Algebra 2 Worksheets | Matrices Worksheets - Free Printable

Let's solve the matrix inverse problems step by step.

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Part 1: State whether each matrix has an inverse

A $ 2 \times 2 $ matrix
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$$
has an inverse if and only if its determinant is non-zero.
The determinant is:
$$
\text{det} = ad - bc
$$

If $ \text{det} \neq 0 $, then the matrix has an inverse.
If $ \text{det} = 0 $, then no inverse exists.

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#### 1)
$$
\begin{bmatrix}
-3 & 5 \\
-2 & 4
\end{bmatrix}
$$
$$
\text{det} = (-3)(4) - (5)(-2) = -12 + 10 = -2 \neq 0
$$
✔ Has an inverse

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#### 2)
$$
\begin{bmatrix}
-5 & -20 \\
2 & 8
\end{bmatrix}
$$
$$
\text{det} = (-5)(8) - (-20)(2) = -40 + 40 = 0
$$
✘ No inverse exists

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#### 3)
$$
\begin{bmatrix}
-6 & -5 \\
5 & 4
\end{bmatrix}
$$
$$
\text{det} = (-6)(4) - (-5)(5) = -24 + 25 = 1 \neq 0
$$
✔ Has an inverse

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#### 4)
$$
\begin{bmatrix}
-5 & -6 \\
6 & -2
\end{bmatrix}
$$
$$
\text{det} = (-5)(-2) - (-6)(6) = 10 + 36 = 46 \neq 0
$$
✔ Has an inverse

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#### 5)
$$
\begin{bmatrix}
4 & -3 \\
-2 & -2
\end{bmatrix}
$$
$$
\text{det} = (4)(-2) - (-3)(-2) = -8 - 6 = -14 \neq 0
$$
✔ Has an inverse

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#### 6)
$$
\begin{bmatrix}
-6 & 0 \\
4 & -6
\end{bmatrix}
$$
$$
\text{det} = (-6)(-6) - (0)(4) = 36 - 0 = 36 \neq 0
$$
✔ Has an inverse

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✔ Summary for Part 1:

1. Yes
2. No
3. Yes
4. Yes
5. Yes
6. Yes

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Part 2: Find the inverse of each matrix, or "No inverse exists"

For a $ 2 \times 2 $ matrix:
$$
A = \begin{bmatrix}
a & b \\
c & d
\end{bmatrix}, \quad
A^{-1} = \frac{1}{ad - bc}
\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}
$$

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#### 7)
$$
\begin{bmatrix}
-1 & -3 \\
3 & 1
\end{bmatrix}
$$
$$
\text{det} = (-1)(1) - (-3)(3) = -1 + 9 = 8
$$
$$
A^{-1} = \frac{1}{8}
\begin{bmatrix}
1 & 3 \\
-3 & -1
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{8} & \frac{3}{8} \\
-\frac{3}{8} & -\frac{1}{8}
\end{bmatrix}
$$

✔ Inverse exists:
$$
\boxed{
\begin{bmatrix}
\frac{1}{8} & \frac{3}{8} \\
-\frac{3}{8} & -\frac{1}{8}
\end{bmatrix}
}
$$

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#### 8)
$$
\begin{bmatrix}
-5 & 4 \\
2 & -1
\end{bmatrix}
$$
$$
\text{det} = (-5)(-1) - (4)(2) = 5 - 8 = -3
$$
$$
A^{-1} = \frac{1}{-3}
\begin{bmatrix}
-1 & -4 \\
-2 & -5
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{3} & \frac{4}{3} \\
\frac{2}{3} & \frac{5}{3}
\end{bmatrix}
$$

Wait — let’s double-check the formula:

Standard formula:
$$
\frac{1}{\text{det}} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
$$

So:
- $ a = -5, b = 4, c = 2, d = -1 $
- $ A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 & -4 \\ -2 & -5 \end{bmatrix} $

Wait:
$$
\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}
=
\begin{bmatrix}
-1 & -4 \\
-2 & -5
\end{bmatrix}
$$

Now divide by $-3$:
$$
= \begin{bmatrix}
\frac{-1}{-3} & \frac{-4}{-3} \\
\frac{-2}{-3} & \frac{-5}{-3}
\end{bmatrix}
= \begin{bmatrix}
\frac{1}{3} & \frac{4}{3} \\
\frac{2}{3} & \frac{5}{3}
\end{bmatrix}
$$

✔ Correct.

$$
\boxed{
\begin{bmatrix}
\frac{1}{3} & \frac{4}{3} \\
\frac{2}{3} & \frac{5}{3}
\end{bmatrix}
}
$$

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#### 9)
$$
\begin{bmatrix}
2 & -5 \\
-4 & -5
\end{bmatrix}
$$
$$
\text{det} = (2)(-5) - (-5)(-4) = -10 - 20 = -30
$$
$$
A^{-1} = \frac{1}{-30}
\begin{bmatrix}
-5 & 5 \\
4 & 2
\end{bmatrix}
=
\begin{bmatrix}
\frac{5}{30} & -\frac{5}{30} \\
-\frac{4}{30} & -\frac{2}{30}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{6} & -\frac{1}{6} \\
-\frac{2}{15} & -\frac{1}{15}
\end{bmatrix}
$$

Wait — let's carefully apply the formula:

Matrix:
$$
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
=
\begin{bmatrix}
2 & -5 \\
-4 & -5
\end{bmatrix}
$$

Then:
$$
A^{-1} = \frac{1}{-30}
\begin{bmatrix}
-5 & 5 \\
4 & 2
\end{bmatrix}
$$

Because:
- $ d = -5 $, $ -b = -(-5) = 5 $
- $ -c = -(-4) = 4 $, $ a = 2 $

So:
$$
A^{-1} = \frac{1}{-30}
\begin{bmatrix}
-5 & 5 \\
4 & 2
\end{bmatrix}
=
\begin{bmatrix}
\frac{-5}{-30} & \frac{5}{-30} \\
\frac{4}{-30} & \frac{2}{-30}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{6} & -\frac{1}{6} \\
-\frac{2}{15} & -\frac{1}{15}
\end{bmatrix}
$$

✔ Final answer:
$$
\boxed{
\begin{bmatrix}
\frac{1}{6} & -\frac{1}{6} \\
-\frac{2}{15} & -\frac{1}{15}
\end{bmatrix}
}
$$

---

#### 10)
$$
\begin{bmatrix}
4 & 3 \\
5 & 0
\end{bmatrix}
$$
$$
\text{det} = (4)(0) - (3)(5) = 0 - 15 = -15 \neq 0
$$
So inverse exists.

$$
A^{-1} = \frac{1}{-15}
\begin{bmatrix}
0 & -3 \\
-5 & 4
\end{bmatrix}
=
\begin{bmatrix}
0 & \frac{3}{15} \\
\frac{5}{15} & -\frac{4}{15}
\end{bmatrix}
=
\begin{bmatrix}
0 & \frac{1}{5} \\
\frac{1}{3} & -\frac{4}{15}
\end{bmatrix}
$$

Wait: Let's compute correctly:

$$
\frac{1}{-15} \cdot
\begin{bmatrix}
0 & -3 \\
-5 & 4
\end{bmatrix}
=
\begin{bmatrix}
\frac{0}{-15} & \frac{-3}{-15} \\
\frac{-5}{-15} & \frac{4}{-15}
\end{bmatrix}
=
\begin{bmatrix}
0 & \frac{1}{5} \\
\frac{1}{3} & -\frac{4}{15}
\end{bmatrix}
$$

✔ Answer:
$$
\boxed{
\begin{bmatrix}
0 & \frac{1}{5} \\
\frac{1}{3} & -\frac{4}{15}
\end{bmatrix}
}
$$

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#### 11)
$$
\begin{bmatrix}
5 & 2 \\
1 & 5
\end{bmatrix}
$$
$$
\text{det} = (5)(5) - (2)(1) = 25 - 2 = 23 \neq 0
$$
So inverse exists.

$$
A^{-1} = \frac{1}{23}
\begin{bmatrix}
5 & -2 \\
-1 & 5
\end{bmatrix}
=
\begin{bmatrix}
\frac{5}{23} & -\frac{2}{23} \\
-\frac{1}{23} & \frac{5}{23}
\end{bmatrix}
$$

✔ Answer:
$$
\boxed{
\begin{bmatrix}
\frac{5}{23} & -\frac{2}{23} \\
-\frac{1}{23} & \frac{5}{23}
\end{bmatrix}
}
$$

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#### 12)
$$
\begin{bmatrix}
-1 & -3 \\
0 & 0
\end{bmatrix}
$$
$$
\text{det} = (-1)(0) - (-3)(0) = 0 - 0 = 0
$$
✘ No inverse exists

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✔ Final Answers:

#### Part 1: Does it have an inverse?
1. Yes
2. No
3. Yes
4. Yes
5. Yes
6. Yes

#### Part 2: Inverses
7. $\boxed{\begin{bmatrix} \frac{1}{8} & \frac{3}{8} \\ -\frac{3}{8} & -\frac{1}{8} \end{bmatrix}}$
8. $\boxed{\begin{bmatrix} \frac{1}{3} & \frac{4}{3} \\ \frac{2}{3} & \frac{5}{3} \end{bmatrix}}$
9. $\boxed{\begin{bmatrix} \frac{1}{6} & -\frac{1}{6} \\ -\frac{2}{15} & -\frac{1}{15} \end{bmatrix}}$
10. $\boxed{\begin{bmatrix} 0 & \frac{1}{5} \\ \frac{1}{3} & -\frac{4}{15} \end{bmatrix}}$
11. $\boxed{\begin{bmatrix} \frac{5}{23} & -\frac{2}{23} \\ -\frac{1}{23} & \frac{5}{23} \end{bmatrix}}$
12. No inverse exists

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