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Chapter 4: Problem 3

Compute the determinants of the following matrices in \(M_{2 \times 2}(C)\). (a) \(\left(\begin{array}{rr}-1+i & 1-4 i \\ 3+2 i & 2-3 i\end{array}\right)\) (b) \(\left(\begin{array}{rc}5-2 i & 6+4 i \\ -3+i & 7 i\end{array}\right)\) (c) \(\left(\begin{array}{rr}2 i & 3 \\ 4 & 6 i\end{array}\right)\)

Short Answer

Expert verified
The short version of the answer is: \(det(A) = (-1+i)(2-3i) - (1-4i)(3+2i) = 11 - 7i\) \(det(B) = (5-2i)(7i) - (6+4i)(-3+i) = -10 - 38i\) \(det(C) = (2i)(6i) - (3)(4) = -12 - 12i\)

Step by step solution

01

Matrix (a) Determinant Calculation

Let matrix (a) be \(A = \begin{pmatrix} -1+i & 1-4i \\ 3+2i & 2-3i \end{pmatrix}\). To find the determinant, we use the formula \(ad-bc\). For this matrix: - \(a = -1 + i\) - \(b = 1 - 4i\) - \(c = 3 + 2i\) - \(d = 2 - 3i\) \(det(A) = (-1+i)(2-3i) - (1-4i)(3+2i)\)
02

Matrix (b) Determinant Calculation

Let matrix (b) be \(B = \begin{pmatrix} 5-2i & 6+4i \\ -3+i & 7i \end{pmatrix}\). To find the determinant, we use the formula \(ad-bc\). For this matrix: - \(a = 5 - 2i\) - \(b = 6 + 4i\) - \(c = -3 + i\) - \(d = 7i\) \(det(B) = (5-2i)(7i) - (6+4i)(-3+i)\)
03

Matrix (c) Determinant Calculation

Let matrix (c) be \(C = \begin{pmatrix} 2i & 3 \\ 4 & 6i \end{pmatrix}\). To find the determinant, we use the formula \(ad-bc\). For this matrix: - \(a = 2i\) - \(b = 3\) - \(c = 4\) - \(d = 6i\) \(det(C) = (2i)(6i) - (3)(4)\)

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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) \(v=(a, b)\).

The vectors \(u_{1}=(1,2,0), u_{2}=(1,3,2), u_{3}=(0,1,3)\) form a basis \(S\) of \(\mathbf{R}^{3}\). Find the coordinate vector \([v]\) of \(v\) relative to \(S\) where (a) \(v=(2,7,-4),\) (b) \(v=(a, b, c)\).

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 \(v_{1}, \ldots, v_{n}\) belong to a vector space \(V\) over \(K,\) and let \(P=\left[a_{i j}\right]\) be an \(n\) -square matrix over \(K\). Let \\[w_{1}=a_{11} v_{1}+a_{12} v_{2}+\cdots+a_{1 n} v_{n}, \quad \ldots, \quad w_{n}=a_{n 1} v_{1}+a_{n 2} v_{2}+\cdots+a_{n n} v_{n}\\] (a) Suppose \(P\) is invertible. Show that \(\left\\{w_{i}\right\\}\) and \(\left\\{v_{i}\right\\}\) span the same space; hence, \(\left\\{w_{i}\right\\}\) is independent if and only if \(\left\\{v_{i}\right\\}\) is independent. (b) Suppose \(P\) is not invertible. Show that \(\left\\{w_{i}\right\\}\) is dependent (c) Suppose \(\left\\{w_{i}\right\\}\) is independent. Show that \(P\) is invertible.

Suppose \(W_{1}, W_{2}, \ldots, W_{r}\) are subspaces of a vector space \(V\). Show that (a) \(\operatorname{span}\left(W_{1}, W_{2}, \ldots, W_{r}\right)=W_{1}+W_{2}+\cdots+W_{r}\). (b) If \(S_{i}\) spans \(W_{i}\) for \(i=1, \ldots, r,\) then \(S_{1} \cup S_{2} \cup \ldots \cup S_{r}\) spans \(W_{1}+W_{2}+\cdots+W_{r}\).
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