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Chapter 4: Problem 8

Prove that if \(A \in \mathrm{M}_{2 \times 2}(F)\) is upper triangular, then \(\operatorname{det}(A)\) equals the product of the diagonal entries of \(A\).

Short Answer

Expert verified
Given a 2x2 upper triangular matrix \(A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}\), calculating its determinant using the formula \(det(A) = ad - bc\), where c=0, yields \(det(A) = ad\). Thus, the determinant of A is the product of its diagonal elements a and d.

Step by step solution

01

Define the matrix A

Let's define the given 2x2 upper triangular matrix A with elements from a field F: \(A = \begin{pmatrix} a & b \\ 0 & d \end{pmatrix}\) where a, b, d are elements of the field F. Because A is upper triangular, the lower left entry is 0.
02

Calculate the determinant of A

To calculate the determinant of the 2x2 matrix A, we can use the formula det(A) = ad - bc. However, since A is an upper triangular matrix, the entry c is equal to 0. Therefore, the determinant formula becomes det(A) = ad - b*0. Now compute the determinant of A: \(det(A) = ad - b * 0 = ad\)
03

Conclude and interpret the result

We have found that the determinant of the upper triangular 2x2 matrix A is given by: \(det(A) = ad\) This result shows that the determinant of A is indeed the product of its diagonal elements a and d. We have proved the required statement for a 2x2 upper triangular matrix.

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Most popular questions from this chapter

Find the coordinate vector of \(A=\left[\begin{array}{ll}2 & 3 \\ 4 & -7\end{array}\right]\) in the real vector space \(\mathbf{M}=\mathbf{M}_{2,2}\) relative to (a) the basis \(S=\left\\{\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right],\left[\begin{array}{rr}1 & -1 \\ 1 & 0\end{array}\right],\left[\begin{array}{rr}1 & -1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right]\right\\}\), (b) the usual basis \(E=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}\)

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 \(K\).

Find a homogeneous system whose solution space is spanned by the following sets of three vectors: (a) \(\quad(1,-2,0,3,-1),(2,-3,2,5,-3),(1,-2,1,2,-2)\); (b) \(\quad(1,1,2,1,1),(1,2,1,4,3),(3,5,4,9,7)\).

Prove that if \(\delta: M_{n \times n}(F) \rightarrow F\) is an alternating \(n\)-linear function, then there exists a scalar \(k\) such that $\delta(A)=k \operatorname{det}(A)\( for all \)A \in \mathrm{M}_{n \times n}(F)$.

Find a basis and the dimension of the subspace \(W\) of \(\mathbf{P}(t)\) spanned by (a) \(u=t^{3}+2 t^{2}-2 t+1, \quad v=t^{3}+3 t^{2}-3 t+4, \quad w=2 t^{3}+t^{2}-7 t-7\), (b) \(u=t^{3}+t^{2}-3 t+2, \quad v=2 t^{3}+t^{2}+t-4, \quad w=4 t^{3}+3 t^{2}-5 t+2\).
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