Chapter 8: Problem 39
Find all \(t\) such that (a) \(\left|\begin{array}{cc}t-4 & 3 \\ 2 & t-9\end{array}\right|=0\) (b) \(\left|\begin{array}{cc}t-1 & 4 \\ 3 & t-2\end{array}\right|=0\)
Short Answer
Expert verified
(a) \(t = 2, 11\)
(b) \(t = 5, -2\)
Step by step solution
01
Apply the determinant formula
For the first matrix, we have the values \(
A = \begin{bmatrix}
t-4 &3 \\
2 &t-9
\end{bmatrix}\).
Applying the matrix determinant formula det(A) = ad - bc, we get:
\(|A| = (t-4)(t-9) - (3)(2)\)
#Step 2: Equating the determinant to zero for the first matrix#
02
Set the determinant equal to zero
Now, we want to find the t values such that the determinant is equal to zero:
\((t-4)(t-9) - (3)(2) = 0\)
#Step 3: Solve for t for the first matrix#
03
Solve the equation
Expanding and simplifying the equation, we get:
\(t^2 -13t +22 = 0\)
Factorizing the quadratic equation, we find:
\((t - 2)(t - 11) = 0\)
So, the values of \(t\) that make the determinant equal to zero are \(t = 2\) and \(t = 11\).
Now let's do the same process for the second matrix.
#Step 4: Calculating the determinant for the second matrix#
04
Apply the determinant formula
For the second matrix, we have the values:
\(
B = \begin{bmatrix}
t-1 &4 \\
3 &t-2
\end{bmatrix}\)
Applying the matrix determinant formula det(B) = ad - bc, we get:
\(|B| = (t-1)(t-2) - (4)(3)\)
#Step 5: Equating the determinant to zero for the second matrix#
05
Set the determinant equal to zero
Now, we want to find the values of `t` such that the determinant is equal to zero:
\((t-1)(t-2) - (4)(3) = 0\)
#Step 6: Solve for t for the second matrix#
06
Solve the equation
Expanding and simplifying the equation, we get:
\(t^2 - 3t -10 = 0\)
Factorizing the quadratic equation, we find:
\((t - 5)(t + 2) = 0\)
So, the values of `t` that make the determinant equal to zero are \(t = 5\) and \(t = -2\).
#Final answer#:
(a) \(t = 2, 11\)
(b) \(t = 5, -2\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Equations
Linear equations are mathematical statements where each term is either a constant or the product of a constant and a single variable. In essence, they are algebraic expressions that form a straight line when graphed on a Cartesian plane. Consider the standard form of a linear equation: \(ax + b = 0\). Here, \(a\) and \(b\) are constants, and \(x\) is the variable we're solving for. A linear equation essentially shows the relationship between two variables.
- Characteristics: The graph is always a straight line.
- The equation has a constant rate of change, known as the slope.
- It can have either one solution, infinite solutions, or no solution.
Quadratic Equations
Quadratic equations are expressions that can be generally written in the form \(ax^2 + bx + c = 0\). They are called quadratic because the term "quad" generally means square, indicating the square of the variable \(x\). Here, \(a\), \(b\), and \(c\) are constants.
- Characteristics: The highest power of the variable is 2.
- These equations typically have two solutions, which can be found by factorization, completing the square, or using the quadratic formula.
- The graph of a quadratic equation is a parabola.
Factorization
Factorization refers to breaking down a complex expression into simpler terms or factors that when multiplied give back the original expression. In the context of quadratic equations, factorization involves rewriting the equation in a product form, where each factor could be set to zero to find the values of the variable.
- Method: Look for two numbers that multiply to the constant term \(c\) and add up to \(b\).
- Rewrite the equation using these numbers to break down the middle term.
- Factor by grouping and solve for the variable.