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?



  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 $-1$.
  • $\begingroup$ 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

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)


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