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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.

Quadratic Equation
A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( x \) is the variable, and \( a \), \( b \), and \( c \) are constants with \( a eq 0 \).Quadratic equations are called such because the highest exponent of the variable is two, making them second-degree polynomials.They generally have two solutions, which can either be real or complex numbers, depending on the discriminant.To solve a quadratic equation, one can use several methods:
  • Factoring: Expressing the quadratic as a product of its factors.
  • Quadratic Formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • Completing the Square: Rewriting the quadratic to reveal the square form.
  • Graphing: Finding the x-intercepts of the quadratic graph.
When we solve a quadratic equation, like in our exercise, we aim to find the values of the variable that satisfy the equation, meaning where it equals zero.The ability of a quadratic equation to express two solutions makes it perfect for applications involving parabolic paths, optimization, and more.
Determinant Formula
The determinant of a matrix is a special number that is calculated from its elements and provides key insights into the matrix's properties.For a 2x2 matrix, applying the determinant formula is straightforward:Given a matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), its determinant \( |A| \) is given by:\[|A| = ad - bc\]In the context of matrices, the determinant can tell us several things:
  • Singularity: If the determinant is zero, the matrix is singular, meaning it does not have an inverse.
  • Relation to Quadratics: In our exercise, the determinant is equated to zero to form a quadratic equation whose solutions define key parameters for the matrix.
  • Geometric Interpretation: The absolute value of a determinant represents the scaling factor for area or volume when matrices are seen as linear transformations.
Understanding the determinant is crucial in linear algebra and is fundamental for matrix manipulations, solving systems of equations, and more.
Factorization
Factorization involves breaking down a complex expression into simpler parts, or "factors," that when multiplied together, recreate the original expression.It's a powerful tool particularly for solving polynomial equations like quadratics.To factorize a quadratic of the form \( ax^2 + bx + c \), our goal is to rewrite it as \( (px + q)(rx + s) \), where \( p, q, r, \) and \( s \) are numbers that satisfy:
  • \( pr = a \)
  • \( qs = c \)
  • \( ps + qr = b \)
Factorization not only simplifies equations for solving but also provides insight into the roots of the polynomial.In our exercise, once the quadratic equations were formed from the determinant expressions, they were factorized to find critical values of \( t \). These roots essentially define where the determinant of the matrix equals zero, indicating singularity.Mastering factorization is crucial for algebraic manipulations, making it a fundamental skill in mathematics education.

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Most popular questions from this chapter

Prove Theorem 8.1: \(\left|A^{T}\right|=|A|\) If \(A=\left[a_{i j}\right]\), then \(A^{T}=\left[b_{i j}\right]\), with \(b_{i j}=a_{j i}\). Hence, $$ \left|A^{T}\right|=\sum_{\sigma \in S_{x}}(\operatorname{sgn} \sigma) b_{1 \sigma(1)} b_{2 \sigma(2)} \cdots b_{n \sigma(n)}=\sum_{\sigma \in S_{\underline{z}}}(\operatorname{sgn} \sigma) a_{\sigma(1), 1} a_{\sigma(2), 2} \cdots a_{\sigma(n), n} $$ Let \(\tau=\sigma^{-1}\). By Problem \(8.21 \operatorname{sgn} \tau=\operatorname{sgn} \sigma\), and \(a_{\sigma(1), 1} a_{\sigma(2), 2} \cdots a_{\sigma(n), n}=a_{1 \tau(1)} a_{2 \tau(2)} \cdots a_{n \tau(n)} .\) Hence, $$ \left|A^{T}\right|=\sum_{\sigma \in S_{n}}(\operatorname{sgn} \tau) a_{1 t(1)} a_{2 \tau(2)} \cdots a_{n z(n)} $$However, as \(\sigma\) runs through all the elements of \(S_{n}, \tau=\sigma^{-1}\) also runs through all the elements of \(S_{n}\). Thus, \(\left|A^{T}\right|=|A|\)

Let \(D\) be a 2 -linear, alternating function. Show that \(D(A, B)=-D(B, A)\)

\(\operatorname{Let} A=\left[\begin{array}{rrrr}2 & 1 & -3 & 4 \\ 5 & -4 & 7 & -2 \\ 4 & 0 & 6 & -3 \\ 3 & -2 & 5 & 2\end{array}\right]\) (a) Find \(A_{23},\) the cofactor (signed minor) of 7 in \(A\) (b) Find the minor and the signed minor of the submatrix \(M=A(2,4 ; 2,3)\) (c) Find the principal minor determined by the first and third diagonal entries- -that is, by \(M=A(1,3 ; \quad 1,3)\)

Let \(\sigma, \tau \in S_{n} .\) Show that \(\operatorname{sgn}(\tau \circ \sigma)=(\operatorname{sgn} \tau)(\operatorname{sgn} \sigma) .\) Thus, the product of two even or two odd permutations is even, and the product of an odd and an even permutation is odd.

For \(k=1,2,3,4\), find the sum \(S_{k}\) of all principal minors of order \(k\) for (a) \(A=\left[\begin{array}{rrrr}1 & 2 & 3 & -1 \\ 1 & -2 & 0 & 5 \\ 0 & 1 & -2 & 2 \\ 4 & 0 & -1 & -3\end{array}\right]\) (b) \(B=\left[\begin{array}{llll}1 & 2 & 1 & 2 \\ 0 & 1 & 2 & 3 \\ 1 & 3 & 0 & 4 \\ 2 & 7 & 4 & 5\end{array}\right]\)
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