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Chapter 16: Problem 30

Find the value of the determinant \(\left[\begin{array}{rrr}3 & 2 & 1 \\ 0 & -1 & -1 \\ 2 & 0 & 2\end{array} \mid\right.\) a) 8 b) 4 d) \(-4\) c) \(-8\)

Short Answer

Expert verified
Question: Find the value of the given 3x3 determinant: \(\left[\begin{array}{rrr}3 & 2 & 1 \\\ 0 & -1 & -1 \\\ 2 & 0 & 2\end{array} \mid\right.\) Options: a) 8 b) 4 c) \(-4\) d) \(-8\) Answer: d) \(-8\)

Step by step solution

01

Write down the given determinant and options

We are given a 3x3 determinant: \(\left[\begin{array}{rrr}3 & 2 & 1 \\\ 0 & -1 & -1 \\\ 2 & 0 & 2\end{array} \mid\right.\) And we need to find its value from the given options: a) 8 b) 4 c) \(-4\) d) \(-8\)
02

Expand along the first row

We will expand the determinant along the first row. So, find the minor of each element in the first row and multiply it by its corresponding cofactor (-1) raised to the power given by the sum of its row and column number. The determinant value is the sum of these products. Determinant = \(3 \cdot M_{11} - 2 \cdot M_{12} + 1 \cdot M_{13}\) Where \(M_{ij}\) represents the minor of the element in the ith row and jth column.
03

Calculate each minor

Minor M_11 is the determinant of the 2x2 matrix that remains after removing the 1st row and 1st column: \(M_{11} = \mathrm{det} \begin{vmatrix} -1 & -1 \\ 0 & 2\end{vmatrix} = (-1) \cdot 2 - (-1) \cdot 0 = -2\) Minor M_12 is the determinant of the 2x2 matrix that remains after removing the 1st row and 2nd column: \(M_{12} = \mathrm{det} \begin{vmatrix} 0 & -1 \\ 2 & 2\end{vmatrix} = 0 \cdot 2 - (-1) \cdot 2 = 2\) Minor M_13 is the determinant of the 2x2 matrix that remains after removing the 1st row and 3rd column: \(M_{13} = \mathrm{det} \begin{vmatrix} 0 & -1 \\ 2 & 0\end{vmatrix} = 0 \cdot 0 - (-1) \cdot 2 = 2\)
04

Calculate the determinant value

Now, substitute the values of \(M_{11}\), \(M_{12}\), and \(M_{13}\) into the expression in Step 2: Determinant = \(3 \cdot (-2) - 2 \cdot (2) + 1 \cdot (2) = -6 - 4 + 2 = -8\)
05

Choose the correct option

The value of the determinant is -8, which corresponds to option d) \(-8\). So, the correct answer is d) \(-8\).

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Most popular questions from this chapter

The expression \(\tan \theta \sec \theta\left(1-\sin ^{2} \theta\right) / \cos \theta \operatorname{simplifies~to~}\) a) \(\sin \theta\) b) \(\cos \theta\) c) \(\tan \theta\) d) \(\sec \theta\)

Express \((3+2 i) e^{2 i t}+(3-2 i) e^{-2 i t}\) in terms of trigonometric functions. a) \(3 \cos 2 t-4 \sin 2 t\) c) \(6 \cos 2 t-4 \sin 2 t\) b) \(3 \cos 2 t-2 \sin 2 t\) d) \(3 \sin 2 t+2 \sin 2 t\)

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Solve for \(\left[x_{i}\right]\) $$ \begin{aligned} 3 x_{1}+2 x_{2} &=-2 \\ x_{1}-x_{2}+x_{3} &=0 \\ 4 x_{1}+2 x_{3} &=4 \end{aligned} $$ a) \(\left[\begin{array}{r}2 \\ 4 \\ -6\end{array}\right]\) \(\left[\begin{array}{l}2 \\ 4 \\ 8\end{array}\right]\) d) \(\left[\begin{array}{r}8 \\ 14\end{array}\right]\)
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