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Chapter 3: Problem 49

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

Short Answer

Expert verified
The solutions to the equation are \(x = -1\) and \(x = -4\).

Step by step solution

01

Find the Determinant

First, calculate the determinant of the matrix. Using the formula for the determinant of a 2x2 matrix, which is \(ad - bc\), where \(a = x + 3\), \(b = 2\), \(c = 1\), and \(d = x + 2\), the determinant would be \((x + 3)(x + 2) - (2 * 1)\).
02

Set the determinant equal to zero

Next, set the determinant equal to zero to solve for \(x\). Therefore, the equation \((x + 3)(x + 2) - 2 = 0\) is obtained.
03

Simplify the Equation

Expand the products and simplify the equation. Hence, \(x^2 + 5x + 6 - 2 = 0\), which simplifies to \(x^2 + 5x + 4 = 0\).
04

Solve for \(x\)

Solve this quadratic equation for \(x\). This equation factors to \((x + 1)(x + 4) = 0\). Therefore, the two possible solutions are \(x = -1\) and \(x = -4\).

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Most popular questions from this chapter

Use a graphing utility or computer software program with matrix capabilities to find the eigenvalues of the matrix. Then find the corresponding eigenvectors. $$\left[\begin{array}{rrrr} 3 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 2 & 5 \\ 0 & 0 & 3 & 0 \end{array}\right]$$

Find the volume of the tetrahedron having the given vertices. $$(3,-1,1),(4,-4,4),(1,1,1),(0,0,1)$$

Prove that if \(|A|=1\) and all entries of \(A\) are integers, then all entries of \(\left|A^{-1}\right|\) must also be integers.

Prove that if \(A\) is an \(n \times n\) invertible matrix, then \(\operatorname{adj}\left(A^{-1}\right)=\) \([\operatorname{adj}(A)]^{-1}\)

Determine whether the points are coplanar $$(-4,1,0),(0,1,2),(4,3,-1),(0,0,1)$$
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