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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?

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# Homework Help: [Linear Algebra] Showing equality via determinant properties

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