Math / Algebra

QuestionWhich of the following is the inverse of the matrix $\left[\begin{array}{ll}1 & 1 \\ 4 & 2\end{array}\right]$ ? $\left[\begin{array}{cc}-2 & 1 \\ 4 & -1\end{array}\right]$ $\left[\begin{array}{cc}\frac{1}{2} & \frac{1}{4} \\ -1 & 1\end{array}\right]$ $\left[\begin{array}{ll}4 & 2 \\ 1 & 1\end{array}\right]$ $\left[\begin{array}{cc}-1 & \frac{1}{2} \\ 2 & -\frac{1}{2}\end{array}\right]$ Previous

Studdy Solution

STEP 1

What is this asking? Which of the given matrices, when multiplied by the original matrix, results in the identity matrix? Watch out! Don't forget that matrix multiplication isn't commutative – the order matters!
Also, remember the specific form of the identity matrix.

STEP 2

1. Calculate the determinant.
2. Find the adjugate matrix.
3. Calculate the inverse matrix.
4. Verify the inverse.

STEP 3

Alright, let's kick things off by finding the determinant of our original matrix!
We'll call our original matrix $A$ .
Remember, the determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $ad - bc$ .

STEP 4

So, for our matrix $A = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ , the determinant is $(1 \cdot 2) - (1 \cdot 4) = 2 - 4 = \mathbf{-2}$ .
This -2 is super important – it's the key to unlocking the inverse!

STEP 5

Next up, we need to find the adjugate of matrix $A$ .
For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the adjugate is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .
Basically, we swap the positions of $a$ and $d$ and change the signs of $b$ and $c$ .

STEP 6

For our matrix $A$ , the adjugate is $\begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix}$ .
See how we swapped the $\mathbf{1}$ and $\mathbf{2}$ and flipped the signs of the other two entries?

STEP 7

Now, the grand finale – calculating the inverse!
The inverse of a matrix $A$ is given by $\frac{1}{\text{det}(A)} \cdot \text{adj}(A)$ , where $\text{det}(A)$ is the determinant of $A$ and $\text{adj}(A)$ is the adjugate of $A$ .

STEP 8

We already found that $\text{det}(A) = \mathbf{-2}$ and $\text{adj}(A) = \begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix}$ .
So, the inverse, $A^{-1}$ , is: $A^{-1} = \frac{1}{-2} \begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{-2} & \frac{-1}{-2} \\ \frac{-4}{-2} & \frac{1}{-2} \end{bmatrix} = \begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}$

STEP 9

Let's double-check our work!
If we multiply our original matrix $A$ by its inverse $A^{-1}$ , we should get the identity matrix, which is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ for a 2x2 matrix.

STEP 10

$\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} -1 + 2 & \frac{1}{2} - \frac{1}{2} \\ -4 + 4 & 2 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ Boom! We got the identity matrix, so our inverse is correct!

STEP 11

The inverse of the matrix $\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ is $\begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}$ .

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Studdy Solution

STEP 1

What is this asking? Which of the given matrices, when multiplied by the original matrix, results in the identity matrix? Watch out! Don't forget that matrix multiplication isn't commutative – the order matters!
Also, remember the specific form of the identity matrix.

STEP 2

1. Calculate the determinant.
2. Find the adjugate matrix.
3. Calculate the inverse matrix.
4. Verify the inverse.

STEP 3

Alright, let's kick things off by finding the determinant of our original matrix!
We'll call our original matrix $A$ .
Remember, the determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $ad - bc$ .

STEP 4

So, for our matrix $A = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ , the determinant is $(1 \cdot 2) - (1 \cdot 4) = 2 - 4 = \mathbf{-2}$ .
This -2 is super important – it's the key to unlocking the inverse!

STEP 5

Next up, we need to find the adjugate of matrix $A$ .
For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ , the adjugate is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .
Basically, we swap the positions of $a$ and $d$ and change the signs of $b$ and $c$ .

STEP 6

For our matrix $A$ , the adjugate is $\begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix}$ .
See how we swapped the $\mathbf{1}$ and $\mathbf{2}$ and flipped the signs of the other two entries?

STEP 7

Now, the grand finale – calculating the inverse!
The inverse of a matrix $A$ is given by $\frac{1}{\text{det}(A)} \cdot \text{adj}(A)$ , where $\text{det}(A)$ is the determinant of $A$ and $\text{adj}(A)$ is the adjugate of $A$ .

STEP 8

STEP 9

Let's double-check our work!
If we multiply our original matrix $A$ by its inverse $A^{-1}$ , we should get the identity matrix, which is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ for a 2x2 matrix.

STEP 10

STEP 11

The inverse of the matrix $\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ is $\begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}$ .

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Studdy Solution

STEP 1

What is this asking? Which of the given matrices, when multiplied by the original matrix, results in the identity matrix? Watch out! Don't forget that matrix multiplication isn't commutative – the order matters!Also, remember the specific form of the identity matrix.

STEP 2

1. Calculate the determinant.2. Find the adjugate matrix.3. Calculate the inverse matrix.4. Verify the inverse.

STEP 3

Alright, let's **kick things off** by finding the determinant of our **original matrix**!We'll call our original matrix AAA.Remember, the determinant of a 2x2 matrix [abcd]\begin{bmatrix} a & b \\ c & d \end{bmatrix}[ac​bd​] is calculated as ad−bcad - bcad−bc.

STEP 4

So, for our matrix A=[1142]A = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}A=[14​12​], the determinant is (1⋅2)−(1⋅4)=2−4=−2(1 \cdot 2) - (1 \cdot 4) = 2 - 4 = \mathbf{-2}(1⋅2)−(1⋅4)=2−4=−2.This **-2** is super important – it's the key to unlocking the inverse!

STEP 5

STEP 6

For our matrix AAA, the adjugate is [2−1−41]\begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix}[2−4​−11​].See how we swapped the 1\mathbf{1}1 and 2\mathbf{2}2 and flipped the signs of the other two entries?

STEP 7

Now, the grand finale – calculating the inverse!The inverse of a matrix AAA is given by 1det(A)⋅adj(A)\frac{1}{\text{det}(A)} \cdot \text{adj}(A)det(A)1​⋅adj(A), where det(A)\text{det}(A)det(A) is the determinant of AAA and adj(A)\text{adj}(A)adj(A) is the adjugate of AAA.

STEP 8

STEP 9

Let's double-check our work!If we multiply our original matrix AAA by its inverse A−1A^{-1}A−1, we should get the identity matrix, which is [1001]\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}[10​01​] for a 2x2 matrix.

STEP 10

STEP 11

The inverse of the matrix [1142]\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}[14​12​] is [−1122−12]\begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}[−12​21​−21​​].

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What is this asking? Which of the given matrices, when multiplied by the original matrix, results in the identity matrix? Watch out! Don't forget that matrix multiplication isn't commutative – the order matters!
Also, remember the specific form of the identity matrix.

1. Calculate the determinant.
2. Find the adjugate matrix.
3. Calculate the inverse matrix.
4. Verify the inverse.

Alright, let's kick things off by finding the determinant of our original matrix!
We'll call our original matrix $A$ .
Remember, the determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $ad - bc$ .

So, for our matrix $A = \begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ , the determinant is $(1 \cdot 2) - (1 \cdot 4) = 2 - 4 = \mathbf{-2}$ .
This -2 is super important – it's the key to unlocking the inverse!

For our matrix $A$ , the adjugate is $\begin{bmatrix} 2 & -1 \\ -4 & 1 \end{bmatrix}$ .
See how we swapped the $\mathbf{1}$ and $\mathbf{2}$ and flipped the signs of the other two entries?

Now, the grand finale – calculating the inverse!
The inverse of a matrix $A$ is given by $\frac{1}{\text{det}(A)} \cdot \text{adj}(A)$ , where $\text{det}(A)$ is the determinant of $A$ and $\text{adj}(A)$ is the adjugate of $A$ .

Let's double-check our work!
If we multiply our original matrix $A$ by its inverse $A^{-1}$ , we should get the identity matrix, which is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ for a 2x2 matrix.

The inverse of the matrix $\begin{bmatrix} 1 & 1 \\ 4 & 2 \end{bmatrix}$ is $\begin{bmatrix} -1 & \frac{1}{2} \\ 2 & -\frac{1}{2} \end{bmatrix}$ .