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Chapter 12: Problem 51

Solve for \(x\) \(\left|\begin{array}{ll}x & x \\ 4 & 3\end{array}\right|=5\)

Short Answer

Expert verified
\(-5\)

Step by step solution

01

- Understanding the Determinant

Given the matrix \(\left|\begin{array}{ll}x & x \ 4 & 3\end{array}\right|=5\),recall that the determinant of a 2x2 matrix \(\left|\begin{array}{cc}a & b \ c & d\end{array}\right|\)can be calculated using the formula:\(ad - bc\).
02

- Apply the Determinant Formula

Substitute the values from the matrix into the determinant formula: \(x \cdot 3 - x \cdot 4 = 5\).
03

- Simplify the Equation

Simplify the equation: \(3x - 4x = 5\).Combine like terms to get: \(-x = 5\).
04

- Solve for x

Divide both sides by \(-1\) to solve for \(x\).\(x = -5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant Formula
The determinant of a matrix is a special number that can help us solve many problems in linear algebra. For a 2x2 matrix, the determinant formula is simple and essential to understand. If you have a 2x2 matrix like this: \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\] then the determinant, which is often denoted as \( \text{det}(A) \) or \( |A| \), can be calculated using the formula: \( ad - bc \).

Here鈥檚 what each variable means:
  • \(a\) is the element in the first row, first column.
  • \(b\) is the element in the first row, second column.
  • \(c\) is the element in the second row, first column.
  • \(d\) is the element in the second row, second column.

By multiplying the elements on the diagonals and subtracting, the determinant gives us a single value that holds many properties of the matrix.
2x2 Matrices
A 2x2 matrix is one of the simplest types of matrices. It has two rows and two columns, making calculations straightforward compared to larger matrices. When working with 2x2 matrices:
  • Always identify the elements accurately.
  • Remember that the order of elements affects the final result: \(ad - bc\) from the determinant formula is a specific order.

In our exercise, the given matrix is: \[\begin{bmatrix} x & x \ 4 & 3 \end{bmatrix}\] This means:
  • \(a = x\)
  • \(b = x\)
  • \(c = 4\)
  • \(d = 3\)

Plugging these values into the determinant formula helps us set up the problem鈥檚 main equation.
Solving Equations
Solving equations is about finding the value of the unknown variable that makes the equation true. For our matrix, we set up the equation using the determinant formula:
\[x \times 3 - x \times 4 = 5\]
Breaking it down step-by-step:
  • First, multiply: \(3x\) and \(4x\) in the equation.
  • Next, calculate their difference: \(3x - 4x\) which simplifies to \(-x\).
  • Finally, set the equation to match the given value: \(-x = 5\).

To solve for \(x\), divide both sides by \(-1\):
\[x = \frac{5}{-1} = -5\]
Therefore, the value of \(x\) that satisfies the determinant equation is \(-5\). Understanding how to manipulate the matrix elements using the determinant formula makes solving these equations clear and straightforward.

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Most popular questions from this chapter

Presale Order A wireless store owner takes presale orders for a new smartphone and tablet. He gets 340 preorders for the smartphone and 250 preorders for the tablet. The combined value of the preorders is \(\$ 486,000 .\) If the price of a smartphone and tablet together is \(\$ 1665,\) how much does each device cost?

Challenge Problem Solve for \(x, y,\) and \(z,\) assuming \(a \neq 0, b \neq 0,\) and \(c \neq 0\) $$ \left\\{\begin{array}{l} a x+b y+c z =a+b+c \\ a^{2} x+b^{2} y+c^{2} z =a c+a b+b c \\ a b x+b c y \quad \quad=b c+a c \end{array}\right. $$

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} x+2 y=4 \\ 2 x+4 y=8 \end{array}\right. $$

Mixing a Solution A chemist wants to make 14 liters of a \(40 \%\) acid solution. She has solutions that are \(30 \%\) acid and \(65 \%\) acid. How much of each must she mix?

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} 2 x-3 y-z=0 \\ -x+2 y+z=5 \\ 3 x-4 y-z=1 \end{array}\right. $$
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