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Chapter 3: Problem 58

Find the values of \(\lambda\) for which the determinant is zero. $$\left|\begin{array}{rrr} \lambda & 0 & 1 \\ 0 & \lambda & 3 \\ 2 & 2 & \lambda-2 \end{array}\right|$$

Short Answer

Expert verified
The values of \( \lambda \) for which the determinant is zero are \( \lambda = 0, -\sqrt{8}, \sqrt{8} \)

Step by step solution

01

Write down the determinant formula for 3x3 matrix

For a 3x3 matrix in the form of \(\left| \begin{matrix}a & b & c \d & e & f \g & h & i \\end{matrix} \right|\), the determinant \(\Delta\) is calculated by the formula \[ \Delta = aei + bfg + cdh - ceg - bdi - afh \]
02

Substitute the values into the formula

Substitute the given values from the matrix into the formula: \[ \Delta = \lambda * \lambda * (\lambda - 2) + 0 * 2 * 2 + 1 * 0 * 2 - 2 * 0* \lambda - 2 * \lambda * 3 - 1 * 0 * (\lambda - 2) \]. After simplification, this gives \[ \Delta = \lambda^3 - 2\lambda - 6\lambda \] which can be further simplified to \[ \Delta = \lambda^3 - 8\lambda \].
03

Set the determinant equal to zero and solve for lambda

In order for the determinant to be zero, set \( \Delta = 0 \). This gives us the equation \( \lambda^3 - 8\lambda = 0 \). It can be rewritten as \( \lambda(\lambda^2 - 8) = 0 \). From this we can find the values of \( \lambda \) which are \( \lambda = 0, -\sqrt{8}, \sqrt{8} \).

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Most popular questions from this chapter

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 18 x_{1}+12 x_{2}=13 \\ 30 x_{1}+24 x_{2}=23 \end{array}$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{aligned} -x_{1}-x_{2} \quad+x_{4}=-8 \\ 3 x_{1}+5 x_{2}+5 x_{3} \quad=24 \\ 2 x_{3}+x_{4} =-6 \\ -2 x_{1}-3 x_{2}-3 x_{3} \quad=-15 \end{aligned}$$

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 20 x_{1}+8 x_{2}=11 \\ 12 x_{1}-24 x_{2}=21 \end{array}$$

Prove that if \(A\) is an orthogonal matrix, then \(|A|=\pm 1\)

Use Cramer's Rule to solve the system of linear equations, if possible. $$\begin{array}{l} 2 x_{1}-x_{2}=-10 \\ 3 x_{1}+2 x_{2}=-1 \end{array}$$
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