$7$ . $7$ Homework $\newline$ Score: $1 / 6 \quad 1 / 6$ answered $\newline$ Question $2$ $\newline$ Objective $2$ . $17$ $\newline$ Given $A=\left[\begin{array}{ccc}-5 & 13 & 1 \\ 3 & -8 & 0 \\ 0 & 0 & 1\end{array}\right], x=$ and $b=\left[\begin{array}{l}350 \\ 150 \\ 400\end{array}\right]$ $\newline$ Solve the matrix equation $A x=b$ for $x$ $\newline$ Submit Question

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1 / 6 \quad 1 / 6

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2

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Objective

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17

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Given

A=\left[\begin{array}{ccc}-5 & 13 & 1 \\ 3 & -8 & 0 \\ 0 & 0 & 1\end{array}\right], x=

and

b=\left[\begin{array}{l}350 \\ 150 \\ 400\end{array}\right]

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Solve the matrix equation

A x=b

for

x

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Find Inverse of Matrix A: First, we need to find the inverse of matrix $A$ , which is $A^{-1}$ , so we can multiply both sides of the equation $Ax=b$ by $A^{-1}$ to get $x=A^{-1}b$ .
Check Determinant of $A$ : To find the inverse of $A$ , we'll use the formula for the inverse of a $3 \times 3$ matrix. But first, let's check if the determinant of $A$ is not zero, because if it is, the inverse doesn't exist.
Calculate Cofactor Matrix: The determinant of $A$ is $|A| = -5(-8)(1) + 13(0)(1) + 1(3)(0) - 1(-8)(0) - 5(0)(1) - 13(3)(1) = 40 - 0 + 0 - 0 - 0 - 39 = 1$ .
Transpose to Adjugate Matrix: Since the determinant is not zero, we can find the inverse. The inverse of $A$ is $A^{-1} = \frac{1}{|A|} \cdot \text{adj}(A)$ , where $\text{adj}(A)$ is the adjugate of $A$ .
Calculate Inverse of $A$ : The adjugate of $A$ , $\text{adj}(A)$ , is found by taking the transpose of the cofactor matrix of $A$ . Let's calculate the cofactor matrix first.
Matrix Multiplication for $x$ : The cofactor matrix of $A$ is: $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} (-1)^{1+1}\cdot(-8)\cdot(1) - (-0)\cdot(0), & (-1)^{1+2}\cdot(3)\cdot(1) - (0)\cdot(0), & (-1)^{1+3}\cdot(3)\cdot(-8) - (0)\cdot(13), \ (-1)^{2+1}\cdot(13)\cdot(1) - (1)\cdot(0), & (-1)^{2+2}\cdot(-5)\cdot(1) - (1)\cdot(0), & (-1)^{2+3}\cdot(-5)\cdot(0) - (1)\cdot(13), \ (-1)^{3+1}\cdot(13)\cdot(0) - (-8)\cdot(1), & (-1)^{3+2}\cdot(-5)\cdot(0) - (3)\cdot(1), & (-1)^{3+3}\cdot(-5)\cdot(-8) - (3)\cdot(13)\end{array}\right] $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} -8, & -3, & 24, \ -13, & -5, & -13, \ 8, & -3, & 40\end{array}\right]
Matrix Multiplication for $x$ : The cofactor matrix of $A$ is: $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} (-1)^{1+1}\cdot(-8)(1) - (-0)(0), & (-1)^{1+2}\cdot(3)(1) - (0)(0), & (-1)^{1+3}\cdot(3)(-8) - (0)(13), \ (-1)^{2+1}\cdot(13)(1) - (1)(0), & (-1)^{2+2}\cdot(-5)(1) - (1)(0), & (-1)^{2+3}\cdot(-5)(0) - (1)(13), \ (-1)^{3+1}\cdot(13)(0) - (-8)(1), & (-1)^{3+2}\cdot(-5)(0) - (3)(1), & (-1)^{3+3}\cdot(-5)(-8) - (3)(13)\ \end{array}\right] $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} -8, & -3, & 24, \ -13, & -5, & -13, \ 8, & -3, & 40\ \end{array}\right]Now, we transpose the cofactor matrix to get the adjugate matrix, $\text{adj}(A)$ . $\newline$ \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\ \end{array}\right]
Matrix Multiplication for $x$ : The cofactor matrix of $A$ is: $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} (-1)^{1+1}\cdot(-8)\cdot(1) - (-0)\cdot(0), & (-1)^{1+2}\cdot(3)\cdot(1) - (0)\cdot(0), & (-1)^{1+3}\cdot(3)\cdot(-8) - (0)\cdot(13), \ (-1)^{2+1}\cdot(13)\cdot(1) - (1)\cdot(0), & (-1)^{2+2}\cdot(-5)\cdot(1) - (1)\cdot(0), & (-1)^{2+3}\cdot(-5)\cdot(0) - (1)\cdot(13), \ (-1)^{3+1}\cdot(13)\cdot(0) - (-8)\cdot(1), & (-1)^{3+2}\cdot(-5)\cdot(0) - (3)\cdot(1), & (-1)^{3+3}\cdot(-5)\cdot(-8) - (3)\cdot(13)\end{array}\right] $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} -8, & -3, & 24, \ -13, & -5, & -13, \ 8, & -3, & 40\end{array}\right]Now, we transpose the cofactor matrix to get the adjugate matrix, $\text{adj}(A)$ . $\newline$ \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]Next, we multiply the adjugate matrix by $\frac{1}{|A|}$ to get the inverse of $A$ , $A^{-1}$ . $\newline$ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]
Matrix Multiplication for $x$ : The cofactor matrix of $A$ is: $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} (-1)^{1+1}\cdot(-8)\cdot(1) - (-0)\cdot(0), & (-1)^{1+2}\cdot(3)\cdot(1) - (0)\cdot(0), & (-1)^{1+3}\cdot(3)\cdot(-8) - (0)\cdot(13), \ (-1)^{2+1}\cdot(13)\cdot(1) - (1)\cdot(0), & (-1)^{2+2}\cdot(-5)\cdot(1) - (1)\cdot(0), & (-1)^{2+3}\cdot(-5)\cdot(0) - (1)\cdot(13), \ (-1)^{3+1}\cdot(13)\cdot(0) - (-8)\cdot(1), & (-1)^{3+2}\cdot(-5)\cdot(0) - (3)\cdot(1), & (-1)^{3+3}\cdot(-5)\cdot(-8) - (3)\cdot(13)\end{array}\right] $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} -8, & -3, & 24, \ -13, & -5, & -13, \ 8, & -3, & 40\end{array}\right]Now, we transpose the cofactor matrix to get the adjugate matrix, $\text{adj}(A)$ . $\newline$ \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]Next, we multiply the adjugate matrix by $\frac{1}{|A|}$ to get the inverse of $A$ , $A^{-1}$ . $\newline$ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]Now we can find $x$ by multiplying $A^{-1}$ with $b$ . $\newline$ x = A^{-1}b = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right] \cdot \left[\begin{array}{c} 350, \ 150, \ 400\end{array}\right]
Matrix Multiplication for $x$ : The cofactor matrix of $A$ is: $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} (-1)^{1+1}\cdot(-8)\cdot(1) - (-0)\cdot(0), & (-1)^{1+2}\cdot(3)\cdot(1) - (0)\cdot(0), & (-1)^{1+3}\cdot(3)\cdot(-8) - (0)\cdot(13), \ (-1)^{2+1}\cdot(13)\cdot(1) - (1)\cdot(0), & (-1)^{2+2}\cdot(-5)\cdot(1) - (1)\cdot(0), & (-1)^{2+3}\cdot(-5)\cdot(0) - (1)\cdot(13), \ (-1)^{3+1}\cdot(13)\cdot(0) - (-8)\cdot(1), & (-1)^{3+2}\cdot(-5)\cdot(0) - (3)\cdot(1), & (-1)^{3+3}\cdot(-5)\cdot(-8) - (3)\cdot(13)\end{array}\right] $\newline$ \text{Cof}(A) = \left[\begin{array}{ccc} -8, & -3, & 24, \ -13, & -5, & -13, \ 8, & -3, & 40\end{array}\right]Now, we transpose the cofactor matrix to get the adjugate matrix, $\text{adj}(A)$ . $\newline$ \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]Next, we multiply the adjugate matrix by $\frac{1}{|A|}$ to get the inverse of $A$ , $A^{-1}$ . $\newline$ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A) = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right]Now we can find $x$ by multiplying $A^{-1}$ with $b$ . $\newline$ x = A^{-1}b = \left[\begin{array}{ccc} -8, & -13, & 8, \ -3, & -5, & -3, \ 24, & -13, & 40\end{array}\right] \cdot \left[\begin{array}{c} 350, \ 150, \ 400\end{array}\right]Let's do the matrix multiplication. $\newline$ x = \left[\begin{array}{c} (-8)(350) + (-13)(150) + (8)(400), \ (-3)(350) + (-5)(150) + (-3)(400), \ (24)(350) + (-13)(150) + (40)(400)\end{array}\right] $\newline$ x = \left[\begin{array}{c} -2800 - 1950 + 3200, \ -1050 - 750 - 1200, \ 8400 - 1950 + 16000\end{array}\right]
Matrix Multiplication for x: The cofactor matrix of A is: $\newline$ Cof(A) = \left[\begin{array}{ccc}$\newline(-1)^{1+1}\cdot(-8)\cdot(1) - (-0)\cdot(0), & (-1)^{1+2}\cdot(3)\cdot(1) - (0)\cdot(0), & (-1)^{1+3}\cdot(3)\cdot(-8) - (0)\cdot(13),(\newline$(-1)^{2+1}\cdot(13)\cdot(1) - (1)\cdot(0), & (-1)^{2+2}\cdot(-5)\cdot(1) - (1)\cdot(0), & (-1)^{2+3}\cdot(-5)\cdot(0) - (1)\cdot(13),(\newline\)(-1)^{3+1}\cdot(13)\cdot(0) - (-8)\cdot(1), & (-1)^{3+2}\cdot(-5)\cdot(0) - (3)\cdot(1), & (-1)^{3+3}\cdot(-5)\cdot(-8) - (3)\cdot(13)\end{array}\right]\) $\newline$ Cof(A) = \left[\begin{array}{ccc}$\newline-8, & -3, & 24,(\newline$-13, & -5, & -13,(\newline\)8, & -3, & 40\end{array}\right]\)Now, we transpose the cofactor matrix to get the adjugate matrix, adj(A). $\newline$ adj(A) = \left[\begin{array}{ccc}$\newline-8, & -13, & 8,(\newline$-3, & -5, & -3,(\newline\)24, & -13, & 40\end{array}\right]\)Next, we multiply the adjugate matrix by $\frac{1}{|A|}$ to get the inverse of A, $A^{-1}$ . $\newline$ A^{-1} = \frac{1}{1} \cdot \text{adj}(A) = \text{adj}(A) = \left[\begin{array}{ccc}$\newline-8, & -13, & 8,(\newline$-3, & -5, & -3,(\newline\)24, & -13, & 40\end{array}\right]\)Now we can find x by multiplying $A^{-1}$ with b. $\newline$ x = A^{-1}b = \left[\begin{array}{ccc}$\newline-8, & -13, & 8,(\newline$-3, & -5, & -3,(\newline\)24, & -13, & 40\end{array}\right] \cdot \left[\begin{array}{c} $\newline$ 350,(\newline\)150,(\newline\)400\end{array}\right]\)Let's do the matrix multiplication. $\newline$ x = \left[\begin{array}{c}$\newline(-8)(350) + (-13)(150) + (8)(400),(\newline$(-3)(350) + (-5)(150) + (-3)(400),(\newline\)(24)(350) + (-13)(150) + (40)(400)\end{array}\right]\) $\newline$ x = \left[\begin{array}{c}$\newline-2800 - 1950 + 3200,(\newline$-1050 - 750 - 1200,(\newline\)8400 - 1950 + 16000\end{array}\right]\)Now, we add the numbers in each row to get the final values for x. $\newline$ x = \left[\begin{array}{ccc}$\newline-8, & -3, & 24,(\newline$-13, & -5, & -13,(\newline\)8, & -3, & 40\end{array}\right]\) $0$