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Chapter 8: Problem 3

Compute the determinant of each of the following matrices: (a) \(A=\left[\begin{array}{lll}2 & 3 & 4 \\ 5 & 6 & 7 \\ 8 & 9 & 1\end{array}\right]\) (b) \(B=\left[\begin{array}{rrrr}4 & -6 & 8 & 9 \\ 0 & -2 & 7 & -3 \\ 0 & 0 & 5 & 6 \\ 0 & 0 & 0 & 3\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{rrr}\frac{1}{2} & -1 & -\frac{1}{3} \\\ \frac{3}{4} & \frac{1}{2} & -1 \\ 1 & -4 & 1\end{array}\right]\)

Short Answer

Expert verified
The determinants of the given matrices are: \( det(A) = 63 \), \( det(B) = -40 \), and \( det(C) = \dfrac{19}{6} \).

Step by step solution

01

Matrix A determinant calculation

Use the standard formula for finding the determinant of a 3x3 matrix. For matrix A, that is: \( det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \) where A = \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} =\begin{bmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \\ 8 & 9 & 1 \\ \end{bmatrix}\)
02

Calculate determinant of A

Plugging the values from matrix A into the determinant formula, we have: \( det(A)= 2(6(1) - 7(9)) - 3(5(1) - 7(8)) + 4(5(9) - 6(8)) \) Now, we can solve for the determinant of A: \( det(A) = 2(-63) - 3(-51) + 4(9) = -126 + 153 + 36 = 63 \)
03

Matrix B determinant calculation

Since matrix B is upper-triangular, its determinant is simply the product of its diagonal elements. This means: \( det(B) = p * q * r * s \) where B=\(\begin{bmatrix} p & . & . & . \\ 0 & q & . & . \\ 0 & 0 & r & . \\ 0 & 0 & 0 & s \end{bmatrix} = \begin{bmatrix} 4 & -6 & 8 & 9 \\ 0 & -2 & 7 & -3 \\ 0 & 0 & 5 & 6 \\ 0 & 0 & 0 & 3 \end{bmatrix}\)
04

Calculate determinant of B

Plugging the values from matrix B into the determinant formula, we have: \( det(B) = 4 * (-2) * 5 * 3 = -40 \)
05

Matrix C determinant calculation

Use the standard formula for finding the determinant of a 3x3 matrix. For matrix C, that is: \( det(C) = a(ei - fh) - b(di - fg) + c(dh - eg) \) where C = \(\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix} =\begin{bmatrix} \frac{1}{2} & -1 & -\frac{1}{3} \\ \frac{3}{4} & \frac{1}{2} & -1 \\ 1 & -4 & 1 \\\end{bmatrix}\)
06

Calculate determinant of C

Plugging the values from matrix C into the determinant formula, we have: \( det(C) = \dfrac{1}{2}\left(\dfrac{1}{2}\left(1\) - (-1)(-4)\) - (-1)\left(\dfrac{3}{4}\left(1\) - (-1)(1)\) - \dfrac{1}{3}(\dfrac{3}{4}(-4) - \dfrac{1}{2}(-1))\) Now, we can solve for the determinant of C: \( det(C) = \dfrac{1}{2}(2 + 1 - 8) - (-1)(4 + 3) + \dfrac{1}{3}(-12 + \dfrac{1}{2}) = -\dfrac{1}{2}(5) + 7 - \dfrac{1}{3}(11\frac{1}{2}) = -\dfrac{5}{2} + 7 - \dfrac{23}{6} \) \( det(C) = -\dfrac{5}{2} - \dfrac{23}{6} + \dfrac{42}{6} = -\dfrac{43}{6} + 42 = \dfrac{19}{6} \) We've found the determinants of the matrices: - \( det(A) = 63 \) - \( det(B) = -40 \) - \( det(C) = \dfrac{19}{6} \)

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Most popular questions from this chapter

Let \(A=\left[\begin{array}{rrrr}1 & 3 & 0 & -1 \\ -4 & 2 & 5 & 1 \\ 1 & 0 & 3 & -2 \\ 3 & -2 & 1 & 4\end{array}\right] .\) Find the number \(N_{k}\) and sum \(S_{k}\) of principal minors of order: (a) \(k=1\) (b) \(k=2\) (c) \(k=3\) (d) \(\quad k=4\)

Prove Theorem \(8.3(\mathrm{i}):\) If two rows (columns) of \(A\) are interchanged, then \(|B|=-|A|\)

Consider a permutation \(\sigma=j_{1} j_{2} \ldots j_{n} .\) Let \(\left\\{e_{i}\right\\}\) be the usual basis of \(K^{n},\) and let \(A\) be the matrix whose \(i\) th row is \(e_{j_{i}}\left[\text { i.c., } A=\left(e_{j_{1}}, e_{j_{2}}, \ldots, e_{j_{n}}\right)\right] .\) Show that \(|A|=\operatorname{sgn} \sigma\)

Find the volume \(V(S)\) of the parallelepiped \(S\) in \(\mathbf{R}^{4}\) determined by the following vectors: $$u_{1}=(1,-2,5,-1), u_{2}=(2,1,-2,1), u_{3}=(3,0,1-2), u_{4}=(1,-1,4,-1)$$

Prove Theorem \(8.10\) (Cramer's rule): The (square) system \(A X=B\) has a unique solution if and only if \(D \neq 0 .\) In this case, \(x_{i}=N_{i} / D\) for each \(i\). By previous results, \(A X=B\) has a unique solution if and only if \(A\) is invertible, and \(A\) is invertible if and only if \(D=|A| \neq 0\) Now suppose \(D \neq 0 .\) By Theorem $8.9, A^{-1}=(1 / D)(\operatorname{adj} A) .\( Multiplying \)A X=B\( by \)A^{-1}$, we obtain $$ X=A^{-1} A X=(1 / D)(\operatorname{adj} A) B $$ Note that the ith row of \((1 / D)(\operatorname{adj} A)\) is $(1 / D)\left(A_{1 i}, A_{2 i}, \ldots, A_{n i}\right) .\( If \)B=\left(b_{1}, b_{2}, \ldots, b_{n}\right)^{T}$, then, by (1), $$ x_{i}=(1 / D)\left(b_{1} A_{1 i}+b_{2} A_{2 i}+\cdots+b_{n} A_{n i}\right) $$ However, as in Problem $8.33, b_{1} A_{1 i}+b_{2} A_{2 i}+\cdots+b_{n} A_{n i}=N_{i}$, the determinant of the matrix obtained by replacing the ith column of \(A\) by the column vector \(B .\) Thus, \(x_{i}=(1 / D) N_{i}\), as required.
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