Matrix Inverse Calculation Explained

To find the inverse of a matrix A, denoted as \(A^{-1}\), we use the formula:

\[A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\]

Where:

• \(\det(A)\) is the determinant of matrix A.
• \(\text{adj}(A)\) is the adjoint of matrix A, which is the transpose of the cofactor matrix.

Determinant of Matrix A

First, let's calculate the determinant of the given matrix A:

\[A = \left( {\begin{array}{} 1&1&3\\ 1&3&{ - 3}\\ { - 2}&{ - 4}&{ - 4} \end{array}} \right)\]

The determinant \(\det(A)\) is calculated as:

\[\det(A) = 1 \cdot \det\left( {\begin{array}{} 3&{ - 3}\\ { - 4}&{ - 4} \end{array}} \right) - 1 \cdot \det\left( {\begin{array}{} 1&{ - 3}\\ { - 2}&{ - 4} \end{array}} \right) + 3 \cdot \det\left( {\begin{array}{} 1&3\\ { - 2}&{ - 4} \end{array}} \right)\]

\[\det(A) = 1 \cdot ((3)(-4) - (-3)(-4)) - 1 \cdot ((1)(-4) - (-3)(-2)) + 3 \cdot ((1)(-4) - (3)(-2))\]

\[\det(A) = 1 \cdot (-12 - 12) - 1 \cdot (-4 - 6) + 3 \cdot (-4 - (-6))\]

\[\det(A) = 1 \cdot (-24) - 1 \cdot (-10) + 3 \cdot (-4 + 6)\]

\[\det(A) = -24 + 10 + 3 \cdot (2)\]

\[\det(A) = -24 + 10 + 6\]

\[\det(A) = -24 + 16\]

\[\det(A) = -8\]

Cofactor Matrix of A

Next, we find the cofactor matrix. The cofactor \(C_{ij}\) for each element \(a_{ij}\) is given by \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the minor determinant obtained by removing the \(i\)-th row and \(j\)-th column.

• \(C_{11} = (-1)^{1+1} \det\left( {\begin{array}{} 3&{ - 3}\\ { - 4}&{ - 4} \end{array}} \right) = 1 \cdot (-12 - 12) = -24\)
• \(C_{12} = (-1)^{1+2} \det\left( {\begin{array}{} 1&{ - 3}\\ { - 2}&{ - 4} \end{array}} \right) = -1 \cdot (-4 - 6) = -1 \cdot (-10) = 10\)
• \(C_{13} = (-1)^{1+3} \det\left( {\begin{array}{} 1&3\\ { - 2}&{ - 4} \end{array}} \right) = 1 \cdot (-4 - (-6)) = 1 \cdot (-4 + 6) = 2\)
• \(C_{21} = (-1)^{2+1} \det\left( {\begin{array}{} 1&3\\ { - 4}&{ - 4} \end{array}} \right) = -1 \cdot (-4 - (-12)) = -1 \cdot (-4 + 12) = -1 \cdot (8) = -8\)
• \(C_{22} = (-1)^{2+2} \det\left( {\begin{array}{} 1&3\\ { - 2}&{ - 4} \end{array}} \right) = 1 \cdot (-4 - (-6)) = 1 \cdot (-4 + 6) = 2\)
• \(C_{23} = (-1)^{2+3} \det\left( {\begin{array}{} 1&1\\ { - 2}&{ - 4} \end{array}} \right) = -1 \cdot (-4 - (-2)) = -1 \cdot (-4 + 2) = -1 \cdot (-2) = 2\)
• \(C_{31} = (-1)^{3+1} \det\left( {\begin{array}{} 1&3\\ 3&{ - 3} \end{array}} \right) = 1 \cdot (-3 - 9) = -12\)
• \(C_{32} = (-1)^{3+2} \det\left( {\begin{array}{} 1&3\\ 1&{ - 3} \end{array}} \right) = -1 \cdot (-3 - 3) = -1 \cdot (-6) = 6\)
• \(C_{33} = (-1)^{3+3} \det\left( {\begin{array}{} 1&1\\ 1&3 \end{array}} \right) = 1 \cdot (3 - 1) = 2\)

The cofactor matrix \(C\) is:

Adjoint of Matrix A

The adjoint of matrix A, \(\text{adj}(A)\), is the transpose of the cofactor matrix \(C\).

\[\text{adj}(A) = C^T\]

Inverse of Matrix A

Finally, we calculate the inverse of matrix A using the formula \(A^{-1} = \frac{1}{\det(A)} \text{adj}(A)\).

\[A^{-1} = \frac{1}{-8} \left( {\begin{array}{} { - 24}&{ - 8}&{ - 12}\\ {10}&2&6\\ 2&2&2 \end{array}} \right)\]

Now, divide each element of the adjoint matrix by -8:

\[A^{-1} = \left( {\begin{array}{} {\frac{-24}{-8}}&{\frac{-8}{-8}}&{\frac{-12}{-8}}\\ {\frac{10}{-8}}&{\frac{2}{-8}}&{\frac{6}{-8}}\\ {\frac{2}{-8}}&{\frac{2}{-8}}&{\frac{2}{-8}} \end{array}} \right)\]

\[A^{-1} = \left( {\begin{array}{} {3}&{1}&{1.5}\\ { - 1.25}&{ - 0.25}&{ - 0.75}\\ { - 0.25}&{-0.25}&{ - 0.25} \end{array}} \right)\]

Comparing this result with the given options, we find that this matches option 3.