- 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
3.6. $\quad$ Exercises
$\quad$ In this set of exercises you may assume existence of a determinant function. Determinants of order three may be computed by Equation (3.2).
\begin{align*}
(3.2) \qquad
\det\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}
&=
a_{11}\begin{vmatrix}
a_{22} & a_{23} \\ a_{32} & a_{33}
\end{vmatrix}
-
a_{12}\begin{vmatrix}
a_{21} & a_{23} \\ a_{31} & a_{33}
\end{vmatrix}
+
a_{13}\begin{vmatrix}
a_{21} & a_{22} \\ a_{31} & a_{32}
\end{vmatrix}
\end{align*}
4. $\quad$ Compute the determinant of each of the following matrices by transforming each of them to an upper triangular matrix. \begin{align} &\text{(a)} \qquad \begin{bmatrix} 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & -1 \end{bmatrix} \\ \\ &\text{(b)} \qquad \begin{bmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \end{bmatrix} \\ \\ &\text{(c)} \qquad \begin{bmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^4 & b^4 & c^4 & d^4 \end{bmatrix} \\ \\ &\text{(d)} \qquad \begin{bmatrix} a & 1 & 0 & 0 & 0 \\ 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \end{bmatrix} \\ \\ &\text{(e)} \qquad \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & 1 & -1 \end{bmatrix} \end{align}
Solution. $\quad$ We recall from Section 4 that when the row operations of the Gauss-Jordan elimination process are applied to a matrix,
$\quad$ (1) $\quad$ Interchanging two equations;
$\quad$ (2) $\quad$ Multiplying all the terms of an equation by a nonzero scalar;
$\quad$ (3) $\quad$ Adding one equation to a multiple of another.
Its determinant changes as follows:
$\quad$ When (1) is applied to a matrix $A,$ the sign of its determinant changes. When (2) is applied to $A,$ for some nonzero scalar $c,$ the determinant is multiplied by $c.$ When (3) is applied, the determinant remains unchanged.
$\quad$ Additionally, we proved in Section 4 that the determinant of an upper triangular matrix is the product of its diagonal elements. As such, we have the following:
(a) \begin{align*} \det \begin{bmatrix} 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & -1 \end{bmatrix} &= \begin{vmatrix} 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & -1 \end{vmatrix} \\ &= -\begin{vmatrix} 1 & -1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & 1 & -1 \\ 1 & -1 & -1 & -1 \end{vmatrix} \\ &= -\begin{vmatrix} 1 & -1 & 1 & 1 \\ 0 & 2 & -2 & -2 \\ 0 & 2 & 0 & -2 \\ 0 & 0 & -2 & -2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & -1 & 1 & 1 \\ 0 & 2 & 0 & -2 \\ 0 & 2 & -2 & -2 \\ 0 & 0 & -2 & -2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & -1 & 1 & 1 \\ 0 & 2 & 0 & -2 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & -2 & -2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & -1 & 1 & 1 \\ 0 & 2 & 0 & -2 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \end{vmatrix} \\ &= 8. \quad \blacksquare \end{align*}
(b) \begin{align*} \det \begin{bmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \end{bmatrix} &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^3 & b^3 & c^3 & d^3 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & c - a & d - a \\ 0 & b^2 - ab & c^2 - ac & d^2 - ad \\ 0 & b^3 - ab^2 & c^3 - ac^2 & d^3 - ad^2 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & c - a & d - a \\ 0 & b(b - a) & c(c - a) & d(d - a) \\ 0 & b^2(b - a) & c^2(c - a) & d^2(d - a) \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & c - a & d - a \\ 0 & 0 & (c - b)(c - a) & (d - b)(d - a) \\ 0 & 0 & (c^2 - bc)(c - a) & (d^2 - bd)(d - a) \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & c - a & d - a \\ 0 & 0 & (c - b)(c - a) & (d - b)(d - a) \\ 0 & 0 & c(c - b)(c - a) & d(d - b)(d - a) \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & c - a & d - a \\ 0 & 0 & (c - b)(c - a) & (d - b)(d - a) \\ 0 & 0 & 0 & (d-c)(d - b)(d - a) \end{vmatrix} \\ &= (b-a)(c-b)(c-a)(d-c)(d-b)(d-a). \quad \blacksquare \end{align*}
(c) \begin{align*} \det \begin{bmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^4 & b^4 & c^4 & d^4 \end{bmatrix} &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ a & b & c & d \\ a^2 & b^2 & c^2 & d^2 \\ a^4 & b^4 & c^4 & d^4 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & (b-a) & (c-a) & (d-a) \\ 0 & b(b - a) & c(c-a) & d(d-a) \\ 0 & b^2(b^2 - a^2) & c^2(c^2 - a^2) & d^2(d^2 - a^2) \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & (b-a) & (c-a) & (d-a) \\ 0 & 0 & (c-b)(c-a) & (d-b)(d-a) \\ 0 & 0 & (c-a)[c^2(c+a) - bc(b+a)] & (d-a)[d^2(d+a) - bd(b+a)] \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & (b-a) & (c-a) & (d-a) \\ 0 & 0 & (c-b)(c-a) & (d-b)(d-a) \\ 0 & 0 & c(c-a)(c-b)(a + b + c) & d(d-a)(d-b)(a + b + d) \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & (b-a) & (c-a) & (d-a) \\ 0 & 0 & (c-b)(c-a) & (d-b)(d-a) \\ 0 & 0 & 0 & (d-b)(d-a)[d(a+b+d)-c(a+b+c)] \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & (b-a) & (c-a) & (d-a) \\ 0 & 0 & (c-b)(c-a) & (d-b)(d-a) \\ 0 & 0 & 0 & (d-c)(d-b)(d-a)(a+b+c+d) \end{vmatrix} \\ &= (b-a)(c-a)(c-b)(d-a)(d-b)(d-c)(a+b+c+d). \quad \blacksquare \end{align*}
(d) \begin{align*} \det \begin{bmatrix} a & 1 & 0 & 0 & 0 \\ 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \end{bmatrix} &= \begin{vmatrix} a & 1 & 0 & 0 & 0 \\ 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \end{vmatrix} \\ &= \begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ a & 1 & 0 & 0 & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 4a & 4 & 0 & 0 & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & (4 - a^2) & -2a & 0 & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & -6a - a(4 - a^2) & -3(4 - a^2) & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & a(a^2 - 10) & 3(a^2 - 4) & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & 2a(a^2 - 10) & 6(a^2 - 4) & 0 \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & 0 & (16a^2 - 24 - a^4) & 4a(10-a^2) \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & 0 & 0 & 4a(10-a^2) - (16a^2 - 24 - a^4) \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & 0 & 0 & (64a - 20a^3 + a^5) \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\begin{vmatrix} 4 & a & 2 & 0 & 0 \\ 0 & 3 & a & 3 & 0 \\ 0 & 0 & 2 & a & 4 \\ 0 & 0 & 0 & 1 & a \\ 0 & 0 & 0 & 0 & a(a^2 - 4)(a^2 - 16) \end{vmatrix} \\ &= \left(\frac{1}{4}\right)\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)(4)(3)(2)(1)a(a^2 - 4)(a^2 - 16) \\ \\ &= a(a^2 - 4)(a^2 - 16). \quad \blacksquare \end{align*}
(e) \begin{align*} \det \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & 1 & -1 \end{bmatrix} &= \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & -1 & 1 \\ 1 & -1 & 1 & -1 & 1 & 1 \\ 1 & -1 & -1 & 1 & 1 & -1 \end{vmatrix} \\ &= \begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & -2 & -2 & -2 \\ 0 & 0 & -2 & -2 & 0 & 0 \\ 0 & -2 & -2 & 0 & -2 & 0 \\ 0 & -2 & 0 & -2 & 0 & 0 \\ 0 & -2 & -2 & 0 & 0 & -2 \end{vmatrix} \\ &= (-32)\begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 & 1 \end{vmatrix} \\ &= (32)\begin{vmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 5 \end{vmatrix} \\ &= 32(-5) = -160. \quad \blacksquare \end{align*}