The classical adjoint (which is also known as an adjugate) of a matrix is found using two steps. The first step is to create a matrix of cofactors. A cofactor for a given entry is found by multiplying (-1) raised to the power of the row plus the column for the entry, multiplied by the minor of that entry. The minor for an entry is found by taking the determinant of the matrix that is formed when the row and column of the given entry is deleted. After a matrix of cofactors is created, we find the transpose of this matrix, which will give us the adjoint of our matrix.

Test Objectives
• Demonstrate the ability to find a matrix of cofactors
• Demonstrate the ability to find the transpose of a matrix
• Demonstrate the ability to find the adjoint of a matrix
Adjoint of a Matrix Practice Test:

#1:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}1 & -2\\ 1 & 4\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}3 & 7\\ 2 & 5\end{array}\right]$$

#2:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}8 & -9\\ 0 & 1\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-6 & 5\\ 1 & -1\end{array}\right]$$

#3:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}3 & -2\\ 1 & 0\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}1 & -1 & 0\\ 5 & -4 & 3 \\-2 & 0 & 1\end{array}\right]$$

#4:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-1 & 3 & 5\\ 6 & 7 & -9 \\1 & 7 & 5\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-4 & 2 & -1\\ 3 & 0 & 1 \\5 & 8 & 7\end{array}\right]$$

#5:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}1 & 0 & 0\\ 15 & -8 & 9 \\-13 & 1 & 2\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}1 & 0 & -1 & 1\\ 2 & 5 & 0 & 3 \\0 & 1 & -1 & -2 \\0 & 3 & 0 & 1\end{array}\right]$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}4 & 2\\ -1 & 1\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}5 & -7\\ -2 & 3\end{array}\right]$$

#2:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}1 & 9\\ 0 & 8\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cc}-1 & -5\\ -1 & -6\end{array}\right]$$

#3:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{cc}0 & 2\\ -1 & 3\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-4 & 1 & -3\\ -11 & 1 & -3 \\ -8 & 2 & 1\end{array}\right]$$

#4:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}98 & 20 & -62\\ -39 & -10 & 21 \\ 35 & 10 & -25\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-8 & -22 & 2\\ -16 & -23 & 1 \\ 24 & 42 & -6\end{array}\right]$$

#5:

Solutions:

$$a)\hspace{.2em}$$ $$\left[ \begin{array}{ccc}-25 & 0 & 0\\ -147 & 2 & -9 \\ -89 & -1 & -8\end{array}\right]$$

$$b)\hspace{.2em}$$ $$\left[ \begin{array}{cccc}4 & -10 & -4 & 18\\ 2 & -1 & -2 & -3 \\ 14 & -7 & 2 & 11 \\-6 & 3 & 6 & -7\end{array}\right]$$