Chapter 3.5, Problem 1AP

For Problems 1-6, evaluate the determinant of the given matrix $A$ by using (a) Definition $\mathrm{3.1.8}$, (b) elementary row operations to reduce $A$ to an upper triangular matrix, and (c) the Cofactor Expansion Theorem.

$A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$.

To determine

a.

To explain:

The value of the determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ using definition $\mathrm{3.1.8}$.

Answer to Problem 1AP

Solution:

The determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ is $18$.

Explanation of Solution

Given:

The matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$.

Approach:

By using the definition of determinants the value of general matrix of second order is $a_{11}{a}_{22}-a_{12}{a}_{21}$.

Calculation:

As, the value of determinant for general matrix of second order is $a_{11}{a}_{22}-a_{12}{a}_{21}$.

Then for matrix $A$ the value of $\det \left(A\right)=6\times 1-\left(6\times \left(-2\right)\right)$.

So, $\det \left(A\right)=18$.

To determine

b.

The value of the determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ using elementary row operations to reduce $A$ to an upper triangular matrix.

Answer to Problem 1AP

Solution:

The determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ is $18$ using elementary row operations to reduce $A$ to an upper triangular matrix.

Explanation of Solution

Given:

The matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$.

Approach:

By converting the given matrix to upper triangular matrix in which the entries below diagonal elements are zero and then taking the product of diagonal elements of the equivalent matrix gives the determinant of the matrix.

Calculation:

Firstly apply the row operation as adding one third of the first row to the second row it gives $A\sim \left[\begin{array}{cc}6& 6\\ 0& 3\end{array}\right]$.

Thus, $\det \left(A\right)=6\times 3$.

That is, $\det \left(A\right)=18$.

To determine

c.

The value of the determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ using cofactor expansion theorem.

Answer to Problem 1AP

Solution:

The determinant of matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$ is $18$ using cofactor expansion theorem.

Explanation of Solution

Given:

The matrix $A=\left[\begin{array}{cc}6& 6\\ -2& 1\end{array}\right]$.

Approach:

According to cofactor expansion theorem the determinant of general second order matrix is given by $a_{11}{C}_{11}+a_{12}{C}_{12}$ where $C_{11}=\left|a_{22}\right|,C_{12}=-\left|a_{21}\right|$

Calculation:

So, by the expansion theorem the $\det \left(A\right)=6\times \left|1\right|+6\times \left(-\left|-2\right|\right)$.

Thus, $\det \left(A\right)=18$.