Matrix Algebra and Definitions: Learning Objectives 7, 8, 9

Look at the formula for $A^{-1}$. What mathematical operation becomes impossible if the determinant is zero?

The formula for the inverse of a $2 \times 2$ matrix, $A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}$, involves multiplying the adjugate matrix by the scalar value $\frac{1}{\det(A)}$. If $\det(A)=0$, this scalar becomes $\frac{1}{0}$, which is an undefined value. Division by zero is not a valid mathematical operation, so the formula for the inverse cannot be computed. This is consistent with the broader theorem that a square matrix is invertible if and only if its determinant is non-zero.

For each matrix, calculate the determinant $ad-bc$. If the determinant is non-zero, use the formula for the inverse of a $2 \times 2$ matrix.

1. Invertible, $A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$
2. Not invertible
3. Invertible, $A^{-1} = \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix}$

a) $\det(A) = (2)(3) - (1)(5) = 6-5 = 1$. Since $\det(A) \neq 0$, it is invertible. \[ A^{-1} = \frac{1}{1}\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} \] b) $\det(A) = (12)(1) - (3)(4) = 12-12 = 0$. Since $\det(A) = 0$, the matrix is not invertible.

c) $\det(A) = (1)(-3) - (-2)(2) = -3 - (-4) = 1$. Since $\det(A) \neq 0$, it is invertible. \[ A^{-1} = \frac{1}{1}\begin{bmatrix} -3 & -(-2) \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 2 \\ -2 & 1 \end{bmatrix} \]

For each pair, compute the product $AB$. If the result is the identity matrix, they are inverses. Also, remember that only square matrices can have inverses.

1. Inverses
2. Not inverses
3. Inverses
4. Not inverses (matrices are not square)

a) $AB = \begin{bmatrix} 1 & 5 \\ -2 & -9 \end{bmatrix}\begin{bmatrix} -9 & -5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} -9+10 & -5+5 \\ 18-18 & 10-9 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$. Yes, they are inverses.

b) $AB = \begin{bmatrix} 2 & 4 \\ 5 & 7 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 9 & 6 \end{bmatrix} = \begin{bmatrix} 6+36 & -4+24 \\ 15+63 & -10+42 \end{bmatrix} = \begin{bmatrix} 42 & 20 \\ 78 & 32 \end{bmatrix} \neq I$. No, they are not inverses.

c) $AB = \begin{bmatrix} 1 & 4 \\ -1 & -3 \end{bmatrix}\begin{bmatrix} -3 & -4 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -3+4 & -4+4 \\ 3-3 & 4-3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$. Yes, they are inverses.

d) Only square matrices can have inverses. Since A and B are not square, they cannot be inverses of each other.

Follow the steps systematically. For the determinant, expand along a row or column with a zero to simplify. The cofactor $C_{ij} = (-1)^{i+j}\det(M_{ij})$, where $M_{ij}$ is the minor matrix for the $ij^{th}$ element. The adjoint matrix (or adjugate) is the transpose of the cofactor matrix.

a) $\det(D) = -84 \neq 0$
b) Cofactor matrix $C = \begin{bmatrix} -16 & 20 & 16 \\ 4 & 16 & -4 \\ -6 & -3 & -15 \end{bmatrix}$
c) $\text{Adj}(D) = \begin{bmatrix} -16 & 4 & -6 \\ 20 & 16 & -3 \\ 16 & -4 & -15 \end{bmatrix}$
d) $D^{-1} = -\frac{1}{84}\begin{bmatrix} -16 & 4 & -6 \\ 20 & 16 & -3 \\ 16 & -4 & -15 \end{bmatrix}$

a) Determinant: Expand along the third row: \[ \det(D) = 4\begin{vmatrix} -1 & -1 \\ -4 & 2 \end{vmatrix} - 0 + 4\begin{vmatrix} 3 & -1 \\ -3 & -4 \end{vmatrix} = 4(-2-4) + 4(-12-3) = 4(-6) + 4(-15) = -24 - 60 = -84 \neq 0 \] b) Cofactor Matrix: $$ \begin{align*} C_{11} & = (-1)^{i+j} \times \det \begin{pmatrix} -4 & 2 \\ 0 & 4 \end{pmatrix} = -16\\ C_{12} & = (-1)^{i+j} \times \det \begin{pmatrix} -3 & 2 \\ 4 & 4 \end{pmatrix} = 20\\ C_{13} & = (-1)^{i+j} \times \det \begin{pmatrix} -3 & -4 \\ 4 & 0 \end{pmatrix} = 16\\ C_{21} & = (-1)^{i+j} \times \det \begin{pmatrix} -1 & -1 \\ 0 & 4 \end{pmatrix} = 4\\ C_{22} & = (-1)^{i+j} \times \det \begin{pmatrix} 3 & -1 \\ 4 & 4 \end{pmatrix} = 16\\ C_{23} & = (-1)^{i+j} \times \det \begin{pmatrix} 3 & -1 \\ 4 & 0 \end{pmatrix} = -4\\ C_{31} & = (-1)^{i+j} \times \det \begin{pmatrix} -1 & -1 \\ -4 & 2 \end{pmatrix} = -6\\ C_{32} & = (-1)^{i+j} \times \det \begin{pmatrix} 3 & -1 \\ -3 & 2 \end{pmatrix} = -3\\ C_{33} & = (-1)^{i+j} \times \det \begin{pmatrix} 3 & -1 \\ -3 & -4 \end{pmatrix} = -15\\ \end{align*} $$ Cofactor Matrix: $C = \begin{bmatrix} -16 & 20 & 16 \\ 4 & 16 & -4 \\ -6 & -3 & -15 \end{bmatrix}$

c) Adjoint Matrix: \[ \text{Adj}(D) = C^T = \begin{bmatrix} -16 & 4 & -6 \\ 20 & 16 & -3 \\ 16 & -4 & -15 \end{bmatrix} \] d) Inverse Matrix: \[ D^{-1} = \frac{1}{-84}\begin{bmatrix} -16 & 4 & -6 \\ 20 & 16 & -3 \\ 16 & -4 & -15 \end{bmatrix} \]

First, calculate the determinant of each matrix. An inverse exists only if the determinant is non-zero. If it is, proceed to find the inverse using the adjugate matrix method.

1. Since $\det(A) = 0$, the inverse does not exist.
2. $B^{-1} = \begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}$

a) For Matrix A: \[ \det(A) = 1\begin{vmatrix} -1 & 3 \\ 1 & 2 \end{vmatrix} - 2\begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} + (-1)\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix} \] \[ = 1(-2-3) - 2(4-9) - 1(2 - (-3)) = -5 - 2(-5) - 1(5) = -5 + 10 - 5 = 0 \] Since the determinant is 0, matrix A is not invertible.

b) For Matrix B: \[ \det(B) = 2\begin{vmatrix} 3 & 1 \\ 2 & 3 \end{vmatrix} - (-1)\begin{vmatrix} -5 & 1 \\ -3 & 3 \end{vmatrix} + 3\begin{vmatrix} -5 & 3 \\ -3 & 2 \end{vmatrix} \] \[ = 2(9-2) + 1(-15 - (-3)) + 3(-10 - (-9)) = 2(7) + (-12) + 3(-1) = 14 - 12 - 3 = -1 \] Since the determinant is non-zero, the inverse exists. The cofactor matrix is: \[ C = \begin{bmatrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{bmatrix} \] The adjugate matrix is the transpose of C: \[ \text{Adj}(B) = C^T = \begin{bmatrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{bmatrix} \] The inverse is: \[ B^{-1} = \frac{1}{\det(B)}\text{Adj}(B) = \frac{1}{-1}\begin{bmatrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{bmatrix} = \begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix} \]

Start with the equation $B = BI$. Use the fact that $AC=I$ and the associative property of matrix multiplication.

Assume $B$ and $C$ are both inverses of $A$. By definition, this means: \[ AB = I \quad \text{and} \quad BA = I \] \[ AC = I \quad \text{and} \quad CA = I \] We want to show that $B=C$. Let's start with $B$: \[ B = BI \quad (\text{Property of identity matrix}) \] Since $AC=I$, we can substitute for $I$: \[ B = B(AC) \] Using the associative property of matrix multiplication: \[ B = (BA)C \] Since $BA=I$, we can substitute again: \[ B = IC = C \] Thus, $B=C$. This implies that if a matrix has an inverse, that inverse is unique.

To prove a property of inverses, show that the proposed inverse, when multiplied by the original matrix, results in the identity matrix ($X \cdot X^{-1} = I$). For false properties, construct a simple $2 \times 2$ counterexample.

1. True
2. False. The correct property is $(AB)^{-1} = B^{-1}A^{-1}$.
3. True
4. False
5. True

a) True. By definition, the inverse of $A^{-1}$ is the matrix $X$ such that $A^{-1}X=I$. We know $A^{-1}A=I$, so by the uniqueness of the inverse, $X=A$.

b) False. The correct property is: $(AB)^{-1} = B^{-1}A^{-1}$. To prove this, we show $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$.

c) True. We want to show $(A^T)^{-1} = (A^{-1})^T$. We know $(A^{-1})^T A^T = (AA^{-1})^T = I^T = I$. Since the inverse is unique, $(A^T)^{-1}$ must be equal to $(A^{-1})^T$.

d) False. Let $A=I$ and $B=I$. Then $(A+B)^{-1} = (2I)^{-1} = \frac{1}{2}I$. But $A^{-1}+B^{-1} = I+I = 2I$. These are not equal.

e) True. We want to show $(cA)^{-1} = c^{-1}A^{-1}$. We check the product: $(cA)(c^{-1}A^{-1}) = (c \cdot c^{-1})(AA^{-1}) = (1)(I) = I$. This confirms the property.

Start with the idempotent property, $A^2 = A$. Since $A$ is non-singular, its inverse $A^{-1}$ exists. Multiply both sides of the equation by $A^{-1}$.

We are given that $A$ is idempotent, so $A^2 = A$.
We are also given that $A$ is non-singular, which means it is invertible and $A^{-1}$ exists.
Let's start with the idempotent equation and multiply both sides on the left by $A^{-1}$: \[ A^{-1}(A^2) = A^{-1}A \] Using the associative property on the left side: \[ (A^{-1}A)A = A^{-1}A \] By definition of an inverse, $A^{-1}A = I$, where $I$ is the identity matrix. \[ IA = I \] By the property of the identity matrix, $IA = A$. \[ A = I \] Thus, an invertible idempotent matrix must be the identity matrix.

Show that the condition for being full rank (linearly independent columns) is equivalent to the condition for being invertible (non-zero determinant).

1. Condition for Invertibility: A $2 \times 2$ matrix $A$ is invertible if and only if its determinant is non-zero. For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, this means $\det(A) = ad-bc \neq 0$.

2. Condition for Full Rank: A $2 \times 2$ matrix is full rank if its rank is 2. This means its two columns (or rows) must be linearly independent.

3. Equivalence: Let the columns be $\mathbf{v}_1 = \begin{bmatrix} a \\ c \end{bmatrix}$ and $\mathbf{v}_2 = \begin{bmatrix} b \\ d \end{bmatrix}$. The columns are linearly independent if one is not a scalar multiple of the other. They are dependent if $\mathbf{v}_2 = k\mathbf{v}_1$ for some scalar $k$, which means $b=ka$ and $d=kc$.
If $a \neq 0$, then $k=b/a$, so $d=(b/a)c \implies ad=bc$.
If $a=0$, then for dependence, $b$ must also be 0. In this case, $ad=0$ and $bc=0$, so $ad=bc$ still holds.
The columns are linearly independent if and only if $ad \neq bc$.

The condition for invertibility ($ad-bc \neq 0$) is identical to the condition for the columns being linearly independent ($ad \neq bc$), which is the condition for being full rank. Thus, a $2 \times 2$ matrix is invertible if and only if it is of full rank.