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Question Solve linear system Ax=bAx=b where AA is 3×33 \times 3 matrix, x=[x1x2x3]x=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right], b=[−2−4−3]b=\left[\begin{array}{l}-2 \\ -4 \\ -3\end{array}\right], and A−1=[14−12−3205−4]A^{-1}=\left[\begin{array}{ccc}1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4\end{array}\right]. Find the value of x3x_{3}.

Studdy Solution

STEP 1

Assumptions
1. The linear system is given by Ax=bA x = b.
2. AA is a 3×33 \times 3 matrix.
3. x=[x1x2x3]x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.
4. b=[−2−4−3]b = \begin{bmatrix} -2 \\ -4 \\ -3 \end{bmatrix}.
5. The inverse of matrix AA is given by A−1=[14−12−3205−4]A^{-1} = \begin{bmatrix} 1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4 \end{bmatrix}.
6. We need to find the value of x3x_3.

STEP 2

To find xx, we can use the inverse of matrix AA, since Ax=bA x = b implies that x=A−1bx = A^{-1} b.

STEP 3

Calculate the product of A−1A^{-1} and bb to find xx.
x=A−1b=[14−12−3205−4][−2−4−3]x = A^{-1} b = \begin{bmatrix} 1 & 4 & -1 \\ 2 & -3 & 2 \\ 0 & 5 & -4 \end{bmatrix} \begin{bmatrix} -2 \\ -4 \\ -3 \end{bmatrix}

STEP 4

Perform the matrix multiplication to find the product.
x=[(1⋅−2)+(4⋅−4)+(−1⋅−3)(2⋅−2)+(−3⋅−4)+(2⋅−3)(0⋅−2)+(5⋅−4)+(−4⋅−3)]x = \begin{bmatrix} (1 \cdot -2) + (4 \cdot -4) + (-1 \cdot -3) \\ (2 \cdot -2) + (-3 \cdot -4) + (2 \cdot -3) \\ (0 \cdot -2) + (5 \cdot -4) + (-4 \cdot -3) \end{bmatrix}

STEP 5

Calculate the entries of the resulting matrix.
x=[(−2)+(−16)+(3)(−4)+(12)+(−6)(0)+(−20)+(12)]x = \begin{bmatrix} (-2) + (-16) + (3) \\ (-4) + (12) + (-6) \\ (0) + (-20) + (12) \end{bmatrix}

STEP 6

Simplify the entries to find the components of vector xx.
x=[−2−16+3−4+12−60−20+12]=[−152−8]x = \begin{bmatrix} -2 - 16 + 3 \\ -4 + 12 - 6 \\ 0 - 20 + 12 \end{bmatrix} = \begin{bmatrix} -15 \\ 2 \\ -8 \end{bmatrix}

STEP 7

Identify the value of x3x_3 from the vector xx.
x3=−8x_3 = -8
The unknown x3x_3 is equal to −8-8, which corresponds to option a.

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