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Chapter 3: Problem 60

Evaluate the determinant, in which the entries are functions. Determinants of this type occur when changes of variables are made in calculus. $$\left|\begin{array}{cc} 3 x^{2} & -3 y^{2} \\ 1 & 1 \end{array}\right|$$

Short Answer

Expert verified
The determinant of the given matrix is \(3x^2 + 3y^2\).

Step by step solution

01

Recall Formula for Determinant of a 2x2 Matrix

The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is calculated by subtracting the product of the elements diagonally from the product of the other diagonal. It is calculated as following: \(ad - bc\). Our matrix is: \(\begin{bmatrix} 3x^2 & -3y^2 \\ 1 & 1 \end{bmatrix}\)
02

Apply Formula to Given Matrix

When we substitute the values from our matrix into the determinant formula, we get: \((3x^2 * 1) - (-3y^2 * 1)\).
03

Simplify the Expression

The formula then simplifies to: \(3x^2 + 3y^2\).

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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. $$\begin{aligned} 3 x_{1}-2 x_{2}+9 x_{3}+4 x_{4} &=35 \\ -x_{1} \quad-9 x_{3}-6 x_{4}=-17 \\ 2 x_{3}+x_{4} =5 \\ 2 x_{1}+2 x_{2}\quad\quad +8 x_{4}=-4 \end{aligned}$$

Use a graphing utility with matrix capabilities to determine whether \(A\) is orthogonal. To test for orthogonality, find (a) \(A^{-1},\) (b) \(A^{T},\) and (c) \(|A|,\) and verify that \(A^{-1}=A^{T}\) and \(|A|=\pm 1\) $$A=\left[\begin{array}{rrr} \frac{2}{3} & -\frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} & -\frac{2}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{2}{3} \end{array}\right]$$

Use a graphing utility with matrix capabilities to determine whether \(A\) is orthogonal. To test for orthogonality, find (a) \(A^{-1},\) (b) \(A^{T},\) and (c) \(|A|,\) and verify that \(A^{-1}=A^{T}\) and \(|A|=\pm 1\) $$A=\left[\begin{array}{llr} \frac{3}{5} & 0 & -\frac{4}{5} \\ 0 & 1 & 0 \\ \frac{4}{5} & 0 & \frac{3}{5} \end{array}\right]$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{array}{l} \frac{5}{6} x_{1}-x_{2}=-20 \\ \frac{4}{3} x_{1}-\frac{7}{2} x_{2}=-51 \end{array}$$

Verify that \(\lambda_{i}\) is an eigenvalue of \(A\) and that \(\mathbf{x}_{i}\) is a corresponding eigenvector. $$\begin{array}{l} A=\left[\begin{array}{ll} 1 & 2 \\ 0 & -3 \end{array}\right] ; \quad \lambda_{1}=1, \quad \mathbf{x}_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \\ \lambda_{2}=-3, \quad \mathbf{x}_{2}=\left[\begin{array}{r} -1 \\ 2 \end{array}\right] \end{array}$$
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