Math / Algebra

Question1. (10 points) Find the inverse of the matrix $A=\left[\begin{array}{ccc} 1 & 2 & -1 \\ -4 & -7 & 3 \\ -2 & -6 & 5 \end{array}\right]$
2. (10 points total) Apply the Invertible Matrix Theorem to determine which of the matrices below are invertible using as few calculations as possible. (a) $(2$ points $)\left[\begin{array}{cc}4 & -2 \\ -6 & 3\end{array}\right]$ ; (b) (2 points) $\left[\begin{array}{ccc}5 & 0 & -3 \\ 2 & 0 & 1 \\ -6 & 0 & 7\end{array}\right]$ ; (c) $(3$ points $)\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & -1 & -1\end{array}\right]$ ; (d) (3 points $)\left[\begin{array}{llll}3 & 4 & 7 & 8 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & 2 & 9 \\ 0 & 0 & 0 & 4\end{array}\right]$ .
3. (10 points) Find the determinant of $A$ by cofactor expansion and row reduction: $A=\left[\begin{array}{cccc} 2 & 3 & 0 & 1 \\ 4 & 7 & 0 & 3 \\ 7 & 9 & -2 & 4 \\ 1 & 2 & 0 & 4 \end{array}\right]$
4. (10 points) Let $U \in \mathbb{R}^{n \times n}$ be such that $U^{T} U=I_{n}$ . Show that either $\operatorname{det} U=1$ or $\operatorname{det} U=-1$ .

Studdy Solution

STEP 1

1. The matrices are square, which is necessary for them to be invertible.
2. The determinant is a key factor in determining invertibility.
3. For a matrix $U$ such that $U^T U = I_n$ , $U$ is orthogonal.

STEP 2

1. Find the inverse of the given matrix $A$ .
2. Determine the invertibility of the given matrices using the Invertible Matrix Theorem.
3. Calculate the determinant of matrix $A$ using cofactor expansion and row reduction.
4. Prove that for an orthogonal matrix $U$ , the determinant is either 1 or -1.

STEP 3

To find the inverse of matrix $A$ , we first need to calculate its determinant. The formula for the determinant of a 3x3 matrix is:
$\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$
For matrix $A$ :
$A = \begin{bmatrix} 1 & 2 & -1 \\ -4 & -7 & 3 \\ -2 & -6 & 5 \end{bmatrix}$
Calculate the determinant:
$\text{det}(A) = 1((-7)(5) - (3)(-6)) - 2((-4)(5) - (3)(-2)) + (-1)((-4)(-6) - (-7)(-2))$
$= 1(-35 + 18) - 2(-20 + 6) - 1(24 - 14)$
$= 1(-17) - 2(-14) - 1(10)$
$= -17 + 28 - 10$
$= 1$
Since the determinant is non-zero, the matrix is invertible.

STEP 4

To find the inverse of matrix $A$ , use the formula for the inverse of a 3x3 matrix:
$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)$
Since $\text{det}(A) = 1$ , we only need to find the adjugate of $A$ .
The adjugate of $A$ is the transpose of the cofactor matrix of $A$ . Calculate the cofactor matrix:
$\text{Cofactor}(A) = \begin{bmatrix} (-7)(5) - (3)(-6) & -((-4)(5) - (3)(-2)) & (-4)(-6) - (-7)(-2) \\ -((-2)(5) - (3)(-1)) & (1)(5) - (-1)(-2) & -((1)(-6) - (2)(-2)) \\ ((-2)(3) - (-6)(-1)) & -((1)(3) - (-1)(-4)) & (1)(-7) - (2)(-4) \end{bmatrix}$
$= \begin{bmatrix} -17 & 14 & 10 \\ 13 & 3 & -2 \\ 0 & 7 & -15 \end{bmatrix}$
Transpose the cofactor matrix to get the adjugate:
$\text{adj}(A) = \begin{bmatrix} -17 & 13 & 0 \\ 14 & 3 & 7 \\ 10 & -2 & -15 \end{bmatrix}$
Thus, the inverse is:
$A^{-1} = \begin{bmatrix} -17 & 13 & 0 \\ 14 & 3 & 7 \\ 10 & -2 & -15 \end{bmatrix}$

STEP 5

(a) For the matrix $\begin{bmatrix}4 & -2 \\ -6 & 3\end{bmatrix}$ , calculate the determinant:
$\text{det} = 4 \cdot 3 - (-2) \cdot (-6) = 12 - 12 = 0$
Since the determinant is zero, the matrix is not invertible.

STEP 6

(b) For the matrix $\begin{bmatrix}5 & 0 & -3 \\ 2 & 0 & 1 \\ -6 & 0 & 7\end{bmatrix}$ , notice that the second column is all zeros. This implies the matrix is not full rank, hence not invertible.

STEP 7

(c) For the matrix $\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & -1 & -1\end{bmatrix}$ , calculate the determinant:
$\text{det} = 1(1 \cdot (-1) - 4 \cdot (-1)) = 1(-1 + 4) = 3$
Since the determinant is non-zero, the matrix is invertible.

STEP 8

(d) For the matrix $\begin{bmatrix}3 & 4 & 7 & 8 \\ 0 & 1 & 4 & 6 \\ 0 & 0 & 2 & 9 \\ 0 & 0 & 0 & 4\end{bmatrix}$ , notice it is an upper triangular matrix. The determinant is the product of the diagonal elements:
$\text{det} = 3 \cdot 1 \cdot 2 \cdot 4 = 24$
Since the determinant is non-zero, the matrix is invertible.

STEP 9

To find the determinant of matrix $A$ using cofactor expansion, expand along the first row:
$A = \begin{bmatrix} 2 & 3 & 0 & 1 \\ 4 & 7 & 0 & 3 \\ 7 & 9 & -2 & 4 \\ 1 & 2 & 0 & 4 \end{bmatrix}$
$\text{det}(A) = 2 \cdot \text{det}\begin{bmatrix} 7 & 0 & 3 \\ 9 & -2 & 4 \\ 2 & 0 & 4 \end{bmatrix} - 3 \cdot \text{det}\begin{bmatrix} 4 & 0 & 3 \\ 7 & -2 & 4 \\ 1 & 0 & 4 \end{bmatrix} + 0 \cdot \text{det}\begin{bmatrix} 4 & 7 & 3 \\ 7 & 9 & 4 \\ 1 & 2 & 4 \end{bmatrix} - 1 \cdot \text{det}\begin{bmatrix} 4 & 7 & 0 \\ 7 & 9 & -2 \\ 1 & 2 & 0 \end{bmatrix}$
Calculate each of the 3x3 determinants and simplify to find the determinant of $A$ .

STEP 10

Use row reduction to simplify the calculation:
Perform row operations to get an upper triangular form and calculate the determinant as the product of the diagonal elements.

STEP 11

Given $U^T U = I_n$ , $U$ is orthogonal. The determinant of an orthogonal matrix satisfies $|\det(U)| = 1$ .
To show $\det(U) = \pm 1$ , note that:
$\det(U^T U) = \det(U^T) \det(U) = (\det(U))^2 = \det(I_n) = 1$
Thus, $(\det(U))^2 = 1$ , implying $\det(U) = \pm 1$ .
The solution to the problem is complete.

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Studdy Solution

STEP 1

1. The matrices are square, which is necessary for them to be invertible.
2. The determinant is a key factor in determining invertibility.
3. For a matrix $U$ such that $U^T U = I_n$ , $U$ is orthogonal.

STEP 2

STEP 3

STEP 4

STEP 5

(a) For the matrix $\begin{bmatrix}4 & -2 \\ -6 & 3\end{bmatrix}$ , calculate the determinant:
$\text{det} = 4 \cdot 3 - (-2) \cdot (-6) = 12 - 12 = 0$
Since the determinant is zero, the matrix is not invertible.

STEP 6

(b) For the matrix $\begin{bmatrix}5 & 0 & -3 \\ 2 & 0 & 1 \\ -6 & 0 & 7\end{bmatrix}$ , notice that the second column is all zeros. This implies the matrix is not full rank, hence not invertible.

STEP 7

(c) For the matrix $\begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & -1 & -1\end{bmatrix}$ , calculate the determinant:
$\text{det} = 1(1 \cdot (-1) - 4 \cdot (-1)) = 1(-1 + 4) = 3$
Since the determinant is non-zero, the matrix is invertible.

STEP 8

STEP 9

STEP 10

Use row reduction to simplify the calculation:
Perform row operations to get an upper triangular form and calculate the determinant as the product of the diagonal elements.

STEP 11

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Studdy Solution

STEP 1

1. The matrices are square, which is necessary for them to be invertible.2. The determinant is a key factor in determining invertibility.3. For a matrix U U U such that UTU=In U^T U = I_n UTU=In​, U U U is orthogonal.

STEP 2

STEP 3

STEP 4

STEP 5

STEP 6

(b) For the matrix [50−3201−607]\begin{bmatrix}5 & 0 & -3 \\ 2 & 0 & 1 \\ -6 & 0 & 7\end{bmatrix}⎣⎡​52−6​000​−317​⎦⎤​, notice that the second column is all zeros. This implies the matrix is not full rank, hence not invertible.

STEP 7

STEP 8

STEP 9

STEP 10

Use row reduction to simplify the calculation:Perform row operations to get an upper triangular form and calculate the determinant as the product of the diagonal elements.

STEP 11

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Start learning now

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1. The matrices are square, which is necessary for them to be invertible.
2. The determinant is a key factor in determining invertibility.
3. For a matrix $U$ such that $U^T U = I_n$ , $U$ is orthogonal.

(b) For the matrix $\begin{bmatrix}5 & 0 & -3 \\ 2 & 0 & 1 \\ -6 & 0 & 7\end{bmatrix}$ , notice that the second column is all zeros. This implies the matrix is not full rank, hence not invertible.

Use row reduction to simplify the calculation:
Perform row operations to get an upper triangular form and calculate the determinant as the product of the diagonal elements.