What value can 𝑘 not take if the
rank of the matrix 𝐴 equal to seven, four, negative 15, 22, 𝑘, negative 24,
negative nine, 15, negative 21 is three?
In this question, we are told that
the rank of a three-by-three matrix is three. We begin by recalling that the rank
of an 𝑛-by-𝑛 matrix can only be equal to 𝑛 if the determinant is nonzero. Since we need to find the value
that 𝑘 cannot take, this is the value where the determinant of 𝐴 is equal to
zero. We can calculate the determinant of
any three-by-three matrix by expanding over any row or column in the matrix. And in this question, we will use
the most common method of expanding over the top row.
We begin by multiplying the first
element in this row, seven, by the determinant of the two-by-two matrix 𝑘, negative
24, 15, negative 21. And this is equal to seven
multiplied by negative 21𝑘 plus 360. To calculate the expression in the
parentheses, we multiply 𝑘 by negative 21 and then subtract negative 24 multiplied
by 15. Next, we multiply negative four by
the determinant of the two-by-two matrix 22, negative 24, negative nine, negative
21. This is equal to negative four
multiplied by negative 462 minus 216. Finally, we add negative 15
multiplied by the determinant of the two-by-two matrix 22, 𝑘, negative nine,
15. This is the same as subtracting 15
multiplied by 330 plus nine 𝑘.
We now have an expression for the
determinant, which we can simplify by distributing the parentheses. The determinant of matrix 𝐴 is
equal to negative 147𝑘 plus 2520 plus 2712 minus 4950 minus 135𝑘, which in turn
simplifies to negative 282𝑘 plus 282. Setting the determinant equal to
zero, we can solve for 𝑘 by adding 282𝑘 to both sides. Dividing through by 282, we have 𝑘
is equal to one.
This means that if 𝑘 is equal to
one, the determinant of matrix 𝐴 is zero. And since this would lead to a rank
that is not equal to three, the value that 𝑘 cannot take is one.