1
$\begingroup$

Suppose $A = \left( \begin{array}{ccc} v_1 &v_2&v_3&x \end{array} \right) $

where $v_1,v_2,v_3$ are fixed vectors in $\mathbb{R}^4$ and $x$ is any vector in $\mathbb{R}^4$. Let $T:\mathbb{R}^4\rightarrow \mathbb{R}$ be the linear transformation defined as $T(x)=\det(A)$. Suppose $T(e_1)=4,T(e_2)=1,T(e_3)=-1,T(e_4)=2$.

(a) What is $T\left( \begin{array}{ccc} 1\\2\\3\\4 \end{array} \right)$?

This is just $4+2(1)+3(-1)+4(2) = 11$.

Now the question I am having much difficulty with is this next one.

(b) If $x= \left( \begin{array}{ccc} 1\\2\\3\\4 \end{array} \right)$ and $B = \left( \begin{array}{ccc} x^T\\2v_1^T\\4v_2^T\\6v_3^T \end{array} \right)$, what is $\det(B)$?

I know we can do some clever manipulation using Gauss-Jordan elimination to find the determinant, but I'm honestly stumped. Can someone please help me here?

$\endgroup$
2
$\begingroup$

Hints:

  1. Transposition preserves determinant
  2. Multiplying a row or column by a scalar multiplies the determinant by that scalar
  3. Interchanging two rows or columns multiplies the determinant by $-1$.
$\endgroup$
1
  • $\begingroup$ For your point 2. is dividing various rows by scalars the same thing as multiplying various rows by scalars? Because I know when you divide rows by scalars, you can multiply the determinant by that scalar. $\endgroup$
    – Parker
    Nov 19 '14 at 5:24
0
$\begingroup$

a) Determinants are linear in columns (or rows), so yes, you just need to do the linear combination.

b) You can use linearity again, together with the property of interchanging rows.

$det(A^T) = det(A) = 11 = det\left(\begin{bmatrix}v_1^T\\v_2^T\\v_3^T\\x^T\end{bmatrix}\right)$

$det\left(\begin{bmatrix}x^T\\v_2^T\\v_3^T\\v_1^T\end{bmatrix}\right) = -11$

$det\left(\begin{bmatrix}x^T\\v_1^T\\v_3^T\\v_2^T\end{bmatrix}\right) = 11$

$det\left(\begin{bmatrix}x^T\\v_1^T\\v_2^T\\v_3^T\end{bmatrix}\right) = -11$

$det\left(\begin{bmatrix}x^T\\2\,v_1^T\\v_2^T\\v_3^T\end{bmatrix}\right) = -2\cdot 11$

$det\left(\begin{bmatrix}x^T\\2\,v_1^T\\4\,v_2^T\\v_3^T\end{bmatrix}\right) = -4\cdot 2\cdot11$

$det\left(\begin{bmatrix}x^T\\2\,v_1^T\\4\,v_2^T\\6\,v_3^T\end{bmatrix}\right) = -6\cdot 4\cdot 2\cdot 11 = \boxed{-528}$ (if there were no mistakes)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.