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Chapter 4: Problem 5

Prove that if \(B\) is the matrix obtained by interchanging the rows of a $2 \times 2\( matrix \)A\(, then \)\operatorname{det}(B)=-\operatorname{det}(A)$.

Short Answer

Expert verified
For a general 2x2 matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), interchanging its rows results in matrix \(B = \begin{pmatrix} c & d \\ a & b \end{pmatrix}\). The determinants of A and B are \(\operatorname{det}(A) = ad - bc\) and \(\operatorname{det}(B) = cb - da\), respectively. By factoring out a -1 from the expression for \(\operatorname{det}(B)\), we find \(\operatorname{det}(B) = - \operatorname{det}(A)\), proving the required relationship.

Step by step solution

01

Define a general 2x2 matrix A and create matrix B by interchanging its rows

Let's consider a general 2x2 matrix A: \[ A= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \] Now, let's create the matrix B by interchanging the rows of matrix A: \[ B= \begin{pmatrix} c & d \\ a & b \end{pmatrix} \]
02

Find the determinant of matrix A

To find the determinant of a 2x2 matrix, we use the formula: \[ \operatorname{det}(A) = ad - bc \] For matrix A: \[ \operatorname{det}(A) = ad - bc \]
03

Find the determinant of matrix B

Now we will find the determinant of matrix B using the same formula: For matrix B: \[ \operatorname{det}(B) = cb - da \]
04

Compare the determinants of matrices A and B and prove the relationship

Now, we need to prove that the determinant of matrix B is equal to the negative determinant of matrix A. Let's compare their determinants: \[ \operatorname{det}(B) = cb - da \] Notice that we can factor out a -1 from the expression: \[ \operatorname{det}(B) = -1(da - cb) \] Now, if we recall the determinant of matrix A: \[ \operatorname{det}(A) = ad - bc \] We can rewrite it as: \[ \operatorname{det}(A) = da - cb \] Comparing the rewritten expression of the determinant of A with the factored expression of the determinant of B, we can see that: \[ \operatorname{det}(B) = - \operatorname{det}(A) \] We have now proven that if we interchange the rows of a 2x2 matrix A to get matrix B, the determinant of B is equal to the negative determinant of A.

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Most popular questions from this chapter

Let \(V\) be the vector space of \(n\) -square matrices over a field \(K\). Show that \(W\) is a subspace of \(V\) if \(W\) consists of all matrices \(A=\left[a_{i j}\right]\) that are (a) symmetric \(\left(A^{T}=A \text { or } a_{i j}=a_{j i}\right)\), (b) (upper) triangular, (c) diagonal, (d) scalar.

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

Let the rows of \(A \in M_{n \times n}(F)\) be \(a_{1}, a_{2}, \ldots, a_{n}\), and let \(B\) be the matrix in which the rows are $a_{n}, a_{n-1}, \ldots, a_{1}\(. Calculate \)\operatorname{det}(B)\( in terms of \)\operatorname{det}(A)$.

Let \(U_{1}, U_{2}, U_{3}\) be the following subspaces of \(\mathbf{R}^{3}\) : \\[U_{1}=\\{(a, b, c): a=c\\}, \quad U_{2}=\\{(a, b, c): a+b+c=0\\}, \quad U_{3}=\\{(0,0, c)\\}\\] Show that \((\mathrm{a}) \mathbf{R}^{3}=U_{1}+U_{2},\) (b) \(\mathbf{R}^{3}=U_{2}+U_{3},(\mathrm{c}) \mathbf{R}^{3}=U_{1}+U_{3} .\) When is the sum direct?

Suppose that \(U\) and \(W\) are subspaces of a vector space \(V\) and that \(S=\left\\{u_{i}\right\\}\) spans \(U\) and \(S^{\prime}=\left\\{w_{j}\right\\}\) spans \(W .\) Show that \(S \cup S^{\prime}\) spans \(U+W .\) (Accordingly, by induction, if \(S_{i}\) spans \(W_{i},\) for \(i=1,2, \ldots, n,\) then \(S_{1} \cup \ldots \cup S_{n}\) spans \(W_{1}+\cdots+W_{n}\)).
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