Chapter 7.2, Problem 68E

Solution

a.

To prove: The determinant of A changes sign f two rows or two columns are interchanged. Start with 3 cross 3 matrix and compare the expansion by the expanding by the same row before and after the interchange.

The statement has proven.

Given Information:

The matrix is defined as,

  $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Calculation:

Consider the given information,

  $A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Calculate the determinate by the first row.

  $\begin{array}{c}\left|A\right|=\left|\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ =a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|-a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|+a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\\ =a_{11}{X}_1-a_{12}{X}_2+a_{13}{X}_3\end{array}$

Now, interchange the second row of the matrix with one and calculate the determinate by the second row.

  $\begin{array}{c}\left|B\right|=\left|\begin{array}{ccc}{a}_{21}& a_{22}& a_{23}\\ a_{11}& a_{12}& a_{13}\\ a_{31}& a_{32}& a_{33}\end{array}\right|\\ =-a_{11}\left|\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right|+a_{12}\left|\begin{array}{cc}{a}_{21}& a_{23}\\ a_{31}& a_{33}\end{array}\right|-a_{13}\left|\begin{array}{cc}{a}_{21}& a_{22}\\ a_{31}& a_{32}\end{array}\right|\\ =-a_{11}{X}_1+a_{12}{X}_2-a_{13}{X}_3\\ =-\left|A\right|\end{array}$

To generalize for  n , expand the matrix  A  using one of the rows which will be interchanged. Then it will expand the matrix obtained by interchanging the two rows using the same row.

The minors will be the same except maybe for a change in rows' order, for which we will apply the change of sign rule also.

Therefore, the given statement has proved.

b.

To prove: The determinant of a square matrix with two identical rows ot two identical columns is zero.

The statement has proven.

Given Information:

The determinant is defined as,

  $A_n={\left[a_{ij}\right]}_{n\times n}$

Explanation:

Consider the given matrix,

It has to be prove that the determinant of a square matrix with two identical rows is zero. Use the induction to prove the property.

Step 1. Prove the property for $n=2$ .

  $\begin{array}{c}\det A_2=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\\ =a_{11}{a}_{22}-a_{12}{a}_{21}\\ =0\end{array}$

Step 2. Now, assume that the property true for $kXk$ matrices and also prove for $\left(k+1\right)X\left(k+1\right)$ .

  $A_{k+1}=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1\left(k+1\right)}\\ & & \cdots & \\ a_{\left(k+1\right)1}& a_{\left(k+1\right)2}& \cdots & a_{\left(k+1\right)\left(k+1\right)}\end{array}\right]$

Two of the matrix rows are identical and can interchange them. So none of the two is the first row.

  $\det A_{k+1}=a_{11}{X}_{11}-a_{12}{X}_{12}+\cdots +{\left(-1\right)}^k{X}_{k+1}$

Here, all the minors  $X_i$ ? contain two equal rows, thus according to the assumption that the property is true for  k , all minors have the determinant zero.

  $\begin{array}{c}\det \left(A_{k+1}\right)=a_{11}\cdot 0-a_{12}\cdot 0+\cdots +{\left(-1\right)}^k\cdot 0\\ =0\end{array}$

Hence, the given statement has proved.

c.

To prove: If a scalar multiple of a row is added to another row, the value of the determinant of a square matrix is unchanged.

The statement has proved.

Given Information:

The matrix is defined as,

  $A_n={\left[a_{ij}\right]}_{n\times n}$

Explanation:

Consider the given information,

  $A_n={\left[a_{ij}\right]}_{n\times n}$

The determinant is defined as,

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

The required determinant is defined as,

  $D_2=\left|\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ & & \cdots & \\ a_{k1}+\alpha a_{j1}& a_{k2}+\alpha a_{j2}& \cdots & a_{kn}+\alpha a_{jn}\\ & & \cdots & \\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right|$

Expend the determinant using the number of k rows.

  $\begin{array}{c}{D}_2=\left(a_{k1}+\alpha a_{j1}\right)X_1-\left(a_{k2}+\alpha a_{j2}\right)X_2+\cdots \\ =a_{k1}{X}_1+\alpha a_{j1}{X}_1-a_{k2}{X}_2+\alpha a_{j2}{X}_2+\cdots \\ =\left(a_{k1}{X}_1-a_{k2}{X}_2+\cdots \right)+\alpha \left(a_{j1}{X}_1-a_{j2}{X}_2+\cdots \right)\\ =D_1+\alpha \cdot 0\\ =D_1\end{array}$

Here, the used the fact that the second parenthesis contains the expansion of the determinant containing two equal rows  j , therefore its value is zero.

Therefore, the required result is explained above.

d.

To explain: The point $\left(x_3,y_3\right)$ lies on the line in part (a) by using a determinant.

The point will not lie on the line, if the determinant do not become 0 for the given point.

Given Information:

The determinant is defined as,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Explanation:

Consider the given information,

  $\left|\begin{array}{ccc}1& x& y\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0$

Substitute the given point $\left(x_3,y_3\right)$ in the obtained determinant in part (a).

  $\begin{array}{c}\left|\begin{array}{ccc}1& x_3& y_3\\ 1& x_1& y_1\\ 1& x_2& y_2\end{array}\right|=0\\ 1\left[x_1{y}_2-y_1{x}_2\right]-x_3\left[y_2-y_1\right]+y_3\left[x_2-x_1\right]=0\\ y_3\left[x_2-x_1\right]=x_3\left[y_2-y_1\right]-\left[x_1{y}_2-y_1{x}_2\right]\\ y_3=x_3\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}\end{array}$

Here, if the condition $y_3\ne x_3\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}-\frac{\left(x_1{y}_2-y_1{x}_2\right)}{\left(x_2-x_1\right)}$ is true then the point $\left(x_3,y_3\right)$ will not lie one the line. The value of determinant cannot be zero for the point $\left(x_3,y_3\right)$ .

Therefore, the required result is explained above.