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Chapter 3: Problem 21

Use expansion by cofactors to find the determinant of the matrix. $$\left[\begin{array}{rrr} 2 & 4 & 6 \\ 0 & 3 & 1 \\ 0 & 0 & -5 \end{array}\right]$$

Short Answer

Expert verified
The determinant of the given matrix is -10.

Step by step solution

01

Identify the formula for 3x3 determinant

The determinant of a 3x3 matrix is given by: \[\begin{bmatrix}a & b & c \d & e & f \g & h & i \end{bmatrix}\]is given by \(a(ei 鈭� fh) 鈭� b(di 鈭� fg) + c(dh 鈭� eg)\)
02

Substitute the given matrix into the formula

Placing the values from the provided matrix into the formula:\[\begin{bmatrix}2 & 4 & 6 \0 & 3 & 1 \0 & 0 & -5 \end{bmatrix}\]We have \(2(3*-5 - 1*0) - 4(0*-5 - 1*0) + 6(0*0 - 3*0)\).
03

Carry out the calculations

Now calculate the expressions inside the brackets, then calculate the final result. To do so, perform the operations: \(2*(-15 - 0) - 4*(-5 - 0) + 6*(0 - 0)\). This gives a determinant of \(-30 - (-20) = -10\).

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Most popular questions from this chapter

Find the volume of the tetrahedron having the given vertices. $$(3,-1,1),(4,-4,4),(1,1,1),(0,0,1)$$

Find (a) the characteristic equation, (b) the eigenvalues, and (c) the corresponding eigenvectors of the matrix. $$\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right]$$

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{array}{l} 4 x_{1}-x_{2}+x_{3}=-5 \\ 2 x_{1}+2 x_{2}+3 x_{3}=10 \\ 5 x_{1}-2 x_{2}+6 x_{3}=1 \end{array}$$

Prove that if \(A\) is an orthogonal matrix, then \(|A|=\pm 1\)

Use a graphing utility or a computer software program with matrix capabilities and Cramer's Rule to solve for \(x_{1}\) if possible. $$\begin{array}{rr} -8 x_{1}+7 x_{2}-10 x_{3}= & -151 \\ 12 x_{1}+3 x_{2}-5 x_{3}= & 86 \\ 15 x_{1}-9 x_{2}+2 x_{3}= & 187 \end{array}$$
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