Chapter 7.2, Problem 69E

Solution

a.

To prove: If every element of a row or column of a matrix is multiplies by the real number c , then the determinant of the matrix is multiplies by c .

The statement has proven.

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right]$ with order $n\times n$ .

Calculation:

Consider the given information,

The first matrix for determinant is defined as,

  $D_1=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ & & \cdots & \\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|$

Write another matrix of constant multiplication.

  $D_2=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ & & \cdots & \\ \alpha a_{k1}& \alpha a_{k2}& \cdots & \alpha a_{kn}\\ & & \cdots & \\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|$

Expend the number of rows up to k .

  $\begin{array}{c}{D}_2=\alpha a_{k1}{X}_1-\alpha a_{k2}{X}_2+\cdots \\ =\alpha \left(a_{k1}{X}_1-a_{k2}{X}_2+\cdots \right)\\ =\alpha D_1\end{array}$

Here used the fact that the parenthesis contains the expansion of the determinant  $D_1$ ?.

Therefore, the given statement has proved.

b.

To prove: If all the entries above the main diagonal of a matrix zero, then the determinant s the product of the elements on the main diagonal.

The statement has proven.

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right]$ with order $n\times n$ .

Explanation:

Consider the given equations,

The determinant matrix D is defined as,

  $D=\left|\begin{array}{ccccc}{a}_{11}& 0& 0& \dots & 0\\ a_{21}& a_{22}& 0& \dots & 0\\ & & & \dots & \\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right|$

And the determinant is defined as,

  $\begin{array}{c}D=a_{11}{X}_{11}-0\cdot X_{12}+0\cdot X_{12}-\cdots +{\left(-1\right)}^{n+1}\cdot 0X_{1n}\\ =a_{11}\left[\begin{array}{cccc}{a}_{22}& 0& \cdots & 0\\ & & \cdots & \\ a_{n2}& a_{n3}& \cdots & a_{nn}\end{array}\right]\end{array}$

Expend the term $X_{11}$ by using the first row.

  $\begin{array}{c}D=a_{11}{a}_{22}\left[\begin{array}{cccc}{a}_{33}& 0& \cdots & 0\\ & & \cdots & \\ a_{n3}& a_{n4}& \cdots & a_{nn}\end{array}\right]-0+0-\cdots \\ =a_{11}{a}_{22}\left[\begin{array}{cccc}{a}_{33}& 0& \cdots & 0\\ & & \cdots & \\ a_{n3}& a_{n4}& \cdots & a_{nn}\end{array}\right]\end{array}$

It can continued and can conclude that $D=a_{11}{a}_{22}\cdots \cdots a_{nn}$ .

Hence, the given statement has proved.