- #1
CentreShifter
- 24
- 0
Problem:
Show, without evaluating directly, that
[tex]
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
a_1t+b_1&a_2t+b_2&a_3t+b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
=
(1-t^2)\left|\begin{matrix}
a_1&a_2&a_3 \\
b_1&b_2&b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
Clearly, here I'm supposed to use the determinant properties, do some row ops on the first array, and end up with the RHS.
1. -tR1-R2 -> Row2 (no coefficient on determinant).
[tex]
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
b_1-b_1t^2&b_2-b_2t^2&b_3-b_3t^2 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
From here I can see that [tex]b_1-b_1t^2=b_1(1-t^2)[/tex]. But multiplying R2 by [tex]\frac{1}{1-t^2}[/tex] means I have to also pull that out as a coefficient to the entire array. So now:
[tex]
\frac{1}{1-t^2}
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
b_1&b_2&b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
Another row op, R1 -> R1-tR2, and I have my RHS, except the coefficient is reciprocated. Am I doing this wrong?
Show, without evaluating directly, that
[tex]
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
a_1t+b_1&a_2t+b_2&a_3t+b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
=
(1-t^2)\left|\begin{matrix}
a_1&a_2&a_3 \\
b_1&b_2&b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
Clearly, here I'm supposed to use the determinant properties, do some row ops on the first array, and end up with the RHS.
1. -tR1-R2 -> Row2 (no coefficient on determinant).
[tex]
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
b_1-b_1t^2&b_2-b_2t^2&b_3-b_3t^2 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
From here I can see that [tex]b_1-b_1t^2=b_1(1-t^2)[/tex]. But multiplying R2 by [tex]\frac{1}{1-t^2}[/tex] means I have to also pull that out as a coefficient to the entire array. So now:
[tex]
\frac{1}{1-t^2}
\left|\begin{matrix}
a_1+b_1t&a_2+b_2t&a_3+b_3t \\
b_1&b_2&b_3 \\
c_1&c_2&c_3 \end{matrix}\right|
[/tex]
Another row op, R1 -> R1-tR2, and I have my RHS, except the coefficient is reciprocated. Am I doing this wrong?