Exercises: Cramer's rule

Construct the extended coefficient matrix.

$(\mathrm{A}|\mathrm{b})=\left(\begin{array}{ccc}2& 4& -5\\ 3& 3& 2\\ -4& -5& 6\end{array}|\begin{array}{c}11\\ 17\\ -17\end{array}\right)$

Now replace the column of $x$ in the matrix $A$ by the last column to get the matrix ${\mathrm{A}}_{\mathrm{x}}$. Calculate the determinant of this matrix.

${\mathrm{detA}}_{\mathrm{x}}=\left|\begin{array}{ccc}11& 4& -5\\ 17& 3& 2\\ -17& -5& 6\end{array}\right|=-66$

Do this also for ${\mathrm{A}}_{\mathrm{y}}$ and ${\mathrm{A}}_{\mathrm{z}}$, and calculate the determinants of those matrices.

${\mathrm{detA}}_{\mathrm{y}}=\left|\begin{array}{ccc}2& 11& -5\\ 3& 17& 2\\ -4& -17& 6\end{array}\right|=-99$

${\mathrm{detA}}_{\mathrm{z}}=\left|\begin{array}{ccc}2& 4& 11\\ 3& 3& 17\\ -4& -5& -17\end{array}\right|=-33$

Now calculate the determinant of $A$.

$\mathrm{detA}=\left|\begin{array}{ccc}2& 4& -5\\ 3& 3& 2\\ -4& -5& 6\end{array}\right|=-33$

Now divide the determinant of ${\mathrm{A}}_{\mathrm{x}}$, ${\mathrm{A}}_{\mathrm{y}}$ and ${\mathrm{A}}_{\mathrm{z}}$ by the determinant of $A$. You get $x$,$y$ and $z$ respectively.

$\mathrm{x}={\displaystyle \frac{{\mathrm{detA}}_{\mathrm{x}}}{\mathrm{detA}}}=2$

$\mathrm{y}={\displaystyle \frac{{\mathrm{detA}}_{\mathrm{y}}}{\mathrm{detA}}}=3$

$\mathrm{z}=\frac{{\det \mathrm{A}}_{\mathrm{z}}}{\det \mathrm{A}}=1$

Construct the extended coefficient matrix.

$(\mathrm{A}|\mathrm{b})=\left(\begin{array}{ccc}2& 1& 1\\ 3& -2& 1\\ 4& -2& 1\end{array}|\begin{array}{c}2\\ -2\\ -1.5\end{array}\right)$

Now replace the column of $x$ in the matrix $A$ by the last column to get the matrix ${\mathrm{A}}_{\mathrm{x}}$. Calculate the determinant of this matrix.

${\mathrm{detA}}_{\mathrm{x}}=\left|\begin{array}{ccc}2& 1& 1\\ -2& -2& 1\\ -1.5& -2& 1\end{array}\right|=1.5$

Do this also for ${\mathrm{A}}_{\mathrm{y}}$ and ${\mathrm{A}}_{\mathrm{z}}$, and calculate the determinants of those matrices.

${\mathrm{detA}}_{\mathrm{y}}=\left|\begin{array}{ccc}2& 2& 1\\ 3& -2& 1\\ 4& -1.5& 1\end{array}\right|=4.5$

${\mathrm{detA}}_{\mathrm{z}}=\left|\begin{array}{ccc}2& 1& 2\\ 3& -2& -2\\ 4& -2& -1.5\end{array}\right|=-1.5$

Now calculate the determinant of $A$.

$\det A=\left|\begin{array}{ccc}2& 1& 1\\ 3& -2& 1\\ 4& -2& 1\end{array}\right|=3$

Now divide the determinant of ${\mathrm{A}}_{\mathrm{x}}$, ${\mathrm{A}}_{\mathrm{y}}$ and ${\mathrm{A}}_{\mathrm{z}}$ by the determinant of $A$. You get $x$,$y$ and $z$ respectively.

$\mathrm{x}=\frac{{\det \mathrm{A}}_{\mathrm{x}}}{\det \mathrm{A}}=0.5$

$\mathrm{y}=\frac{{\det \mathrm{A}}_{\mathrm{y}}}{\det \mathrm{A}}=1.5$

$\mathrm{z}=\frac{{\det \mathrm{A}}_{\mathrm{z}}}{\det \mathrm{A}}=-0.5$