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Chapter 3: Problem 50

Solve for \(x\) $$\left|\begin{array}{rr} x-2 & -1 \\ -3 & x \end{array}\right|=0$$

Short Answer

Expert verified
The solutions are \(x = 3\) and \(x = -1\).

Step by step solution

01

Identify the elements of the matrix

Here, our matrix is \(\begin{bmatrix} x-2 & -1 \ -3 & x \end{bmatrix}\). So, \(a = x-2, b = -1, c = -3, \) and \(d = x\).
02

Apply the determinant formula

Substitute \(a, b, c, d\) into the formula \[ \text{{det}}(A) = a*d - b*c \] and solve for \(x\). So, \((x-2)*x - (-1)*(-3) = 0\). This simplifies to \(x^2 -2x - 3 = 0\).
03

Solve for \(x\)

The equation \(x^2 -2x - 3 = 0\) can be factored into \((x-3)(x+1) = 0\). Thus, \(x = 3\) or \(x = -1\) are the possible solutions.

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Most popular questions from this chapter

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} 3 x_{1}-2 x_{2}+x_{3} &=-29 \\ -4 x_{1}+x_{2}-3 x_{3} &=37 \\ x_{1}-5 x_{2}+x_{3} &=-24 \end{aligned}$$

Determine whether the points are collinear. $$(-1,0),(1,1),(3,3)$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} -\frac{1}{4} x_{1}+\frac{3}{8} x_{2} &=-2 \\ \frac{3}{2} x_{1}+\frac{3}{4} x_{2} &=-12 \end{aligned}$$

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 3 x_{1}+2 x_{2}=1 \\ 2 x_{1}+10 x_{2}=6 \end{array}$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} -x_{1}-x_{2} \quad+x_{4}=-8 \\ 3 x_{1}+5 x_{2}+5 x_{3} \quad=24 \\ 2 x_{3}+x_{4} =-6 \\ -2 x_{1}-3 x_{2}-3 x_{3} \quad=-15 \end{aligned}$$
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