Math / Algebra

Question $A=\left[\begin{array}{ll} 3 & 6 \\ 4 & x \end{array}\right] \quad B=\left[\begin{array}{cc} 5 & -2 \\ 3 & 4 \end{array}\right]$
Find $\operatorname{det}\left(5 A^{-1} B^{t}\right)=5$

Studdy Solution

STEP 1

1. The problem involves matrix operations including finding the inverse, transpose, and determinant.
2. The determinant of a matrix product can be expressed as the product of the determinants.
3. The inverse of a matrix $A$ exists if and only if $\det(A) \neq 0$ .

STEP 2

1. Find the inverse of matrix $A$ .
2. Find the transpose of matrix $B$ .
3. Calculate the determinant of the product $5A^{-1}B^t$ .
4. Solve for $x$ .

STEP 3

Calculate the determinant of matrix $A$ to ensure it is invertible:
$A = \begin{bmatrix} 3 & 6 \\ 4 & x \end{bmatrix}$
The determinant of $A$ is given by:
$\det(A) = 3x - 6 \times 4 = 3x - 24$
For $A$ to be invertible, $\det(A) \neq 0$ .

STEP 4

Assuming $\det(A) \neq 0$ , find the inverse of $A$ :
The inverse of a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is:
$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$
Thus, for matrix $A$ :
$A^{-1} = \frac{1}{3x - 24} \begin{bmatrix} x & -6 \\ -4 & 3 \end{bmatrix}$

STEP 5

Find the transpose of matrix $B$ :
$B = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}$
The transpose of $B$ , denoted $B^t$ , is:
$B^t = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}$

STEP 6

Calculate the determinant of the product $5A^{-1}B^t$ .
Using the property of determinants, $\det(kA) = k^n \det(A)$ for a scalar $k$ and an $n \times n$ matrix $A$ , and $\det(AB) = \det(A)\det(B)$ :
$\det(5A^{-1}B^t) = 5^2 \cdot \det(A^{-1}) \cdot \det(B^t)$
Since $\det(A^{-1}) = \frac{1}{\det(A)}$ and $\det(B^t) = \det(B)$ :
$\det(5A^{-1}B^t) = 25 \cdot \frac{1}{3x - 24} \cdot \det(B)$
Calculate $\det(B)$ :
$\det(B) = 5 \times 4 - (-2) \times 3 = 20 + 6 = 26$
Thus:
$\det(5A^{-1}B^t) = 25 \cdot \frac{26}{3x - 24}$
Set the determinant equal to 5:
$25 \cdot \frac{26}{3x - 24} = 5$

STEP 7

Solve for $x$ :
Simplify the equation:
$\frac{650}{3x - 24} = 5$
Multiply both sides by $3x - 24$ :
$650 = 5(3x - 24)$
Expand and solve for $x$ :
$650 = 15x - 120$
Add 120 to both sides:
$770 = 15x$
Divide by 15:
$x = \frac{770}{15} = \frac{154}{3}$
The value of $x$ is:
$\boxed{\frac{154}{3}}$

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Question $A=\left[\begin{array}{ll} 3 & 6 \\ 4 & x \end{array}\right] \quad B=\left[\begin{array}{cc} 5 & -2 \\ 3 & 4 \end{array}\right]$
Find $\operatorname{det}\left(5 A^{-1} B^{t}\right)=5$

Studdy Solution

STEP 1

1. The problem involves matrix operations including finding the inverse, transpose, and determinant.
2. The determinant of a matrix product can be expressed as the product of the determinants.
3. The inverse of a matrix $A$ exists if and only if $\det(A) \neq 0$ .

STEP 2

1. Find the inverse of matrix $A$ .
2. Find the transpose of matrix $B$ .
3. Calculate the determinant of the product $5A^{-1}B^t$ .
4. Solve for $x$ .

STEP 3

Calculate the determinant of matrix $A$ to ensure it is invertible:
$A = \begin{bmatrix} 3 & 6 \\ 4 & x \end{bmatrix}$
The determinant of $A$ is given by:
$\det(A) = 3x - 6 \times 4 = 3x - 24$
For $A$ to be invertible, $\det(A) \neq 0$ .

STEP 4

STEP 5

Find the transpose of matrix $B$ :
$B = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}$
The transpose of $B$ , denoted $B^t$ , is:
$B^t = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}$

STEP 6

STEP 7

Solve for $x$ :
Simplify the equation:
$\frac{650}{3x - 24} = 5$
Multiply both sides by $3x - 24$ :
$650 = 5(3x - 24)$
Expand and solve for $x$ :
$650 = 15x - 120$
Add 120 to both sides:
$770 = 15x$
Divide by 15:
$x = \frac{770}{15} = \frac{154}{3}$
The value of $x$ is:
$\boxed{\frac{154}{3}}$

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QuestionA=[364x]B=[5−234]A=\left[\begin{array}{ll} 3 & 6 \\ 4 & x \end{array}\right] \quad B=\left[\begin{array}{cc} 5 & -2 \\ 3 & 4 \end{array}\right]A=[34​6x​]B=[53​−24​]Find det⁡(5A−1Bt)=5\operatorname{det}\left(5 A^{-1} B^{t}\right)=5det(5A−1Bt)=5

Studdy Solution

STEP 1

1. The problem involves matrix operations including finding the inverse, transpose, and determinant.2. The determinant of a matrix product can be expressed as the product of the determinants.3. The inverse of a matrix A A A exists if and only if det⁡(A)≠0 \det(A) \neq 0 det(A)=0.

STEP 2

1. Find the inverse of matrix A A A.2. Find the transpose of matrix B B B.3. Calculate the determinant of the product 5A−1Bt 5A^{-1}B^t 5A−1Bt.4. Solve for x x x.

STEP 3

STEP 4

STEP 5

Find the transpose of matrix B B B:B=[5−234] B = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix} B=[53​−24​]The transpose of B B B, denoted Bt B^t Bt, is:Bt=[53−24] B^t = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix} Bt=[5−2​34​]

STEP 6

STEP 7

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Question $A=\left[\begin{array}{ll} 3 & 6 \\ 4 & x \end{array}\right] \quad B=\left[\begin{array}{cc} 5 & -2 \\ 3 & 4 \end{array}\right]$
Find $\operatorname{det}\left(5 A^{-1} B^{t}\right)=5$

1. The problem involves matrix operations including finding the inverse, transpose, and determinant.
2. The determinant of a matrix product can be expressed as the product of the determinants.
3. The inverse of a matrix $A$ exists if and only if $\det(A) \neq 0$ .

1. Find the inverse of matrix $A$ .
2. Find the transpose of matrix $B$ .
3. Calculate the determinant of the product $5A^{-1}B^t$ .
4. Solve for $x$ .

Find the transpose of matrix $B$ :
$B = \begin{bmatrix} 5 & -2 \\ 3 & 4 \end{bmatrix}$
The transpose of $B$ , denoted $B^t$ , is:
$B^t = \begin{bmatrix} 5 & 3 \\ -2 & 4 \end{bmatrix}$