1 & 3 \end{array}\right] \text { and } A=\left[\begin{array}{ll} -1 & 3 \\ -1 & 0 \end{array}\right]

\newline

Let

\mathrm{H}=\mathrm{BA}

. Find

\mathrm{H}

\newline

\mathbf{H}=

Write Matrices B and A: First, let's write down the matrices B and A to visualize them better. $\newline$ Matrix B is given as $B = \begin{bmatrix} 2 & 3 \ 1 & 3 \end{bmatrix}$ and matrix A is given as $A = \begin{bmatrix} -1 & 3 \ -1 & 0 \end{bmatrix}$ . $\newline$ To find the product of two matrices, we multiply the rows of the first matrix by the columns of the second matrix.
Calculate Element H[ $1$ , $1$ ]: Now, let's calculate the elements of the first row of the matrix H. The element at the first row and first column of H ( $H[1,1]$ ) is calculated by multiplying the first row of B by the first column of A and summing the results: $(2 \times -1) + (3 \times -1)$ .
Calculate Element $H[1,2]$ : Performing the calculation for $H[1,1]$ : $(2 \times -1) + (3 \times -1) = -2 - 3 = -5$ . $\newline$ So, the first element of matrix $H$ is $-5$ .
Calculate Element H[ $2$ , $1$ : Next, we calculate the element at the first row and second column of H ( $H[1,2]$ ) by multiplying the first row of B by the second column of A and summing the results: ( $2 \times 3$ ) + ( $3 \times 0$ ).
Calculate Element $H[2,2]$ : Performing the calculation for $H[1,2]$ : $(2 \times 3) + (3 \times 0) = 6 + 0 = 6$ . $\newline$ So, the second element of the first row of matrix $H$ is $6$ .
Calculate Element H[ $2$ , $2$ ]: Performing the calculation for H[ $1$ , $2$ ]: $(2 \times 3) + (3 \times 0) = 6 + 0 = 6$ . So, the second element of the first row of matrix H is $6$ .Now, let's calculate the elements of the second row of the matrix H. The element at the second row and first column of H (H[ $2$ , $1$ ]) is calculated by multiplying the second row of B by the first column of A and summing the results: $(1 \times -1) + (3 \times -1)$ .
Calculate Element H[ $2$ , $2$ ]: Performing the calculation for H[ $1$ , $2$ ]: $(2 \times 3) + (3 \times 0) = 6 + 0 = 6$ . So, the second element of the first row of matrix H is $6$ .Now, let's calculate the elements of the second row of the matrix H. The element at the second row and first column of H (H[ $2$ , $1$ ]) is calculated by multiplying the second row of B by the first column of A and summing the results: $(1 \times -1) + (3 \times -1)$ .Performing the calculation for H[ $2$ , $1$ ]: $(1 \times -1) + (3 \times -1) = -1 - 3 = -4$ . So, the first element of the second row of matrix H is $-4$ .
Calculate Element H[ $2$ , $2$ ]: Performing the calculation for H[ $1$ , $2$ ]: $(2 \times 3) + (3 \times 0) = 6 + 0 = 6$ . So, the second element of the first row of matrix H is $6$ .Now, let's calculate the elements of the second row of the matrix H. The element at the second row and first column of H (H[ $2$ , $1$ ]) is calculated by multiplying the second row of B by the first column of A and summing the results: $(1 \times -1) + (3 \times -1)$ .Performing the calculation for H[ $2$ , $1$ ]: $(1 \times -1) + (3 \times -1) = -1 - 3 = -4$ . So, the first element of the second row of matrix H is $-4$ .Finally, we calculate the element at the second row and second column of H (H[ $2$ , $2$ ]) by multiplying the second row of B by the second column of A and summing the results: $(1 \times 3) + (3 \times 0)$ .
Calculate Element H[ $2$ , $2$ ]: Performing the calculation for H[ $1$ , $2$ ]: $(2 \times 3) + (3 \times 0) = 6 + 0 = 6$ . So, the second element of the first row of matrix H is $6$ . Now, let's calculate the elements of the second row of the matrix H. The element at the second row and first column of H (H[ $2$ , $1$ ]) is calculated by multiplying the second row of B by the first column of A and summing the results: $(1 \times -1) + (3 \times -1)$ . Performing the calculation for H[ $2$ , $1$ ]: $(1 \times -1) + (3 \times -1) = -1 - 3 = -4$ . So, the first element of the second row of matrix H is $-4$ . Finally, we calculate the element at the second row and second column of H (H[ $2$ , $2$ ]) by multiplying the second row of B by the second column of A and summing the results: $(1 \times 3) + (3 \times 0)$ . Performing the calculation for H[ $2$ , $2$ ]: $(1 \times 3) + (3 \times 0) = 3 + 0 = 3$ . So, the second element of the second row of matrix H is $3$ .