- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.11. $\quad$ Exercises
7. $\quad$ Use Theorem 3.6 to determine whether the following sets of vectors are linearly dependent or independent.
(a) $\quad$ $A_1 = (1, -1, 0),$ $A_2 = (0, 1, -1),$ $A_3 = (2, 3, -1).$
(b) $\quad$ $A_1 = (1, -1, 2, 1),$ $A_2 = (-1, 2, -1, 0),$ $A_3 = (3, -1, 1, 0),$ $A_4 = (1, 0, 0, 1).$
(c) $\quad$ $A_1 = (1, 0, 0, 0, 1),$ $A_2 = (1, 1, 0, 0, 0),$ $A_3 = (1, 0, 1, 0, 1),$ $A_4 = (1, 1, 0, 1, 1),$ $A_5 = (0, 1, 0, 1, 0).$
Solution. $\quad$ We recall from Theorem 3.6 that a set of $n$ vectors $A_1, \dots, A_n$ in $n$-space is independent if and only if $d(A_1, \dots, A_n) \neq 0.$
(a) $\quad$ $A_1 = (1, -1, 0),$ $A_2 = (0, 1, -1),$ $A_3 = (2, 3, -1).$
We can write $d(A_1, A_2, A_3)$ as
\begin{align*}
\det \begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
2 & 3 & -1 \\
\end{bmatrix}
\end{align*}
This is equivalent to
\begin{align*}
- \det \begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
-2 & -3 & 1 \\
\end{bmatrix} \\ \\
- \det \begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & -5 & 1 \\
\end{bmatrix} \\ \\
- \det \begin{bmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & -4 \\
\end{bmatrix} &= 4.
\end{align*}
From this, we can see that the set of vectors $A_1, A_2, A_3$ form an upper triangular matrix with determinant $4.$ Thus, the set is independent.
(b) $\quad$ $A_1 = (1, -1, 2, 1),$ $A_2 = (-1, 2, -1, 0),$ $A_3 = (3, -1, 1, 0),$ $A_4 = (1, 0, 0, 1).$ We can express $d(A_1, A_2, A_3, A_4)$ as \begin{align*} \det \begin{bmatrix} 1 & -1 & 2 & 1 \\ -1 & 2 & -1 & 0 \\ 3 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} \\ \\ \det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 3 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} \\ \\ \det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ -3 & 1 & -1 & 0 \\ -1 & 0 & 0 & -1 \end{bmatrix} \\ \\ \det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & -2 & 5 & 3 \\ 0 & -1 & 2 & 0 \end{bmatrix} \\ \\ \det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 7 & 5 \\ 0 & 0 & 3 & 1 \end{bmatrix} \\ \\ \frac{-1}{21}\det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 21 & 15 \\ 0 & 0 & -21 & -7 \end{bmatrix} \\ \\ \frac{-1}{21}\det \begin{bmatrix} 1 & -1 & 2 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 21 & 15 \\ 0 & 0 & 0 & 8 \end{bmatrix} &= -8. \end{align*} As we can see, this set of vectors forms an upper-triangular matrix with determinant $-8.$ Thus, the set is independent.
(c) $\quad$ $A_1 = (1, 0, 0, 0, 1),$ $A_2 = (1, 1, 0, 0, 0),$ $A_3 = (1, 0, 1, 0, 1),$ $A_4 = (1, 1, 0, 1, 1),$ $A_5 = (0, 1, 0, 1, 0).$ We can express $d(A_1, A_2, A_3, A_4, A_5)$ as \begin{align*} \det \begin{bmatrix} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 \\ \end{bmatrix} \\ \\ \det \begin{bmatrix} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ \end{bmatrix} \end{align*} As we can see, the fourth and fifth rows of the equivalent matrix are equal. But by Axiom 3' of the determinant function, this means that $d(A_1, A_2, A_3, A_4, A_5) = 0.$ Thus, the set of vectors is dependent.