- 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*}
2. $\quad$ If $\det \begin{bmatrix} x & y & z \\ 3 & 0 & 2 \\ 1 & 1 & 1 \end{bmatrix} = 1,$ compute the determinant of each of the following matrices:
\begin{align*}
\text{(a)} \quad \begin{bmatrix} 2x & 2y & 2z \\ \frac{3}{2} & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix},
\quad
\text{(b)} \quad \begin{bmatrix} x & y & z \\ 3x + 3 & 3y & 3z + 2 \\ x + 1 & y + 1 & z + 1 \end{bmatrix},
\quad
\text{(c)} \quad \begin{bmatrix} x - 1 & y - 1 & z - 1 \\ 4 & 1 & 3 \\ 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.
Accordingly, if $\det \begin{bmatrix} x & y & z \\ 3 & 0 & 2 \\ 1 & 1 & 1 \end{bmatrix} = 1,$ we have the following:
(a) $\quad$ $\det \begin{bmatrix} 2x & 2y & 2z \\ \frac{3}{2} & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix} = 1.$
(b) $\quad \begin{bmatrix} x & y & z \\ 3x + 3 & 3y & 3z + 2 \\ x + 1 & y + 1 & z + 1 \end{bmatrix} = 1.$
(c) $\quad$ $\begin{bmatrix} x - 1 & y - 1 & z - 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix} = 1.$
$\quad$ As we can see in all three examples, the determinant remains unchanged. In part (a), the two row-wise scalar multiplications cancel eachother, and in parts (b) and (c), only the third row operation is applied to the matrices, leaving the determinant unchanged.