Video: Finding the Product of Two Given Matrices

Consider the matrices 𝐴 = [11, βˆ’2 and βˆ’4, 4 and 7, 7], 𝐡 = [βˆ’8, βˆ’9, 6 and βˆ’4, 8, 9]. Find 𝐴𝐡, if possible.

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Video Transcript

Consider the matrices 𝐴 and 𝐡. Find 𝐴𝐡, if possible.

Let’s first establish whether the multiplication of these two matrices is possible. Matrix 𝐴 has three rows and two columns, and matrix 𝐡 has two rows and three columns. Seen is the number of columns in matrix 𝐴 is the same as the number of rows in matrix 𝐡, we know that the resultant matrix exists. Additionally, we can tell the dimensions of the resultant matrix by seeing that matrix 𝐴 has three rows and matrix 𝐡 has three columns. So the resultant matrix will be a three-by-three matrix. We find the first element of 𝐴𝐡 by multiplying the top row of matrix 𝐴 by the left-hand column of matrix 𝐡. Remember that this is just the same as finding the dot product of this first row of matrix 𝐴 with the left-hand column of matrix 𝐡.

That is 11 multiplied by negative eight add negative two multiplied by negative four. To get the top middle element, we multiply the top row of matrix 𝐴 with the middle column of matrix 𝐡. That is 11 multiplied by negative nine add negative two multiplied by eight. To get the top right element, we multiply the top row of matrix 𝐴 with the right-hand column of matrix 𝐡. That is 11 multiplied by six add negative two multiplied by nine. We can then find the middle left component by multiplying the middle row of matrix 𝐴 with the left-hand column of matrix 𝐡.

We find the middle component by multiplying the middle row of matrix 𝐴 with the middle column of matrix 𝐡. And we find the middle right component by multiplying together the middle row of matrix 𝐴 with the right-hand column of matrix 𝐡. And we follow the same pattern for the bottom left component, the bottom middle component, and the bottom right component. We can then simplify each component. And that gives us our final answer, which is just as we worked out a three-by-three matrix.

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