Question Video: Finding a Three-by-Three Determinant Mathematics

Which of the following is equal to the determinant |𝑏 βˆ’ 8𝑐, 𝑐 βˆ’ 7π‘Ž, π‘Ž βˆ’ 7𝑏 and 8𝑐, 7π‘Ž, 7𝑏 and βˆ’6, βˆ’6, βˆ’6|? [A] 6(π‘Ž βˆ’ 𝑐)(7𝑏 βˆ’ 8𝑐) + 42(π‘Ž βˆ’ 𝑏)Β² [B] 6(π‘Ž βˆ’ 𝑐)(7𝑏 βˆ’ 8𝑐) βˆ’ 7(π‘Ž βˆ’ 𝑏)Β² [C] 6(π‘Ž βˆ’ 𝑐)(7𝑏 βˆ’ 8𝑐) βˆ’ 42(π‘Ž βˆ’ 𝑏)Β² [D] βˆ’42(π‘Ž βˆ’ 𝑏)Β² βˆ’ 6(π‘Ž βˆ’ 𝑐)(7𝑏 βˆ’ 8𝑐)

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Video Transcript

Which of the following is equal to the determinant of the matrix with elements 𝑏 minus eight 𝑐, 𝑐 minus seven π‘Ž, π‘Ž minus seven 𝑏, eight 𝑐, seven π‘Ž, seven 𝑏, negative six, negative six, and negative six? Is it (A) six multiplied by π‘Ž minus 𝑐 multiplied by seven 𝑏 minus eight 𝑐 plus 42 multiplied by π‘Ž minus 𝑏 all squared? Or (B) six multiplied by π‘Ž minus 𝑐 multiplied by seven 𝑏 minus eight 𝑐 minus seven π‘Ž minus 𝑏 all squared. Or (C) six multiplied by π‘Ž minus 𝑐 times seven 𝑏 minus eight 𝑐 minus 42 times π‘Ž minus 𝑏 all squared. Or is it (D) negative 42 multiplied by π‘Ž minus 𝑏 all squared minus six times π‘Ž minus 𝑐 multiplied by seven 𝑏 minus eight 𝑐?

Recall that to find the determinant of an π‘š-by-𝑛 matrix, we either expand using a single row or column and the associated matrix minors. And recall that a matrix minor uppercase 𝐴 subscript 𝑖𝑗 is the matrix 𝐴 minus row 𝑖 and minus column 𝑗. So, for example, for a three-by-three matrix, if we want to find the matrix minor 𝐴 one, two, we eliminate the first row and the second column. Our matrix minor is then the two-by-two matrix with elements π‘Ž two, one; π‘Ž two, three; π‘Ž three, one; and π‘Ž three, three. Then to find the determinant of an 𝑛-by-𝑛 matrix expanding along row 𝑖, we use the formula the determinant of the matrix 𝐴 is given by the sum from 𝑗 is one to 𝑛 of negative one raised to the power 𝑖 plus 𝑗 multiplied by the element π‘Ž subscript 𝑖𝑗 multiplied by the determinant of the minor uppercase 𝐴 subscript 𝑖𝑗.

Remember that this formula differs slightly if we choose to expand down a column in our matrix. Either way, a rule of thumb in choosing the row or column to work with is either to choose the row or column with the most zeros or failing that the least complicated row or column. In our case, we have no zero elements. However, the final row is simply negative six, negative six, negative six, so we can use this to expand along.

Our matrix is a three-by-three matrix so that 𝑛 is equal to three. We’re going to expand along the third row so that 𝑖 is equal to three. In our formula for the determinant, therefore, we have negative one raised to the power three plus one multiplied by the element π‘Ž three, one multiplied by the determinant of the matrix minor 𝐴 three, one plus negative one raised to the power three plus two multiplied by the element π‘Ž three, two multiplied by the determinant of the matrix minor 𝐴 three, two plus negative one raised to the power three plus three multiplied by the element π‘Ž three, three multiplied by the determinant of the matrix minor 𝐴 three, three.

Our exponents of negative one are three plus one is four, three plus two is five, and three plus three is six. And negative one to an even power is one; negative one to an odd power is negative one. Now, if we label the rows and columns in our matrix, then the element π‘Ž three, one is the element in row three, column one, and that is negative six. Similarly, the element π‘Ž three, two is also negative six, as is the element π‘Ž three, three so that our determinant is negative six times the determinant of the minor 𝐴 three, one minus negative six times the determinant of the minor 𝐴 three, two minus six times the determinant of the minor 𝐴 three, three.

Now recalling that the minor 𝐴 three, one is the matrix 𝐴 without row three and column one, then our first term is negative six times the determinant of the matrix with elements 𝑐 minus seven π‘Ž, π‘Ž minus seven 𝑏, seven π‘Ž, and seven 𝑏. Similarly, the minor 𝐴 three, two is obtained by eliminating row three and column two. Our second term is therefore six times the determinant of the two-by-two matrix with elements 𝑏 minus eight 𝑐, π‘Ž minus seven 𝑏, eight 𝑐, and seven 𝑏. And similarly, we find our third term negative six multiplied by the determinant of the two-by-two matrix with elements 𝑏 minus eight 𝑐, 𝑐 minus seven π‘Ž, eight 𝑐, and seven π‘Ž.

Now making some room, now to find the determinant of our three-by-three matrix, we need to find the determinants of the three two-by-two matrices. And remember that for the two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, and 𝑑, the determinant of the matrix is π‘Žπ‘‘ minus 𝑏𝑐. Applying this to our first term, π‘Žπ‘‘ is 𝑐 minus seven π‘Ž multiplied by seven 𝑏 and 𝑏𝑐 is π‘Ž minus seven 𝑏 multiplied by seven π‘Ž. And then to our second term, π‘Žπ‘‘ is 𝑏 minus eight 𝑐 times seven 𝑏. And 𝑏𝑐 is π‘Ž minus seven 𝑏 multiplied by eight 𝑐. And similarly, for our third term, we have 𝑏 minus eight 𝑐 multiplied by seven π‘Ž minus 𝑐 minus seven π‘Ž multiplied by eight 𝑐.

Now, let’s expand each of our parentheses and see if we can simplify things a little. Distributing inside the parentheses of our first term, we have negative six multiplied by seven 𝑏𝑐 minus 49π‘Žπ‘ minus seven π‘Ž squared plus 49π‘Žπ‘ and similarly for our second and third terms. In our first term, we have negative 49π‘Žπ‘ and plus 49π‘Žπ‘, and that’s equal to zero. Similarly, in our second term, we have 56𝑏𝑐 and negative 56𝑏𝑐, and in our third term, 56π‘Žπ‘ and negative 56π‘Žπ‘. So now rewriting, we have negative six multiplied by seven 𝑏𝑐 minus seven π‘Ž squared plus six multiplied by seven 𝑏 squared minus eight π‘Žπ‘ minus six times seven π‘Žπ‘ minus eight 𝑐 squared.

So now we’re getting a little closer to a possible solution. Examining the possible solutions (A), (B), (C), and (D), we see that all four contain the expression π‘Ž minus 𝑐, similarly, the expression seven 𝑏 minus eight 𝑐 and π‘Ž minus 𝑏 squared. So let’s see if we can find these expressions in our determinant.

Let’s consider first the expression π‘Ž minus 𝑏 squared. Distributing the parentheses gives us π‘Ž squared minus two π‘Žπ‘ plus 𝑏 squared. In our determinant, π‘Ž squared appears, 𝑏 squared appears, and π‘Žπ‘ appears. And in fact, we have 42π‘Ž squared plus 42𝑏 squared minus 42π‘Žπ‘, and this is almost the same as the expansion of π‘Ž minus 𝑏 squared. A difference is that we have one more instance of π‘Žπ‘. So in fact, the expression inside our parentheses is π‘Ž minus 𝑏 all squared plus π‘Žπ‘. This means that our three terms correspond to 42 multiplied by π‘Ž minus 𝑏 squared plus π‘Žπ‘. And we can rewrite this as 42 π‘Ž minus 𝑏 squared plus 42π‘Žπ‘.

So now if we rearrange our determinant, we have our 42 π‘Ž minus 𝑏 squared at the beginning plus six times seven π‘Žπ‘, that’s 42π‘Žπ‘, minus six times seven 𝑏𝑐, which is the same as six times negative seven 𝑏𝑐, plus six times negative eight π‘Žπ‘ minus six times negative eight 𝑐 squared. So we have 42 times π‘Ž minus 𝑏 squared plus six multiplied by seven π‘Žπ‘ minus seven 𝑏𝑐 minus eight π‘Žπ‘ plus eight 𝑐 squared.

We’ve covered the π‘Ž minus 𝑏 squared term in each of the possible solutions. So now let’s see if we can find an π‘Ž minus 𝑐. If we look at our term seven π‘Žπ‘ minus seven 𝑏𝑐, we have a common factor of seven 𝑏. Similarly, negative eight π‘Žπ‘ plus eight 𝑐 squared is actually negative eight 𝑐 multiplied by π‘Ž minus 𝑐. So we have a common factor of negative eight 𝑐. And now we have a common factor of π‘Ž minus 𝑐. And so we have π‘Ž minus 𝑐 multiplied by seven 𝑏 minus eight 𝑐. So not forgetting the six, our determinant is then six multiplied by π‘Ž minus 𝑐 times seven 𝑏 minus eight 𝑐 plus 42 multiplied by π‘Ž minus 𝑏 all squared. This corresponds to option (A), so option (A) is equal to the determinant given.

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