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Chapter 4: Problem 5

In Exercises 5- 12, evaluate the determinant of the given matrix by cofactor expansion along the indicated row. $$ \left(\begin{array}{rrr} 0 & 1 & 2 \\ -1 & 0 & -3 \\ 2 & 3 & 0 \end{array}\right) $$ along the first row

Short Answer

Expert verified
The determinant of the given matrix using cofactor expansion along the first row is: $$\det(A) = -12$$

Step by step solution

01

Apply cofactor expansion formula

We have: $$\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$ Our matrix is given by: $$ A = \left(\begin{array}{rrr} 0 & 1 & 2 \\\ -1 & 0 & -3 \\\ 2 & 3 & 0 \end{array}\right) $$ Let's identify the elements: $$a_{11} = 0, a_{12} = 1, a_{13} = 2$$ $$a_{21} = -1, a_{22} = 0, a_{23} = -3$$ $$a_{31} = 2, a_{32} = 3, a_{33} = 0$$ Now we can plug these elements into the formula: $$\det(A) = 0(0\cdot0 - -3\cdot3) - 1(-1\cdot0 - -3\cdot2) + 2(-1\cdot3 - 0\cdot2)$$
02

Calculate the determinant

Now we calculate the result: $$\det(A) = 0(0 - 9) - 1(0 - 6) + 2(-3 - 0)$$ $$\det(A) = 0 - 6 + 2(-3)$$ $$\det(A) = -6 - 6$$ $$\det(A) = -12$$ So, the determinant of the given matrix is -12.

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Most popular questions from this chapter

Suppose that \(A_{1}, A_{2}, \ldots\) are linearly independent sets of vectors, and that \(A_{1} \subseteq A_{2} \subseteq \ldots .\) Show that the union \(A=A_{1} \cup A_{2} \cup \ldots\) is also linearly independent.

Let \(r=\operatorname{rank}(A+B) .\) Find \(2 \times 2\) matrices \(A\) and \(B\) such that (a) \(r < \operatorname{rank}(A), \operatorname{rank}(\mathrm{B})\); (b) \(r=\operatorname{rank}(A)=\operatorname{rank}(B)\); (c) \(r > \operatorname{rank}(A), \operatorname{rank}(\mathrm{B})\).

Let \(K\) be a subfield of a field \(L,\) and let \(L\) be a subfield of a field \(E .\) (Thus, \(K \subseteq L \subseteq E\), and \(K\) is a subfield of \(E\).) Suppose \(E\) is of dimension \(n\) over \(L\), and \(L\) is of dimension \(m\) over \(K\). Show that \(E\) is of dimension \(m n\) over \(K\).

Let \(V\) be the set of infinite sequences \(\left(a_{1}, a_{2}, \ldots\right)\) in a field \(K .\) Show that \(V\) is a vector space over \(K\) with addition and scalar multiplication defined by \\[\left(a_{1}, a_{2}, \ldots\right)+\left(b_{1}, b_{2}, \ldots\right)=\left(a_{1}+b_{1}, a_{2}+b_{2}, \ldots\right) \quad \text { and } \quad k\left(a_{1}, a_{2}, \ldots\right)=\left(k a_{1}, k a_{2}, \ldots\right)\\]

Let (a) \(A=\left[\begin{array}{rrrrrr}1 & 2 & 1 & 3 & 1 & 6 \\ 2 & 4 & 3 & 8 & 3 & 15 \\ 1 & 2 & 2 & 5 & 3 & 11 \\ 4 & 8 & 6 & 16 & 7 & 32\end{array}\right]\) (b) \(B=\left[\begin{array}{llllll}1 & 2 & 2 & 1 & 2 & 1 \\ 2 & 4 & 5 & 4 & 5 & 5 \\ 1 & 2 & 3 & 4 & 4 & 6 \\ 3 & 6 & 7 & 7 & 9 & 10\end{array}\right]\) For each matrix (where \(C_{1}, \ldots, C_{6}\) denote its columns) (i) Find its row canonical form \(M\). (ii) Find the columns that are linear combinations of preceding columns. (iii) Find columns (excluding \(C_{6}\) ) that form a basis for the column space. (iv) Express \(C_{6}\) as a linear combination of the basis vectors obtained in (iii).
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