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Show that

LHS = $$\begin{vmatrix}a_1+b_1t & a_2+b_2t & a_3+b_3t \\ a_1t+b_1 & a_2t+b_2 & a_3t+b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$

RHS = (1-t^2) multiply$$\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$

I will perform it from the RHS to LHS.

First, I let detA = detB = $$\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$ Then, I multiply "t" to 1st and 2nd row of detB in order to create t^2 multiply with detB. Next, I swap the 1st and 2nd row of detB to turn t^2 to -t^2. Call this detC.

Finally, I operate detA + detC in order to look identical to the LHS.

But the problem is the 3rd row of my RHS still remains $$\begin{vmatrix}a_1+b_1t & a_2+b_2t & a_3+b_3t \\ a_1t+b_1 & a_2t+b_2 & a_3t+b_3 \\2c_1 & 2c_2 & 2c_3 \\\end{vmatrix}$$

How can I eliminate those "2" in order to get my RHS = LHS ?

Thank you

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2 Answers 2

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In your approach you added two equal determinants, and therefore got twice the expected result. Indeed there now appears as a factor $2$ in the bottom row. To divide that bottom row by two will cut the value of the determinant in half (every term in the determinant's full expansion contains exactly one factor from each row).

Note that the determinant of the sum of two matrices is usually not the same as the sum of the two determinants. For example:

$$ \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 $$

but $\begin{vmatrix} 2 & 2 \\ 2 & 2 \end{vmatrix} = 0$.

However you are not very detailed in how exactly the sum of two determinants can be expressed as a single determinant. In general the determinant is a "multilinear" function of its rows (or columns), so perhaps with more details your approach could be made to work. Instead I offer a simpler approach based on $2\times 2$ cofactors.


Let's compare the cofactors of the entries in the bottom row, $c_1,c_2,c_3$. The determinants are expressed by signed cofactor expansion.

Let's do the first entry in the bottom row. In the LHS we have, by subtracting $t$ times the first row from the second:

$$ \begin{vmatrix} a_2+b_2t & a_3+b_3t \\ a_2t+b_2 & a_3t+b_3 \end{vmatrix} = \begin{vmatrix} a_2+b_2t & a_3+b_3t \\ b_2(1-t^2) & b_3(1-t^2) \end{vmatrix} = (1-t^2) \begin{vmatrix} a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} $$

That is, the cofactor of $c_1$ in the LHS is $1-t^2$ times the cofactor of $c_1$ in the RHS.

Since the same relationship holds for all three cofactors, the LHS determinant is $1-t^2$ times the RHS determinant.

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if you put $t = 1,$ you see that the first two rows are identical, therefore the determinant is zero. now if you put $t = -1,$ the two rows are opposites of each other, so the determinant is zero. note that the determinant is a quadratic in $t$, therefore $det = k(1-t^2)$. set $t=0$ to find the constant $k$ and that should give you the answer.

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  • $\begingroup$ You said det=k(1−t2). Do you mean det refers to detA/detB ? (below)$$\begin{vmatrix}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \\\end{vmatrix}$$ $\endgroup$
    – piggy
    Jan 25, 2015 at 15:25
  • $\begingroup$ there is a constant $k$ there. by the det i meant the determinant of the $LHS$ with all the $t$'s in it. and the constant $k$ is the determinant of the same matrix without the $t$'s. $\endgroup$
    – abel
    Jan 25, 2015 at 16:09

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