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Section 1.1 Determinants and Applications of Determinants

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Subsection 1.1.1 Determinants and their properties

Determinant is a very useful mathematical concept. In linear algebra we use determinants to

  • Determine if a matrix is singular, i.e., it has no inverse.
  • Determine if a system has no solution.
  • Solve systems of equations using Cramer’s Rule.
  • Find the inverse of a matrix.
Definition 1.1.1. First order determinant.

The determinant of a scalar is called a first order determinant and it is equal to the scalar:

\begin{equation*} \text{ if } \boldsymbol{A} = [a], \text{ then } \text{det}(\boldsymbol{A}) = a. \end{equation*}
Definition 1.1.2. Second order determinant.

A second order determinant is the determinant of a \(2 \times 2 \)matrix, and it is defined as

\begin{equation*} \boldsymbol{A} = \left[ \begin{array}{cc} a_{11} \amp a_{12}\\ a_{21} \amp a_{22} \end{array} \right], \hspace{3cm} \text{det}(\boldsymbol{A}) = a_{11} a_{22} - a_{21} a_{22}. \end{equation*}

Find the determinant of the given matrix.

\begin{equation*} \boldsymbol{A} = \begin{array}{lrrr} \lceil \amp {-7} \amp {5} \amp \rceil \\ \lfloor \amp {-6} \amp {8} \amp \rfloor \\ \end{array} \end{equation*}

\(\detA{\bA{A}}\) =

\begin{equation*} \, \end{equation*}
Answer

\(-26\)

Solution
\begin{equation*} \detA{\bA{A}} = \begin{array}{lrrr} \vert \amp {-7} \amp {5} \amp \vert \\ \vert \amp {-6} \amp {8} \amp \vert \\ \end{array} = ({-7})\times ({8}) - ({5}) \times ({-6}) = {-26} \end{equation*}

The inverse of a \(2 \times 2\) matrix

\begin{equation*} \bA{A} = \begin{array}{lccr} \lceil\amp a \amp b \amp \rceil\\ \lfloor \amp c \amp d \amp \rfloor \end{array}, \end{equation*}

can be found as

\begin{equation*} \bA{A}^{-1} = \frac{1}{\detA{\bA{A}}} \begin{array}{lccr} \lceil\amp d \amp -b \amp \rceil\\ \lfloor \amp -c \amp a \amp \rfloor \end{array}. \end{equation*}

Find the inverse of the given matrix.

\begin{equation*} \boldsymbol{A} = \begin{array}{lrrr} \lceil \amp {-5} \amp {-2} \amp \rceil \\ \lfloor \amp {2} \amp {2} \amp \rfloor \\ \end{array} \end{equation*}

\(\lceil\)

\(\rceil\)
\(\boldsymbol{A}^{-1}=\) \(\lfloor\)

\(\rfloor\)
\begin{equation*} \, \end{equation*}
Answer 1

\(-0.333333\)

Answer 2

\(-0.333333\)

Answer 3

\(0.333333\)

Answer 4

\(0.833333\)

Solution
\begin{equation*} \detA{\bA{A}} = \begin{array}{lrrr} \vert \amp {-5} \amp {-2} \amp \vert \\ \vert \amp {2} \amp {2} \amp \vert \\ \end{array} = ({-5})\times ({2}) - ({-2}) \times ({2}) = {-6} \end{equation*}
\begin{equation*} \bA{A}^{-1} = \frac{1}{{-6}} \begin{array}{lrrr} \vert \amp {2} \amp -({-2}) \amp \vert \\ \vert \amp -({2}) \amp {-5} \amp \vert \\ \end{array} = \begin{array}{lrrr} \vert \amp {-0.333333} \amp {-0.333333} \amp \vert \\ \vert \amp {0.333333} \amp {0.833333} \amp \vert \\ \end{array} \end{equation*}

Subsubsection 1.1.1.1 Properties of determinants

  • A determinant is a number associated with a square matrix \(\boldsymbol{A} \in \boldsymbol{F}^{n\times n}\text{.}\)
  • Given a matrix \(\boldsymbol{A}\text{,}\) \(\text{det}(\boldsymbol{A})\) is a unique number.
  • Determinants are defined only for square matrices.
  • Multiplicativity
    \begin{equation*} \detA{A\,B} = \detA{A}\, \detA{B} \end{equation*}
  • Invariance under transpose
    \begin{equation*} \text{det}(\boldsymbol{A}) = \text{det}(\boldsymbol{A^{\mathsf{T}}}) \end{equation*}

  • Invariance under row operations.

    If \(\boldsymbol{A} \in \boldsymbol{F}^{n\times n} \text{,}\) for \(n >1\) and \(\boldsymbol{B}\) is the matrix obtained from \(\boldsymbol{A}\) by summing up the multiple of any row/column to another row/column, then

    \begin{equation*} \detA{B} = \detA{A} \end{equation*}
    \begin{equation*} \begin{array}{ccccc} \boldsymbol{A} = \left[ \begin{array}{cc} 3 \amp 1\\ 0 \amp -2 \end{array} \right], \amp \hspace{0.5cm}\amp \boldsymbol{B} = \left[ \begin{array}{cc} 3\amp1 \\ 9 \amp 1\end{array} \right], \amp \hspace{0.5cm}\amp \boldsymbol{C} = \left[ \begin{array}{cc} 3 \amp 4\\ 0 \amp -2 \end{array} \right]. \\ \amp\amp\\ \text{det}(\boldsymbol{A}) = 3\cdot (-2) - 1 \cdot 0 = -6, \amp\hspace{0.5cm}\amp \text{det}(\boldsymbol{B}) = 3\cdot 1 - 1 \cdot 9 = -6, \amp \hspace{0.5cm}\amp \text{det}(\boldsymbol{C}) = 3\cdot (-2) - 4 \cdot 0 = -6. \end{array} \end{equation*}

  • There is a change of sign under row/column interchange.

    If \(\boldsymbol{A} \in \boldsymbol{F}^{n\times n} \text{,}\) for \(n >1\) and \(\boldsymbol{B}\) is the matrix obtained from \(\boldsymbol{A}\) by interchanging two rows or two columns, then

    \begin{equation*} \detA{B} = - \detA{A} \end{equation*}
    \begin{equation*} \begin{array}{ccccc} \boldsymbol{A} = \left[ \begin{array}{cc} 3 \amp 1\\ 0 \amp -2 \end{array} \right], \amp \hspace{0.5cm}\amp \boldsymbol{B} = \left[ \begin{array}{cc} 0 \amp -2 \\3 \amp 1\end{array} \right], \amp \hspace{0.5cm}\amp \boldsymbol{C} = \left[ \begin{array}{cc} 1 \amp 3 \\-2 \amp 0\end{array} \right]. \\ \amp\amp\\ \text{det}(\boldsymbol{A}) = 3\cdot (-2) - 1 \cdot 0 = -6, \amp\hspace{0.5cm}\amp \text{det}(\boldsymbol{B}) = 0\cdot 1 - (-2) \cdot 3 = 6, \amp \hspace{0.5cm}\amp \text{det}(\boldsymbol{C}) = 1\cdot 0 - 3 \cdot (-2) = 6. \end{array} \end{equation*}
  • Determinant of an identity matrix
    \begin{equation*} \detA{I} = 1 \end{equation*}
  • Determinant of a inverse matrix
    \begin{equation*} \detA{A^{-1}}= \frac{1}{\detA{A}} \end{equation*}
    \begin{equation*} \begin{array}{ccc} \boldsymbol{A} = \left[ \begin{array}{cc} 1 \amp -2 \\ -1 \amp 4 \end{array}\right] \amp\hspace{0.5cm} \amp \boldsymbol{B} = \left[ \begin{array}{cc} 2 \amp 1 \\ \frac{1}{2} \amp \frac{1}{2} \end{array}\right]\\ \amp\amp\\ \detA{A} = 2 \amp\hspace{0.5cm} \amp \detA{B} = \frac{1}{2} \end{array} \end{equation*}
    \begin{equation*} \begin{array}{ccc} \boldsymbol{A} \boldsymbol{B} \amp=\amp \left[ \begin{array}{cc} 1 \amp -2 \\ -1 \amp 4 \end{array}\right] \left[ \begin{array}{cc} 2 \amp 1 \\ \frac{1}{2} \amp \frac{1}{2} \end{array}\right] = \left[ \begin{array}{cc} 1 \amp 0\\ 0 \amp 1 \end{array}\right]\\ \amp\amp\\ \boldsymbol{B} \boldsymbol{A} \amp=\amp \left[ \begin{array}{cc} 2 \amp 1 \\ \frac{1}{2} \amp \frac{1}{2} \end{array}\right] \left[ \begin{array}{cc} 1 \amp -2 \\ -1 \amp 4 \end{array}\right]= \left[ \begin{array}{cc} 1 \amp 0\\ 0 \amp 1 \end{array}\right] \end{array} \end{equation*}

Subsection 1.1.2 Minors and cofactors

Subsubsection 1.1.2.1 Minors and matrix of minors

Definition 1.1.8. Minors of a determinant.

The minor of an entry in a \(n \times n\) determinant is the \((n-1) \times (n-1)\) determinant found by eliminating the row and column in the \(n \times n\) determinant that contains the entry.\\

Consider the \(3 \times 3\) matrix,

\begin{equation*} \left[ \begin{array}{ccc} a_{11} \amp a_{12} \amp a_{13}\\ a_{21} \amp a_{22} \amp a_{23}\\ a_{31} \amp a_{31} \amp a_{33} \end{array}\right], \end{equation*}

we denote its determinant as

\begin{equation*} \left| \begin{array}{ccc} a_{11} \amp a_{12} \amp a_{13}\\ a_{21} \amp a_{22} \amp a_{23}\\ a_{31} \amp a_{31} \amp a_{33} \end{array}\right|. \end{equation*}

So that,

Definition 1.1.10. Matrix of minors.

Given a square matrix \(\bA{A}_{n\times n}\text{,}\) the matrix of minors is the square matrix \(\bA{M}_{n \times n}\) where each element, \(m_{ij},\) corresponds to the minor of \(a_{ij}\text{.}\)

Find the matrix of minors of the following matrix

\begin{equation*} \bA{A} = \left[ \begin{array}{ccc} 3 \amp -2\amp 1\\ 3 \amp -1\amp -2\\ 3\amp -2 \amp -3 \end{array}\right]. \end{equation*}

We first find the minors of each entry:

\begin{equation*} \begin{array}{lll} M_{11} = \left| \begin{array}{cc} -1 \amp -2\\ -2 \amp -3 \end{array}\right| = -1 \hspace{0.6cm} \amp M_{12} = \left| \begin{array}{cc} 3 \amp -2\\ 3 \amp -3 \end{array}\right| = -3 \amp M_{13} = \left| \begin{array}{cc} 3 \amp -1\\ 3 \amp -2 \end{array}\right| = -3\\ \amp \amp \\ M_{21} = \left| \begin{array}{cc} -2 \amp 1\\ -2 \amp -3 \end{array}\right| = 8 \amp M_{22} = \left| \begin{array}{cc} 3 \amp 1\\ 3 \amp -3 \end{array}\right| = -12 \hspace{0.5cm} \amp M_{23} = \left| \begin{array}{cc} 3 \amp -2\\ 3 \amp -2 \end{array}\right| = 0\\ \amp \amp \\ M_{31} = \left| \begin{array}{cc} -2 \amp 1\\ -1\amp -2 \end{array}\right| = 5 \amp M_{32} = \left| \begin{array}{cc} 3 \amp 1\\ 3 \amp -2 \end{array}\right| = -9 \amp M_{33} = \left| \begin{array}{cc} 3 \amp -2\\ 3 \amp -1 \end{array}\right| = 3 \end{array} \end{equation*}

The matrix of minors is

\begin{equation*} \bA{M} = \left[ \begin{array}{ccc} -1 \amp -3\amp-3\\ 8\amp-12\amp 0\\5\amp-9\amp 3 \end{array}\right]. \end{equation*}

Subsubsection 1.1.2.2 Cofactors and cofactor matrix

Definition 1.1.12. Cofactors.

A Cofactor of a matrix element is defined as

\begin{equation*} C_{ij} = (-1)^{i+j} M_{ij}, \end{equation*}

where \(M_{ij} \) is the element's minor.

Definition 1.1.13. Cofactor matrix.

A cofactor matrix is a matrix having the cofactors as the elements of the matrix.

A simple way to find the cofactors and the cofactor matrix is to consider the corresponding minors and change the sign based on the checkerboard pattern below,

\begin{equation*} \left| \begin{array}{ccc} + \amp -\amp +\\ -\amp +\amp -\\ +\amp -\amp+ \end{array}\right|_{3\times 3} \hspace{1cm} \left| \begin{array}{cccc} + \amp -\amp + \amp -\\ -\amp +\amp - \amp +\\ +\amp -\amp+ \amp - \\ -\amp +\amp - \amp + \end{array}\right|_{4\times 4} \hspace{1cm} \left| \begin{array}{ccccc} + \amp -\amp + \amp - \amp +\\ -\amp +\amp - \amp + \amp -\\ +\amp -\amp+ \amp - \amp +\\ -\amp +\amp - \amp + \amp -\\ +\amp -\amp+ \amp - \amp + \end{array}\right|_{5\times 5} \cdots \end{equation*}

Consider the matrix from the previous example and its matrix of minors,

\begin{equation*} \bA{A} = \left[ \begin{array}{ccc} 3 \amp -2\amp 1\\ 3 \amp -1\amp -2\\ 3\amp -2\amp-3 \end{array}\right], \hspace{2cm} \bA{M} = \left[ \begin{array}{ccc} -1 \amp -3\amp -3\\ 8\amp -12\amp 0\\5\amp -9\amp 3 \end{array}\right]. \end{equation*}

Then the Cofactor matrix is simply,

\begin{equation*} \bA{C} = \left[ \begin{array}{ccc} -1 \amp 3\amp -3\\ -8\amp -12\amp 0\\5\amp 9 \amp3 \end{array}\right]. \end{equation*}

Subsection 1.1.3 Determinants of \(n \times n\) matrices with \(n >2 \)

To find the determinant of a \(3 \times 3\) matrix, we multiply each entry of a given row or column by their corresponding cofactors.

Find the determinant of \(\displaystyle \left[ \begin{array}{ccc} 2 \amp -3 \amp -1 \\ 3 \amp 2\amp 0\\ -1 \amp -1\amp -2 \end{array}\right] \)

  • We could find the determinant using the cofactors of the first row:

    \begin{equation*} \begin{array}{ccc} \left| \begin{array}{ccc} 2 \amp -3 \amp -1 \\ 3 \amp 2\amp 0\\ -1 \amp -1\amp -2 \end{array}\right| \amp =\amp 2 \left| \begin{array}{cc} 2 \amp 0 \\ -1 \amp -2 \end{array}\right| - (-3) \left| \begin{array}{cc} 3 \amp 0 \\ -1 \amp -2 \end{array}\right| + (-1) \left| \begin{array}{cc} 3\amp 2 \\ -1 \amp -1 \end{array}\right|\\ \amp =\amp 2 \left(-4 -0\right) + 3 \left( -6+0\right) - 1 \left(-3 +2\right)\\ \amp = \amp -25 \end{array} \end{equation*}

  • We could also use the second row and its cofactors. Note that the last entry in the row is zero, which will save us the time to calculate its minor.

    \begin{equation*} \begin{array}{ccc} \left| \begin{array}{ccc} 2 \amp -3 \amp -1 \\ 3 \amp 2\amp 0\\ -1 \amp -1\amp -2 \end{array}\right| \amp =\amp - 3 \left| \begin{array}{cc} -3 \amp -1 \\ -1 \amp -2 \end{array}\right| + 2 \left| \begin{array}{cc} 2 \amp -1 \\ -1 \amp -2 \end{array}\right| + 0\\ \amp =\amp -3 \left(6 -1\right) + 2 \left( -4-1\right)\\ \amp = \amp -25 \end{array} \end{equation*}

  • Alternatively, we can use a column. For example we can use the entries in column 3 and their cofactors,

    \begin{equation*} \begin{array}{ccc} \left| \begin{array}{ccc} 2 \amp -3 \amp -1 \\ 3 \amp 2\amp 0\\ -1 \amp -1\amp -2 \end{array}\right| \amp =\amp - 1 \left| \begin{array}{cc} 3 \amp 2 \\ -1 \amp -1 \end{array}\right| + 0 + (-2) \left| \begin{array}{cc} 2 \amp -3 \\ 3 \amp 2 \end{array}\right|\\ \amp =\amp -1 \left(-3 +2\right) - 2 \left( 4+ 9\right)\\ \amp =\amp -25 \end{array} \end{equation*}

Find the determinant of the given matrix.

\begin{equation*} \boldsymbol{A} = \begin{array}{lrrrr} \lceil \amp {-3} \amp {-8} \amp {-9} \amp \rceil \\ \vert \amp {-3} \amp {7} \amp {-2} \amp \vert \\ \lfloor \amp {3} \amp {4} \amp {6} \amp \rfloor \\ \end{array} \end{equation*}

\(\detA{\bA{A}}\) =

\begin{equation*} \, \end{equation*}
Answer

\(51\)

Solution
\begin{equation*} \detA{\bA{A}} = {-3} \times\begin{array}{lrrr} \vert \amp {7} \amp {-2} \amp \vert \\ \vert \amp {4} \amp {6} \amp \vert \\ \end{array} -({-8}) \times\begin{array}{lrrr} \vert \amp {-3} \amp {-2} \amp \vert \\ \vert \amp {3} \amp {6} \amp \vert \\ \end{array} +({-9}) \times \begin{array}{lrrr} \vert \amp {-3} \amp {7} \amp \vert \\ \vert \amp {3} \amp {4} \amp \vert \\ \end{array} = {51} \end{equation*}
Insight 1.1.17. Determinants larger than \(3 \times 3 \).

For \(n \times n \) matrices where \(n>3 \) the procedure consists of first using the minors to create \(3 \times 3\) determinants and then use minors again to get \(2 \times 2\) determinants.

Find the determinant of \(\displaystyle \left[ \begin{array}{cccc} 1 \amp -3 \amp 0 \amp 2\\ 0 \amp 1 \amp 1 \amp 0\\ 0 \amp 4 \amp 0 \amp 3\\ 0 \amp 0 \amp 1 \amp 0 \end{array} \right]\)

We use the first column and its cofactors to reduce the problem to calculating a \(3 \times 3 \) determinant:

\begin{equation*} \begin{array}{ccc} \left| \begin{array}{cccc} 1 \amp -3 \amp 0 \amp 2\\ 0 \amp 1 \amp 1 \amp 0\\ 0 \amp 4 \amp 0 \amp 3\\ 0 \amp 0 \amp 1 \amp 0 \end{array} \right| \amp = \amp 1 \left| \begin{array}{ccc} 1 \amp 1 \amp 0\\ 4 \amp 0 \amp 3\\ 0 \amp 1 \amp 0 \end{array} \right| \end{array} \end{equation*}

Next we use the last row and its cofactors and reduce the problem to calculating a \(2 \times 2 \) determinant:

\begin{equation*} \begin{array}{ccc} \left| \begin{array}{ccc} 1 \amp 1 \amp 0\\ 4 \amp 0 \amp 3\\ 0 \amp 1 \amp 0 \end{array} \right| \amp = \amp - 1 \left| \begin{array}{cc} 1 \amp 0\\ 4 \amp 3\\ \end{array} \right| \\ \amp = \amp -(3 -0) = -3. \end{array} \end{equation*}

Find the determinant of the given matrix.

\begin{equation*} \boldsymbol{A} = \begin{array}{lrrrrr} \lceil \amp 1 \amp {1} \amp {-2} \amp {2} \amp \rceil \\ \vert \amp 0 \amp 0 \amp 1 \amp 0 \amp \vert \\ \vert \amp 0 \amp {7} \amp 0 \amp 1 \amp \vert \\ \lfloor \amp 0 \amp 1 \amp {1} \amp {-4} \amp \rfloor \\ \end{array} \end{equation*}

\(\detA{\bA{A}}\) =

\begin{equation*} \, \end{equation*}
Answer

\(29\)

Solution

We use row 1 to reduce the problem to a \(3 \times 3\) determinant:

\begin{equation*} \detA{\bA{A}} = \begin{array}{lrrrr} \vert \amp 0 \amp 1 \amp 0 \amp \vert \\ \vert \amp {7} \amp 0 \amp 1 \amp \vert \\ \vert \amp 1 \amp {1} \amp {-4} \amp \vert \\ \end{array} \end{equation*}

We use row 1 to reduce the problem to a \(2 \times 2\) determinant:

\begin{equation*} \detA{\bA{A}} = \begin{array}{lrrr} \vert \amp {7} \amp 1 \amp \vert \\ \vert \amp 1 \amp {-4} \amp \vert \\ \end{array} = ({7}) \times ({-4}) - (1) \times (1) = {29} \end{equation*}

Subsection 1.1.4 Adjoint and inverse matrices

Definition 1.1.20. Adjugate matrix.

Given a square matrix \(\bA{A}\text{,}\) its adjugate matrix also known as adjoint matrix, \(\text{adj}(\bA{A}),\) is defined as the transpose of the cofactor matrix of \(\bA{A}\text{.}\)

Insight 1.1.21. Using adjoint matrix to find a matrix inverse.
\begin{equation*} \bA{A}^{-1} = \frac{\text{adj}(\bA{A})}{\detA{\bA{A}}}. \end{equation*}
Steps to find the inverse of a matrix using determinants:
  1. Determine the matrix of minors.
  2. Transform the matrix of minors into a matrix of cofactors.
  3. Find the adjoint of the matrix as the transpose of the cofactor matrix.
  4. Divide the adjoint by the determinant

Find the inverse of \(\bA{A} = \left[ \begin{array}{ccc} 2 \amp -3 \amp -1\\ 3 \amp 2 \amp 0\\ -1 \amp 0 \amp -2 \end{array}\right]\)

First we check whether the matrix has an inverse by calculating its determinant,

\begin{equation*} \begin{array}{ccc} \left| \begin{array}{ccc} 2 \amp -3 \amp -1\\ 3 \amp 2 \amp 0\\ -1 \amp 0 \amp -2 \end{array}\right| \amp = \amp -3 \left| \begin{array}{ccc} -3 \amp -1\\ 0 \amp -2 \end{array}\right| + 2 \left| \begin{array}{ccc} 2 \amp -1\\ -1 \amp -2 \end{array}\right| \\ \amp = \amp -3 (6) + 2 (-5) \\ \amp = \amp -28. \end{array} \end{equation*}

Since the \(\detA{A} \ne 0 \) the matrix is invertible.

Next, we find the matrix of minors:

\begin{equation*} \begin{array}{lll} M_{11} = \left| \begin{array}{ccc} 2 \amp 0\\ 0 \amp -2 \end{array}\right| = -4 \hspace{0.6cm} \amp M_{12} = \left| \begin{array}{ccc} 3 \amp 0\\ -1 \amp -2 \end{array}\right| = -6 \hspace{0.6cm} \amp M_{13} = \left| \begin{array}{ccc} 3 \amp 2 \\ -1 \amp 0 \end{array}\right| = 2\\ \amp \amp\\ M_{21} = \left| \begin{array}{ccc} -3 \amp -1\\ 0 \amp -2 \end{array}\right| = 6 \hspace{0.6cm} \amp M_{22} = \left| \begin{array}{ccc} 2 \amp -1\\ -1 \amp -2 \end{array}\right| = -5 \hspace{0.6cm} \amp M_{23} = \left| \begin{array}{ccc} 2 \amp -3 \\ -1 \amp 0 \end{array}\right| = -3\\ \amp \amp \\ M_{31} = \left| \begin{array}{ccc} -3 \amp -1\\ 2 \amp 0\\ \end{array}\right| = 2 \hspace{0.6cm} \amp M_{32} = \left| \begin{array}{ccc} 2 \amp -1\\ 3 \amp 0\\ \end{array}\right| = 3 \hspace{0.6cm} \amp M_{33} = \left| \begin{array}{ccc} 2 \amp -3 \\ 3 \amp 2 \\ \end{array}\right| = 13 \end{array} \end{equation*}
\begin{equation*} \bA{M} = \left[ \begin{array}{ccc} -4 \amp -6 \amp 2\\ 6 \amp -5 \amp -3\\ 2 \amp 3 \amp 13 \end{array}\right]. \end{equation*}

Next we find the cofactor matrix:

\begin{equation*} \bA{C} = \left[ \begin{array}{ccc} -4 \amp 6 \amp 2\\ -6 \amp -5 \amp 3\\ 2 \amp -3 \amp 13 \end{array}\right]. \end{equation*}

And the adjoint matrix:

\begin{equation*} \text{adj}\left(\bA{A}\right) = \left[ \begin{array}{ccc} -4 \amp -6 \amp 2\\ 6 \amp -5 \amp -3\\ 2 \amp 3 \amp 13 \end{array}\right]. \end{equation*}

Finally we find the inverse:

\begin{equation*} \bA{A}^{-1} = \frac{\text{adj}\left(\bA{A}\right)}{\detA{\bA{A}}} = \frac{1}{-28} \left[ \begin{array}{ccc} -4 \amp -6 \amp 2\\ 6 \amp -5 \amp -3\\ 2 \amp 3 \amp 13 \end{array}\right]. \end{equation*}

Find the inverse of the given matrix.

\begin{equation*} \boldsymbol{A} = \begin{array}{lrrrr} \lceil \amp 1 \amp {-1} \amp {2} \amp \rceil \\ \vert \amp {1} \amp {2} \amp {-1} \amp \vert \\ \lfloor \amp -1 \amp {6} \amp 0 \amp \rfloor \\ \end{array} \end{equation*}

\(\,\)

First find its determinant: \(\hspace{1cm} \detA{\bA{A}}\) =

\begin{equation*} \, \end{equation*}

Next, find the matrix of minors >>


 \(\lceil\)


\(\rceil\)
\(\boldsymbol{M}=\) \(\vert\)


\(\vert\)

\(\lfloor\)


\(\rfloor\)

<<

Next, find the cofactor matrix: >>


 \(\lceil\)


\(\rceil\)
\(\boldsymbol{C}=\) \(\vert\)


\(\vert\)

\(\lfloor\)


\(\rfloor\)

<<

Next, find the adjoint matrix: >>


 \(\lceil\)


\(\rceil\)
\(\text{adj}(\boldsymbol{A})=\) \(\vert\)


\(\vert\)

\(\lfloor\)


\(\rfloor\)

<<

Finally, divide the adjoint matrix by the determinant: >>


 \(\lceil\)


\(\rceil\)
\(\boldsymbol{A}^{-1}=\) \(\vert\)


\(\vert\)

\(\lfloor\)


\(\rfloor\)

<<

Answer 1

\(21\)

Answer 2

\(6\)

Answer 3

\(-1\)

Answer 4

\(8\)

Answer 5

\(-12\)

Answer 6

\(2\)

Answer 7

\(5\)

Answer 8

\(-3\)

Answer 9

\(-3\)

Answer 10

\(3\)

Answer 11

\(6\)

Answer 12

\(1\)

Answer 13

\(8\)

Answer 14

\(12\)

Answer 15

\(2\)

Answer 16

\(-5\)

Answer 17

\(-3\)

Answer 18

\(3\)

Answer 19

\(3\)

Answer 20

\(6\)

Answer 21

\(12\)

Answer 22

\(-3\)

Answer 23

\(1\)

Answer 24

\(2\)

Answer 25

\(3\)

Answer 26

\(8\)

Answer 27

\(-5\)

Answer 28

\(3\)

Answer 29

\(0.285714\)

Answer 30

\(0.571429\)

Answer 31

\(-0.142857\)

Answer 32

\(0.047619\)

Answer 33

\(0.0952381\)

Answer 34

\(0.142857\)

Answer 35

\(0.380952\)

Answer 36

\(-0.238095\)

Answer 37

\(0.142857\)

Subsection 1.1.5 Cramer's rule and systems of equations

Cramer's rule is used to solve systems of linear equations.

\begin{equation*} \bA{A}\, \bA{x} = \bA{b}, \end{equation*}

where the coefficient matrix is square, \(\bA{A} \in \mathbb{F}^{n\times n}, \) and \(\detA{\bA{A}} \ne 0\text{.}\)

If the above conditions are met, Cramer's rule states that the solution of the system is given by

\begin{equation*} x_i = \frac{\detA{\bA{A}_i}}{\detA{\bA{A}}}, \end{equation*}

where \(\bA{A}_i \) is the matrix formed by replacing the \(i\)th column of \(\bA{A}\) with the \(\bA{b}\) vector.

Solve the following system of equations using Cramer's rule,

\begin{equation*} \begin{array}{rrrrrrr} 4x_1 \amp+\amp 5x_2 \amp - \amp 2x_3 \amp = \amp 1\\ 7x_1 \amp +\amp 12x_2 \amp - \amp 6x_3 \amp = \amp -1\\ 2x_1 \amp +\amp 6x_2 \amp \amp \amp = \amp 2 \end{array} \end{equation*}

We first determine \(\bA{A} \) and \(\detA{\bA{A}}\text{,}\)

\begin{equation*} \bA{A} = \left[\begin{array}{ccc} 4 \amp 5 \amp -2 \\ 7 \amp 12 \amp -6\\ 2 \amp 6 \amp 0 \end{array}\right] \end{equation*}
\begin{equation*} \detA{\bA{A}} = \left|\begin{array}{ccc} 4 \amp 5 \amp -2 \\ 7 \amp 12 \amp -6\\ 2 \amp 6 \amp 0 \end{array}\right| = 2 \left|\begin{array}{ccc} 5 \amp -2 \\ 12 \amp -6 \end{array}\right| - 6 \left|\begin{array}{ccc} 4 \amp -2 \\ 7 \amp -6 \end{array}\right| = 2(-6) -6 (-10)= -12 + 60 = 48. \end{equation*}

Next, we find the determinants of the \(\bA{A}_i's\text{,}\)

\begin{equation*} \detA{\bA{A}_1} = \left|\begin{array}{ccc} 1 \amp 5 \amp -2 \\ -1 \amp 12 \amp -6\\ 2 \amp 6 \amp 0 \end{array}\right| = 2 \left|\begin{array}{ccc} 5 \amp -2 \\ 12 \amp -6 \end{array}\right| - 6 \left|\begin{array}{ccc} 1 \amp -2 \\ -1\amp -6 \end{array}\right| = 2(-6) - 6 (-8) = -12 + 48 = 36. \end{equation*}
\begin{equation*} \detA{\bA{A}_2} = \left|\begin{array}{ccc} 4 \amp 1 \amp -2 \\ 7 \amp -1 \amp -6\\ 2 \amp 2 \amp 0 \end{array}\right| = 2 \left|\begin{array}{ccc} 1 \amp -2 \\ -1 \amp -6 \end{array}\right| - 2 \left|\begin{array}{ccc} 4 \amp -2 \\ 7\amp -6 \end{array}\right| = 2(-8) -2 (-10) = -16 + 20 = 4. \end{equation*}
\begin{equation*} \detA{\bA{A}_3} = \left|\begin{array}{ccc} 4 \amp 5 \amp 1 \\ 7 \amp 12 \amp -1\\ 2 \amp 6 \amp 2 \end{array}\right| = 4 \left|\begin{array}{ccc} 12 \amp -1\\ 6 \amp 2 \end{array}\right| - 5 \left|\begin{array}{ccc} 7 \amp -1\\ 2 \amp 2 \end{array}\right| + 1 \left|\begin{array}{ccc} 7 \amp 12 \\ 2 \amp 6 \end{array}\right| = 4 (30) -5 (16) + 1 (18) =58. \end{equation*}

Finally, we find the solution as,

\begin{equation*} \begin{array}{ccc} x_1 \amp = \amp \displaystyle \frac{\detA{\bA{A}_1}}{\detA{\bA{A}}} = \frac{36}{48} = \frac{3}{4},\\ \amp \amp\\ x_2 \amp = \amp \displaystyle\frac{\detA{\bA{A}_2}}{\detA{\bA{A}}} = \frac{4}{48} = \frac{1}{12},\\ \amp \amp \\ x_3 \amp = \amp \displaystyle \frac{\detA{\bA{A}_3}}{\detA{\bA{A}}} = \frac{58}{48} = \frac{29}{24}. \end{array} \end{equation*}

Solve the given system of equations using Cramer’s rule.

\begin{equation*} \begin{array}{ccc} 1x_1 + {1} x_2 + {-2} x_3 \amp = \amp {2} \\ {-1}x_1 + {2} x_2 + {3} x_3 \amp = \amp {1} \\ -1x_1 + {8} x_2 + 0 x_3 \amp = \amp {2} \\ \end{array} \end{equation*}

\(\,\)

First find the determinant of the matrix of coefficients \(\hspace{1cm} \detA{\bA{A}}\) =

\begin{equation*} \, \end{equation*}

Next, find the determinant of the matrix corresponding to \(x_1\)

\(\hspace{1cm} \detA{\bA{A}_1}\) =,

\begin{equation*} \, \end{equation*}

and use that to find \(\hspace{1cm} x_1\) =

\begin{equation*} \, \end{equation*}

Next, find the determinant of the matrix corresponding to \(x_2\)

\(\hspace{1cm} \detA{\bA{A}_2} =\) ,

\begin{equation*} \, \end{equation*}

and use that to find \(\hspace{1cm} x_2\) =

\begin{equation*} \, \end{equation*}

Finally, find the determinant of the matrix corresponding to \(x_3\)

\(\hspace{1cm} \detA{\bA{A}_3} =\),

\begin{equation*} \, \end{equation*}

and use that to find \(\hspace{1cm} x_3\) =

\begin{equation*} \, \end{equation*}
Answer 1

\(-15\)

Answer 2

\(-50\)

Answer 3

\(3.33333\)

Answer 4

\(-10\)

Answer 5

\(0.666667\)

Answer 6

\(-15\)

Answer 7

\(1\)