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We know that 2 × 2s can always be multiplied together, and we know that the outcome must always also be a 2 × 2. Let's visualize it:
Looking at where the intersections are, we can see which row of the first matrix and which column of the second matrix collide to come up with the product's entries. Below, we can see that X marks the spot where row one of matrix one and column one of matrix two crash:
We know those four numbers need to crunch into one; we also know we read rows left to right, like a book, and we read columns top to bottom. We multiply the first number from the first matrix's first row with the first number from the second matrix's first column: (2)(4) = 8. Then we multiply the second number from the first matrix's first row with the second number from the second matrix's first column: (3)(6) = 18. Finally, we add those together: 8 + 18 = 26.
Bam: the upper-left entry in our product matrix is 26.
Next? We rubberneck at the collision between row one of matrix one and column two of matrix two.
Right as usual, King Friday.
(2)(1) + (3)(0) = 2 + 0 = 2
Plop that 2 into the upper-right entry of our product. Here's where we're at so far:
We're so almost there. X marks the spot, and we do the same thing. We read the row left to right, we read the column top to bottom. This time we're doing the second row in the product, so we multiply the second row from the first matrix with the first column from the second matrix:
(1)(4) + (2)(6) = 4 + 12 = 16
Throw that 16 into our new matrix:
We do the same thing one last time, this time with the second row in the
first matrix and the second column in the second matrix.
(2)(0) = 1 + 0 = 1
Our answer is:
Find the product:
Whenever we're multiplying by an identity matrix, the answer is always the same as the non-identity matrix:
Find the product:
We know they can be multiplied together because we've got three columns in the first matrix and three rows in the second matrix. We also know that the product matrix will have to be a 2 × 2 zero matrix because whenever we multiply by a zero matrix, the answer is always a zero matrix with the same number of rows as the first matrix has and the same number of columns as the second matrix.