Matrices may be long.

Algebra Level 4

If

A = [ 1 0 0 2 1 0 3 2 1 ] A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \\ \end{bmatrix}

and U 1 , U 2 U_1 , U_2 and U 3 U_3 are column matrices satisfying

A U 1 = [ 1 0 0 ] AU_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}

A U 2 = [ 2 3 0 ] AU_2 = \begin{bmatrix} 2 \\ 3 \\ 0 \\ \end{bmatrix}

A U 3 = [ 2 3 1 ] AU_3 = \begin{bmatrix} 2 \\ 3 \\ 1 \\ \end{bmatrix}

and U U is a 3 × 3 3 \times 3 matrix whose columns C 1 , C 2 , C 3 C_1 , C_2 , C_3 are U 1 , U 2 , U 3 U_1 , U_2 , U_3 respectively, then the value of U |U| is

Try my set
3 -3 3 2 \frac{3}{2} 2 2 3 3

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

John Gilling
Jun 9, 2015

The columns of the matrix A U AU are given by A U 1 , A U 2 , AU_{1}, AU_{2}, and A U 3 AU_{3} , so A U = ( 1 2 2 0 3 3 0 0 1 ) . AU = \left( \begin{array}{ccc} 1 & 2 & 2 \\ 0 & 3 & 3 \\ 0 & 0 & 1 \end{array} \right).

By multiplicativity of the determinant, U = A U A = 3 1 = 3. \lvert U \rvert = \frac{\lvert AU \rvert}{\lvert A \rvert} = \frac{3}{1}=3.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...