Answer to Question #87346 in Linear Algebra for Daniel

Question

3. A necessary and sufficient condition for a matrix (square) A to be invertible is that
a.A is not equal zero
b.|A|≠0
c.|A|>0
d.A<0

4. Given that x + 2y + 3z =1, 3x + 2y + z = 4, x + 3y + 2z = 0. What is x,y and z?
a.(7/4,-3/4,1/4)
b.(5/4,-2/4,1/4)
c.(1,-3/4,1/4)
d.(1,-2,3)

Answer 1

3b - it is a known theorem

4a

$$x=\frac{\begin{vmatrix} 1 & 2 & 3 \\ 4 & 2 & 1\\ 0&3&2 \end{vmatrix}}{\begin{vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1\\ 1&3&2 \end{vmatrix}}=\frac{4+36-16-3}{4+2+27-6-12-3}=\frac{7}{4}$$

$$y=\frac{\begin{vmatrix} 1 & 1 & 3 \\ 3 & 4 & 1\\ 1&0&2 \end{vmatrix}}{\begin{vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1\\ 1&3&2 \end{vmatrix}}=\frac{8+1-12-6}{4+2+27-6-12-3}=-\frac{3}{4}$$

$$z=\frac{\begin{vmatrix} 1 & 2 & 1 \\ 3 & 2 & 4\\ 1&3&0 \end{vmatrix}}{\begin{vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1\\ 1&3&2 \end{vmatrix}}=\frac{8+9-2-12}{4+2+27-6-12-3}=\frac{1}{4}$$