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Chapter 4: Problem 13

In Exercises 13-22, evaluate the determinant of the given matrix by any legitimate method. $$ \left(\begin{array}{lll} 0 & 0 & 1 \\ 0 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right) $$

Short Answer

Expert verified
The determinant of the given 3x3 matrix is \(-8\).

Step by step solution

01

Choose a row or a column for Laplace expansion

For simplicity, we'll use the first row to perform the Laplace expansion. The first row contains a "1" in the third column and "0" in the first two columns, so it will simplify the calculations.
02

Apply the Laplace expansion using the first row

Apply the Laplace expansion using the first row, and remember the rule of alternating signs (positive, negative, positive): Determinant = \((+1)\times0\times\text{minor}_{11}\) \(- (+1)\times0\times\text{minor}_{12}\) \(+ (+1)\times1\times\text{minor}_{13}\) Since the first two terms have a multiplication with 0, they will result in 0: Determinant = \(1\times\text{minor}_{13}\)
03

Find the 2x2 submatrix for minor_{13}

To find the 2x2 submatrix for minor_{13}, remove the first row and the third column from the given matrix: Minor for element 1,3 = \(\begin{pmatrix} 0 & 2\\ 4 & 5 \end{pmatrix}\)
04

Calculate the determinant of the 2x2 submatrix

To find the determinant of the 2x2 submatrix minor_{13}, use the formula: \(\text{det}(minor_{13}) = (0\times5) - (4\times2) = -8\)
05

Multiply the factor with the determinant of the 2x2 submatrix

Now, multiply the factor found in Step 2 with the determinant of the 2x2 submatrix found in Step 4: Determinant = \(1\times(-8) = -8\) Therefore, the determinant of the given 3x3 matrix is -8.

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Most popular questions from this chapter

Relative to the basis \(S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,1),(2,3)\\}\) of \(\mathbf{R}^{2},\) find the coordinate vector of \(v,\) where (a) \(v=(4,-3)\), (b) \(v=(a, b)\).

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Prove that $\operatorname{det}(A B)=\operatorname{det}(A) \cdot \operatorname{det}(B)\( for any \)A, B \in \mathrm{M}_{2 \times 2}(F)$.

Find the value of \(k\) that satisfies the following equation: $$ \operatorname{det}\left(\begin{array}{ccc} 2 a_{1} & 2 a_{2} & 2 a_{3} \\ 3 b_{1}+5 c_{1} & 3 b_{2}+5 c_{2} & 3 b_{3}+5 c_{3} \\ 7 c_{1} & 7 c_{2} & 7 c_{3} \end{array}\right)=k \operatorname{det}\left(\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right) . $$
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