Chapter 10.6, Problem 22E

To determine

The determinant of the given matrix is to be calculated and tell that does it contain the inverse matrix or not:

  $\left[\begin{array}{ccc}-2& -\frac{3}{2}& \frac{1}{2}\\ 2& 4& 0\\ \frac{1}{2}& 2& 1\end{array}\right]$

Answer to Problem 22E

The determinant of $\left[\begin{array}{ccc}-2& -\frac{3}{2}& \frac{1}{2}\\ 2& 4& 0\\ \frac{1}{2}& 2& 1\end{array}\right]$ is $-4$ and it contains the inverse matrix.

Explanation of Solution

Given:

  $\left[\begin{array}{ccc}-2& -\frac{3}{2}& \frac{1}{2}\\ 2& 4& 0\\ \frac{1}{2}& 2& 1\end{array}\right]$

Concept Used:

The determinant of every square matrix with the order $n$ is defined as $A=\left[a_{ij}\right]$ where,

  $a={\left(i,j\right)}^{th}$ of element $A$ .

The given matrix has the inverse matrix when:

• A matrix must contain the same number of rows and column
• its determinant value must not be zero value

Calculations:

The determinant of the given matrix is calculated as:

  $\begin{array}{l}D=\left[\begin{array}{ccc}-2& -\frac{3}{2}& \frac{1}{2}\\ 2& 4& 0\\ \frac{1}{2}& 2& 1\end{array}\right]\\ D=-2\left[\left(4\times 1\right)-\left(0\times 2\right)\right]-\left(-\frac{3}{2}\right)\left[\left(2\times 1\right)-\left(\frac{1}{2}\times 0\right)\right]+\frac{1}{2}\left[\left(2\times 2\right)-\left(\frac{1}{2}\times 4\right)\right]\\ D=-2\left(4\right)+\frac{3}{2}\left(2\right)+\frac{1}{2}\left(4-2\right)\\ D=-8+3+1\\ D=-4\end{array}$

Now, the given matrix is $\left(3\times 3\right)$ and its determinant value is $D=-4$ or non-zero value which clearly shows that this matrix contains the inverse matrix.

Conclusion:

Hence, the determinant of the given matrix is calculated as $D=-4$ .