Answer to Question #349894 in Linear Algebra for Busi

Question

Solve for x in the given matrix equalities

det (x −3 4

3 −5 x

1 2 −1)= 45

Answer 1

$$\begin{vmatrix} x & -3 & 4 \\ 3 & -5 & x \\ 1 & 2 & -1 \\ \end{vmatrix}=x\begin{vmatrix} -5 & x \\ 2 & -1 \end{vmatrix}-(-3)\begin{vmatrix} 3 & x \\ 1 & -1 \end{vmatrix}$$

$$+4\begin{vmatrix} 3 & -5 \\ 1 & 2 \end{vmatrix}=x(5-2x)+3(-3-x)+4(6+5)$$

$$=5x-2x^2-9-3x+44$$

$$=-2x^2+2x+35=45$$

$$2x^2-2x+10=0$$

$$x^2-x+5=0$$

$$D=(-1)^2-4(1)(5)=-19<0$$

There is no solution for $x.$

$$\begin{vmatrix} 2 & -3 & 4 \\ 3 & -5 & x \\ 1 & 2 & -1 \\ \end{vmatrix}=2\begin{vmatrix} -5 & x \\ 2 & -1 \end{vmatrix}-(-3)\begin{vmatrix} 3 & x \\ 1 & -1 \end{vmatrix}$$

$$+4\begin{vmatrix} 3 & -5 \\ 1 & 2 \end{vmatrix}=2(5-2x)+3(-3-x)+4(6+5)$$

$$=10-4x-9-3x+44=-7x+45=45$$

$$x=0$$