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Honors: 8. Matrices
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Grade 912

Honors: 8.04 Inverse matrices and the determinant

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Concept summary

Given a matrix A, the inverse matrix, A^{-1} is a matrix such that AA^{-1}=I where I is the identity matrix. Any matrix where A^{-1} exists is said to be invertible. Only square matrices will have an inverse.

We can use elementary row operations in an augmented matrix to find the inverse of any n\times n square matrix by transforming \left[ \begin{array}{c|c} A & I \end{array} \right] into \left[ \begin{array}{c|c} I & A^{-1} \end{array} \right].

The inverse of a 2\times 2 matrix A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} can also be found using the formula:

\displaystyle A^{-1}=\dfrac{1}{\text{det}(A)}\begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11}\end{bmatrix}
\bm{a_{ij}}
the element in row i and column j

Where \text{det}(A) is the determinant of the matrix. The determinant is a real number associated with any square matrix that can be used in a variety of applications which will be investigated in the next subtopic. The determinant can also be denoted as \left\vert A \right\vert=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}.

The determinant for a 2\times 2 matrix, A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}, can be found with the formula:

\displaystyle \text{det}(A)=\left\vert A \right\vert = a_{11}a_{22}-a_{12}a_{21}
\bm{a_{ij}}
the element in row i and column j

The determinant of a 3\times 3 matrix, A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix}, can be found with the formula:

\displaystyle \text{det}(A)=\left\vert A \right\vert =a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix} +a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}
\bm{a_{ij}}
the element in row i and column j

In the formula for the determinant of a 3\times 3 matrix, the elements a_{11}, a_{12}, and a_{13} are referred to as cofactors and the determinants of the 2\times 2 matrices formed by deleting the row and column of the cofactor are referred to as minors. This alternating pattern of +(\text{cofactor})(\text{minor})-(\text{cofactor})(\text{minor})... can be expanded to any size square matrix.

Worked examples

Example 1

Find the determinant of the following matrices:

a

\begin{bmatrix} 8 & 3 \\ -4 & 2\end{bmatrix}

Approach

The determinant of a 2 \times 2 matrix is the difference between the products of the diagonals.

That is, if A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}, then \left\vert A\right\vert=\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}.

Solution

The determinant of \begin{bmatrix} 8 & 3 \\ -4 & 2\end{bmatrix} is:

\displaystyle \begin{vmatrix}8 & 3 \\ -4 & 2\end{vmatrix}\displaystyle =\displaystyle (8)(2)-(3)(-4)Apply the formula
\displaystyle =\displaystyle 16-(-12)Evaluate the products
\displaystyle =\displaystyle 28Evaluate the difference
b

\begin{bmatrix} 2 & 0 & -3 \\ 1 & -2 & 5\\ 0 & 2 & -1 \end{bmatrix}

Approach

If A=\begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{bmatrix}, then \left\vert A \right\vert=\begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33}\end{vmatrix}=a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}-a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix} +a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix}.

Solution

The determinant of \begin{bmatrix} 2 & 0 & -3 \\ 1 & -2 & 5\\ 0 & 2 & -1 \end{bmatrix} is:

\displaystyle \begin{vmatrix}2 & 0 & -3 \\ 1 & -2 & 5\\ 0 & 2 & -1 \end{vmatrix}\displaystyle =\displaystyle 2\begin{vmatrix} -2 & 5 \\ 2 & -1\end{vmatrix}-0\begin{vmatrix} 1 & 5 \\ 0 & -1\end{vmatrix} +(-3)\begin{vmatrix} 1 & -2 \\ 0 & 2\end{vmatrix}Apply the formula
\displaystyle =\displaystyle 2\left((-2)(-1)-(5)(2)\right)-0\left((1)(-1)-(5)(0)\right)\\ +(-3)\left((1)(2)-(-2)(0)\right) Apply the formula
\displaystyle =\displaystyle 2(2-10)-0(-1-0)-3(2-0)Evaluate the products
\displaystyle =\displaystyle 2(-8)-0(-1)-3(2)Evaluate the difference
\displaystyle =\displaystyle -22Simplify

Reflection

Be sure to alternate the signs between each (\text{cofactor})(\text{minor}) product.

Example 2

Consider the matrix A=\begin{bmatrix} 9 & -6 \\ -11 & 8\end{bmatrix}.

a

Find the inverse of A.

Approach

The inverse of matrix A=\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} is the matrix A^{-1}=\dfrac{1}{\text{det}(A)}\begin{bmatrix} a_{22} & -a_{12} \\ -a_{21} & a_{11}\end{bmatrix}. We will first find the value of the determinant, then find the inverse matrix.

Solution

The determinant of A=\begin{bmatrix} 9 & -6 \\ -11 & 8\end{bmatrix} is:

\displaystyle \begin{vmatrix} 9 & -6 \\ -11 & 8\end{vmatrix}\displaystyle =\displaystyle (9)(8)-(-6)(-11)Apply the formula
\displaystyle =\displaystyle 72-66Evaluate the products
\displaystyle =\displaystyle 6Evaluate the difference

Then, the inverse of A is given by

\displaystyle A^{-1}\displaystyle =\displaystyle \dfrac{1}{6}\begin{bmatrix} 8 & -(-6) \\ -(-11) & 9\end{bmatrix}Apply the formula
\displaystyle =\displaystyle \dfrac{1}{6}\begin{bmatrix} 8 & 6 \\ 11 & 9\end{bmatrix}Simplify the elements
\displaystyle =\displaystyle \begin{bmatrix} \dfrac{8}{6} & \dfrac{6}{6} \\ \\ \dfrac {11}{6} & \dfrac{9}{6}\end{bmatrix}Distribute \dfrac{1}{6}
\displaystyle =\displaystyle \begin{bmatrix} \dfrac{4}{3} & 1 \\ \\ \dfrac {11}{6} & \dfrac{3}{2}\end{bmatrix}Simplify
b

Find AA^{-1}.

Solution

From part (a), we get A^{-1}=\begin{bmatrix} \dfrac{4}{3} & 1 \\ \\ \dfrac {11}{6} & \dfrac{3}{2}\end{bmatrix}. The product AA^{-1} is:

\displaystyle AA^{-1}\displaystyle =\displaystyle \begin{bmatrix} 9 & -6 \\ -11 & 8\end{bmatrix}\begin{bmatrix} \dfrac{4}{3} & 1 \\ \\ \dfrac {11}{6} & \dfrac{3}{2}\end{bmatrix}Apply the formula
\displaystyle =\displaystyle \begin{bmatrix} (9)\left(\dfrac{4}{3}\right)+(-6)\left(\dfrac {11}{6}\right) & (9)(1)+(-6)\left(\dfrac{3}{2}\right) \\ \\ (-11)\left(\dfrac{4}{3}\right)+(8)\left(\dfrac {11}{6}\right)&(-11)(1)+(8)\left(\dfrac{3}{2}\right)\end{bmatrix}Multiply
\displaystyle =\displaystyle \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}Simplify

The matrix AA^{-1} is an identity matrix.

Reflection

Multiplying an inverse to the original matrix is a good way to check that the inverse is correct. The product of any invertible matrix and its inverse is always an identity matrix, I.

Example 3

Find the inverse of the matrix A=\begin{bmatrix} -2 & 3 & 1 \\ 3 &5 &4\\ -4 & -1 & 2\end{bmatrix}.

Approach

For matrices larger than a 2\times 2, we'll use an augmented matrix and row operations.

Use the given matrix to create an augmented matrix in the form of \left[ \begin{array}{c|c} A & I \end{array} \right] and use row operations to transform A into I on the left hand side of the augmented matrix.

Solution

Begin with the augmented matrix:\left[ \begin{array}{ccc|ccc} -2 & 3 & 1 & 1 & 0 & 0 \\ 3 & 5 & 4 & 0 & 1 & 0 \\ -4 & -1 & 2 & 0 & 0 & 1 \end{array} \right]

To make a_{11}=1: \left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 3 & 5 & 4 & 0 & 1 & 0 \\ -4 & -1 & 2 & 0 & 0 & 1 \end{array} \right]; R_1 \to R_2+R_1

To make a_{21}=0 and a_{31}=0: \left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 0 & -19 & -11 & -3 & -2 & 0 \\ 0 & 31 & 22 & 4 & 4 & 1 \end{array} \right];\begin{aligned}R_2 &\to R_2 - 3R_1\\R_3 &\to R_3+4R_1\end{aligned}

To make a_{22}=1:\left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 0 & -7 & 0 & -2 & 0 & 1 \\ 0 & 31 & 22 & 4 & 4 & 1 \end{array} \right];R_2 \to 2R_2 + R_3

\left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 0 & 1 & 0 & \frac{2}{7} & 0 & -\frac{1}{7} \\ 0 & 31 & 22 & 4 & 4 & 1 \end{array} \right];R_2 \to -\dfrac{1}{7}R_2

To make a_{32}=0:\left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 0 & 1 & 0 & \frac{2}{7} & 0 & -\frac{1}{7} \\ 0 & 0 & 22 & \frac{-34}{7} & 4 & \frac{38}{7} \end{array} \right];R_3 \to R_3 -31R_2

To make a_{33}=1:\left[ \begin{array}{ccc|ccc} 1 & 8 & 5 & 1 & 1 & 0 \\ 0 & 1 & 0 & \frac{2}{7} & 0 & -\frac{1}{7} \\ 0 & 0 & 1 & \frac{-17}{77} & \frac{2}{11} & \frac{19}{77} \end{array} \right];R_3 \to \dfrac{1}{22}R_3

To make a_{12}=0:\left[ \begin{array}{ccc|ccc} 1 & 0 & 5 & -\frac{9}{7} & 1 & \frac{8}{7} \\ 0 & 1 & 0 & \frac{2}{7} & 0 & -\frac{1}{7} \\ 0 & 0 & 1 & \frac{-17}{77} & \frac{2}{11} & \frac{19}{77} \end{array} \right];R_1 \to R_1-8R_2

To make a_{13}=0:\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & -\frac{2}{11} & \frac{1}{11} & -\frac{1}{11} \\ 0 & 1 & 0 & \frac{2}{7} & 0 & -\frac{1}{7} \\ 0 & 0 & 1 & \frac{-17}{77} & \frac{2}{11} & \frac{19}{77} \end{array} \right];R_1 \to R_1-5R_3

Therefore, A^{-1}=\begin{bmatrix} -\dfrac{2}{11} & \dfrac{1}{11} & -\dfrac{1}{11} \\ \\ \dfrac {2}{11} & 0 & -\dfrac{1}{7} \\ \\ -\dfrac {17}{77} & \dfrac{2}{11} & \dfrac{19}{77}\end{bmatrix}

Outcomes

MA.912.NSO.4.1

Given a mathematical or real-world context, represent and manipulate data using matrices.

MA.912.NSO.4.3

Solve mathematical and real-world problems involving addition, subtraction and multiplication of matrices.

MA.912.NSO.4.4

Solve mathematical and real-world problems using the inverse and determinant of matrices.

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