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Linear Algebra

Inverse by Cofactors

If \(A\in M_{n\times n}(\mathbb{R})\) is invertible, then \((A^{-1})_{ij}=\frac{1}{|A|}C_{ji}\). \(C_{ji}\) is known as the adjugate of \(A\).

The adjugate, adj(A), is the matrix of cofactors,

Example: Determine the adjugate and inverse of \(A=\left[\begin{array}{rrr} 2&4&-1\\0&3&1\\6&-2&5\end{array}\right]\)

Solution: The nine cofactors are

\(C_{11}=(1)\left|\begin{array}{rr} 3&1\\-2&5\end{array}\right|=17\) \(C_{12}=(-1)\left|\begin{array}{rr} 0&1\\6&5\end{array}\right|=6\) \(C_{13}=(1)\left|\begin{array}{rr} 0&3\\6&-2\end{array}\right|=-18\)
\(C_{21}=(-1)\left|\begin{array}{rr} 4&-1\\-2&5\end{array}\right|=-18\) \(C_{22}=(1)\left|\begin{array}{rr} 2&-1\\6&5\end{array}\right|=16\) \(C_{23}=(-1)\left|\begin{array}{rr} 2&4\\6&-2\end{array}\right|=28\)
\(C_{31}=(1)\left|\begin{array}{rr} 4&-1\\4&1\end{array}\right|=7\) \(C_{32}=(-1)\left|\begin{array}{rr} 2&-1\\0&1\end{array}\right|=-2\) \(C_{33}=(1)\left|\begin{array}{rr} 2&4\\0&3\end{array}\right|=6\)

\[adj(A)=\left[\begin{array}{rrr} C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\end{array}\right]=\left[\begin{array}{rrr} 17&-18&7\\6&16&-2\\-18&28&6\end{array}\right]\]

We can verify by cofactors that the determinant \(|A|=76\). Therefore, we can calculate the inverse

\[A^{-1}=\frac{1}{|A|}adj(A)=\frac{1}{76}\left[\begin{array}{rrr} 17&-18&7\\6&16&-2\\-18&28&6\end{array}\right]\]

Cramer's Rule

We can apply Cramer's rule to solve a system of linear equations in n variables, \(A\bar{x}=\bar{b}\). If \(|A|\neq 0\) so that \(A\) is invertible. Let \(N_i\) be the matrix obtained from \(A\) by replacing the i-th column of \(A\) by \(\bar{b}\). Then the i-th component of \(\bar{x}\) in the solution of \(A\bar{x}=\bar{b}\) is

\[x_i=\frac{|N_i|}{|A|}\]

The matrix \(N_i\) is defined by replacing \(\bar{b}\) by the column i.

For example, if 

\[A=\left[\begin{array}{rrr} 3&2&-1\\0&1&3\\-2&4&0\end{array}\right] \qquad \bar{x}=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} \qquad \bar{b}=\begin{bmatrix} 4\\2\\1\end{bmatrix}\]

Then in \(N_1\), \(\bar{b}\) will replace column 1

\[N_1=\left[\begin{array}{rrr} \mathbf{4}&2&-1\\\mathbf{2}&1&3\\\mathbf{1}&4&0\end{array}\right] \]

In \(N_2\), \(\bar{b}\) will replace column 2

\[N_2=\left[\begin{array}{rrr} 3&\mathbf{4}&-1\\0&\mathbf{2}&3\\-2&\mathbf{1}&0\end{array}\right] \]

In \(N_3\), \(\bar{b}\) will replace column 3

\[N_3=\left[\begin{array}{rrr} 3&2&\mathbf{4}\\0&1&\mathbf{2}\\-2&4&\mathbf{1}\end{array}\right] \]


Example: Solve the following system of equations using Cramer's Rule

\begin{align} 7x+y-4z&=3\\-6x-4y+z&=0\\4x-y-2z&=6\end{align}

Solution

\[A=\left[\begin{array}{rrr} 7&1&-4\\-6&-4&1\\4&-1&-2\end{array}\right] \qquad \bar{b}=\begin{bmatrix}3\\0\\6\end{bmatrix} \]

To solve for variable x, we need to find the matrix \(N_x\) by replacing the x or 1st column with \(\bar{b}\)

\[N_x=\left[\begin{array}{rrr} \mathbf{3}&1&-4\\\mathbf{0}&-4&1\\\mathbf{6}&-1&-2\end{array}\right] \]

We repeat to find \(N_y\) and \(N_z\)

\[N_y=\left[\begin{array}{rrr} 7&\mathbf{3}&-4\\-6&\mathbf{0}&1\\4&\mathbf{6}&-2\end{array}\right] \]

\[N_z=\left[\begin{array}{rrr} 7&1&\mathbf{3}\\-6&-4&\mathbf{0}\\4&-1&\mathbf{6}\end{array}\right] \]

Now we apply Cramer's Rule

\[x=\frac{|N_x|}{|A|} = \frac{-63}{-33}=\frac{21}{11} \qquad y=\frac{|N_y|}{|A|} = \frac{78}{-33}=-\frac{26}{11} \qquad z=\frac{|N_z|}{|A|} = \frac{-66}{-33}=2\]

The solutions check out when plugging them back into each equation.

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