[Linear Algebra] Showing equality via determinant properties

[CentreShifter]

Problem:

Show, without evaluating directly, that
<br /> \left|\begin{matrix}<br /> a_1+b_1t&amp;a_2+b_2t&amp;a_3+b_3t \\<br /> a_1t+b_1&amp;a_2t+b_2&amp;a_3t+b_3 \\<br /> c_1&amp;c_2&amp;c_3 \end{matrix}\right|<br /> =<br /> (1-t^2)\left|\begin{matrix}<br /> a_1&amp;a_2&amp;a_3 \\<br /> b_1&amp;b_2&amp;b_3 \\<br /> c_1&amp;c_2&amp;c_3 \end{matrix}\right|<br />

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).

<br /> \left|\begin{matrix}<br /> a_1+b_1t&amp;a_2+b_2t&amp;a_3+b_3t \\<br /> b_1-b_1t^2&amp;b_2-b_2t^2&amp;b_3-b_3t^2 \\<br /> c_1&amp;c_2&amp;c_3 \end{matrix}\right|<br />

From here I can see that b_1-b_1t^2=b_1(1-t^2). But multiplying R2 by \frac{1}{1-t^2} means I have to also pull that out as a coefficient to the entire array. So now:
<br /> \frac{1}{1-t^2}<br /> \left|\begin{matrix}<br /> a_1+b_1t&amp;a_2+b_2t&amp;a_3+b_3t \\<br /> b_1&amp;b_2&amp;b_3 \\<br /> c_1&amp;c_2&amp;c_3 \end{matrix}\right|<br />

Another row op, R1 -> R1-tR2, and I have my RHS, except the coefficient is reciprocated. Am I doing this wrong?

 

[ehild]

If you multiply a row or column by some t, the value of the determinant becomes t times the original value. So you need to divide the whole determinant by t so as to keep it unchanged.

ehild