91Ó°ÊÓ

Use Cramer's rule to solve the given linear system. $$\begin{aligned} 3 x_{1}+x_{2}+2 x_{3} &=-1, \\ 2 x_{1}-x_{2}+x_{3} &=-1, \\ 5 x_{2}+5 x_{3} &=-5. \end{aligned}$$

Without expanding the determinant, determine all values of \(x\) for which \(\operatorname{det}(A)=0\) if $$A=\left[\begin{array}{rrr} 1 & -1 & x \\ 2 & 1 & x^{2} \\ 4 & -1 & x^{3} \end{array}\right]$$

Use determinants to prove that if \(A\) is invertible and \(B\) and \(C\) are matrices with \(A B=A C,\) then \(B=C\)

If \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right],\) compute the matrix product \(A \cdot \operatorname{adj}(A) .\) What can you conclude about \(\operatorname{det}(A) ?\)

If \(A\) is an \(n \times n\) skew-symmetric matrix and \(n\) is odd, prove that \(\operatorname{det}(A)=0\)

Find all solutions to the system $$ \begin{array}{l} (b+c) x_{1}+a\left(x_{2}+x_{3}\right)=a, \\ (c+a) x_{2}+b\left(x_{3}+x_{1}\right)=b, \\ (a+b) x_{3}+c\left(x_{1}+x_{2}\right)=c, \end{array} $$ where \(a, b, c\) are constants. Make sure you consider all cases (that is, those when there is a unique solution, an infinite number of solutions, and no solutions).