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Chapter 3: Problem 68

Evaluate the determinants to verify the equation. $$\left|\begin{array}{ll} w & c x \\ y & c z \end{array}\right|=c\left|\begin{array}{ll} w & x \\ y & z \end{array}\right|$$

Short Answer

Expert verified
The determinant equation is valid based on the distributive property of determinants. Hence, the equation is verified.

Step by step solution

01

Understand the distributive property

Firstly, we examine the property of determinants, where a scalar multiples a determinant. If a scalar \(c\) multiples one entire row or column of a matrix \(A\), the determinant \(|cA|\) equals \(c |A|\).
02

Apply the property to the matrix

Here the matrix \(\begin{array}{ll} w & c x \\ y & c z \end{array}\) can indeed be expressed as \(c\) times the matrix \(\begin{array}{ll} w & x \\ y & z \end{array}\).
03

Verify the equation

So if we evaluate the determinant of the two matrices, we get \(c\) times the determinant of the matrix \(\begin{array}{ll} w & x \\ y & z \end{array}\) equals to the determinant of the matrix \(\begin{array}{ll} w & c x \\ y & c z \end{array}\). Hence, the equation is verified.

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Most popular questions from this chapter

Find an equation of the plane passing through the three points. $$(0,0,0),(1,-1,0),(0,1,-1)$$

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 \(|A|=\pm 1\) $$A=\left[\begin{array}{llr} \frac{3}{5} & 0 & -\frac{4}{5} \\ 0 & 1 & 0 \\ \frac{4}{5} & 0 & \frac{3}{5} \end{array}\right]$$

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Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 4 x_{1}-2 x_{2}+3 x_{3}=-2 \\ 2 x_{1}+2 x_{2}+5 x_{3}=16 \\ 8 x_{1}-5 x_{2}-2 x_{3}=4 \end{array}$$
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