Question-2

Consider the two matrices:

\[ A=\begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ -1 & 0 & 2 \end{bmatrix},\,\,\,\,B=\begin{bmatrix}1 & 3 & -2\\ 0 & 1 & -1\\ 3 & 4 & 2 \end{bmatrix} \]

Find \(\text{det}(A),\text{det}(B),\text{det}(AB),\text{det}(BA)\).

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The following solution to find the determinant is unnecessarily long. We can perform a sequence of row operations on \(A\):

\(R_{3}\rightarrow R_{3}+R_{2}\)

\[ \begin{aligned} \begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ -1 & 0 & 2 \end{bmatrix} & \rightarrow\begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ 0 & -3 & 6 \end{bmatrix} \end{aligned} \]

This operation doesn’t change the determinant.

\(R_{3}\rightarrow\frac{-1}{3}R_{3}\)

\[ \begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ 0 & -3 & 6 \end{bmatrix}\rightarrow\begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ 0 & 1 & -2 \end{bmatrix} \]

This operation scales the determinant by \(\frac{-1}{3}\).

\(R_{1}\rightarrow R_{1}-2R_{2}\)

\[ \begin{bmatrix}2 & -1 & 1\\ 1 & -3 & 4\\ 0 & 1 & -2 \end{bmatrix}\rightarrow\begin{bmatrix}0 & 5 & -7\\ 1 & -3 & 4\\ 0 & 1 & -2 \end{bmatrix} \]

This operation doesn’t change the determinant.

\(R_{1}\leftrightarrow R_{2}\)

\[ \begin{bmatrix}0 & 5 & -7\\ 1 & -3 & 4\\ 0 & 1 & -2 \end{bmatrix}\rightarrow\begin{bmatrix}1 & -3 & 4\\ 0 & 5 & -7\\ 0 & 1 & -2 \end{bmatrix} \]

This operation scales the determinant by \(-1\).

\(R_{2}\rightarrow R_{2}-5R_{3}\)

\[ \begin{bmatrix}1 & -3 & 4\\ 0 & 5 & -7\\ 0 & 1 & -2 \end{bmatrix}\rightarrow\begin{bmatrix}1 & -3 & 4\\ 0 & 0 & 3\\ 0 & 1 & -2 \end{bmatrix} \]

This operation doesn’t change the determinant.

\(R_{2}\rightarrow\frac{1}{3}R_{2}\)

\[ \begin{bmatrix}1 & -3 & 4\\ 0 & 0 & 3\\ 0 & 1 & -2 \end{bmatrix}\rightarrow\begin{bmatrix}1 & -3 & 4\\ 0 & 0 & 1\\ 0 & 1 & -2 \end{bmatrix} \]

This operation scales the determinant by \(\frac{1}{3}\).

\(R_{2}\leftrightarrow R_{3}\)

\[ \begin{bmatrix}1 & -3 & 4\\ 0 & 0 & 1\\ 0 & 1 & -2 \end{bmatrix}\rightarrow\begin{bmatrix}1 & -3 & 4\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{bmatrix} \]

This operation scales the determinant by \(-1\).

The final matrix that we have is an upper triangular matrix. The determinant of an upper triangular matrix is the product of its diagonal entries, which is \(1\) in this case. So we have:

\[ \begin{aligned} 1 & =\text{det}(A)\times\frac{-1}{3}\times(-1)\times\frac{1}{3}\times(-1)\\ \text{det}(A) & =-9 \end{aligned} \]

For \(\text{det}(B)\), we will expand the determinant along the first column:

\[ \begin{vmatrix}1 & 3 & -2\\ 0 & 1 & -1\\ 3 & 4 & 2 \end{vmatrix}=1(2+4)+3(-3+2)=3 \]

For \(\text{det}(AB)\), we just use the property that \(\text{det}(AB)=\text{det}(A)\cdot\text{det}(B)\). Thus, we get \(\text{det}(AB)=-27\). \(\text{det}(BA)\) is going to be the same.