Answer to Question #182815 in Linear Algebra for Elaine

Question

Which of the following matrix (or given expression) results in a non-singular matrix

1) 2 0 0

0 SINà ½SINà

0 -1 -COSà

2) A v B where A = 1 2 -1

3 1 0

-2 -4 2

and B = 1 0 -1

3 1 0

-2 -4 2

3) 2 0 0

3 -4 0

0 -1 0

4) 2 0 0

3 -4 4

0 1 -1

Answer 1

.A square matrix is non singular iff its determinant is non zero.

Consider matrices A and B in 2

det A=$\begin{vmatrix} 1 & 2 & -1 \\ 3 & 1 & 0\\ -2 & -4 & 2 \end{vmatrix}$ = 1·1·2 + 2·0·(-2) + (-1)·3·(-4) - (-1)·1·(-2) - 1·0·(-4) - 2·3·2 = 2 + 0 + 12 - 2 - 0 - 12 = 0

det В = $\begin{vmatrix} 1 & 0 & -1 \\ 3 & 1 & 0\\ -2 & -4 & 2 \end{vmatrix}$ = ·(-1)·2 + 0·0·(-2) + (-2)·3·(-4) - (-2)·(-1)·(-2) - 1·0·(-4) - 0·3·2 = -2 + 0 + 24 + 4 - 0 - 0 = 26

Since det A=0 and det B=26 then det(A$\lor$ B)=26

so matrix A$\lor$ B is non-singular