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Chapter 2: Problem 2

Use determinants to determine whether the following \(2 \times 2\) matrices are nonsingular: (a) \(\left(\begin{array}{ll}3 & 5 \\ 2 & 4\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 6 \\ 2 & 4\end{array}\right)\) (c) \(\left(\begin{array}{rr}3 & -6 \\ 2 & 4\end{array}\right)\)

Short Answer

Expert verified
For the given matrices: (a) The determinant is 2, which is nonzero, thus the matrix is nonsingular. (b) The determinant is 0, which means the matrix is singular. (c) The determinant is 24, which is nonzero, thus the matrix is nonsingular.

Step by step solution

01

(a) Find the determinant of the matrix

For matrix (a) \(\left(\begin{array}{ll}3 & 5 \\\ 2 & 4\end{array}\right)\), we have \(a=3\), \(b=5\), \(c=2\), and \(d=4\). The determinant is: \[\text{Det}(a) = ad - bc = (3)(4) - (5)(2) = 12 - 10 = 2\]
02

(a) Determine if it is nonsingular

Since the determinant of matrix (a) is nonzero (\(2 \neq 0\)), the matrix is nonsingular.
03

(b) Find the determinant of the matrix

For matrix (b) \(\left(\begin{array}{ll}3 & 6 \\\ 2 & 4\end{array}\right)\), we have \(a=3\), \(b=6\), \(c=2\), and \(d=4\). The determinant is: \[\text{Det}(b) = ad - bc = (3)(4) - (6)(2) = 12 - 12 = 0\]
04

(b) Determine if it is nonsingular

Since the determinant of matrix (b) is zero (\(0 = 0\)), the matrix is singular.
05

(c) Find the determinant of the matrix

For matrix (c) \(\left(\begin{array}{rr}3 & -6 \\\ 2 & 4\end{array}\right)\), we have \(a=3\), \(b=-6\), \(c=2\), and \(d=4\). The determinant is: \[\text{Det}(c) = ad - bc = (3)(4) - (-6)(2) = 12 + 12 = 24\]
06

(c) Determine if it is nonsingular

Since the determinant of matrix (c) is nonzero (\(24 \neq 0\)), the matrix is nonsingular.

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Most popular questions from this chapter

Use Cramer's rule to solve each of the following systems: (a) \(\quad x_{1}+2 x_{2}=3\) (b) \(2 x_{1}+3 x_{2}=2\) \(3 x_{1}-x_{2}=1\) \(3 x_{1}+2 x_{2}=5\) (c) \(2 x_{1}+x_{2}-3 x_{3}=0\) \(4 x_{1}+5 x_{2}+x_{3}=8\) \(-2 x_{1}-x_{2}+4 x_{3}=2\) (d) \(\quad x_{1}+3 x_{2}+x_{3}=1\) \(2 x_{1}+x_{2}+x_{3}=5\) \(-2 x_{1}+2 x_{2}-x_{3}=-8\) (e) \(x_{1}+x_{2}\) \(=0\) \(x_{2}+x_{3}-2 x_{4}=1\) \(x_{1}+2 x_{3}+x_{4}=0\) \(x_{1}+x_{2}+x_{4}=0\)

Consider the \(3 \times 3\) Vandermonde matrix \\[ V=\left(\begin{array}{lll} 1 & x_{1} & x_{1}^{2} \\ 1 & x_{2} & x_{2}^{2} \\ 1 & x_{3} & x_{3}^{2} \end{array}\right) \\] (a) Show that \(\operatorname{det}(V)=\left(x_{2}-x_{1}\right)\left(x_{3}-x_{1}\right)\left(x_{3}-x_{2}\right)\) [Hint: Make use of row operation III. (b) What conditions must the scalars \(x_{1}, x_{2},\) and \(x_{3}\) satisfy in order for \(V\) to be nonsingular?

Let \(A\) be a nonsingular \(n \times n\) matrix with \(n>1\) Show that \\[ \operatorname{det}(\operatorname{adj} A)=(\operatorname{det}(A))^{n-1} \\]

Let \(A\) and \(B\) be \(n \times n\) matrices. Prove that if \(A B=I,\) then \(B A=I .\) What is the significance of this result in terms of the definition of a nonsingular matrix?

Let \(E_{1}, E_{2},\) and \(E_{3}\) be \(3 \times 3\) elementary matrices of types I, II, and III, respectively, and let \(A\) be a \(3 \times 3\) matrix with \(\operatorname{det}(A)=6 .\) Assume, additionally, that \(E_{2}\) was formed from \(I\) by multiplying its second row by 3. Find the values of each of the following: (a) \(\operatorname{det}\left(E_{1} A\right)\) (b) \(\operatorname{det}\left(E_{2} A\right)\) (c) \(\operatorname{det}\left(E_{3} A\right)\) (d) \(\operatorname{det}\left(A E_{1}\right)\) (e) \(\operatorname{det}\left(E_{1}^{2}\right)\) (f) \(\operatorname{det}\left(E_{1} E_{2} E_{3}\right)\)
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