Solving AX = B with Adjugate Matrix
Given the system $AX = B$, where $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is the solution vector. We are provided with the adjugate matrix $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and the vector $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$. The goal is to find the value of $|x + y + z|$.
The solution $X$ can be expressed as $X = \frac{1}{\det A} (\text{adj } A) B$. To use this formula, we first need to determine the determinant of matrix $A$, denoted as $\det A$.
Determine $\det A$
We utilize the property relating the determinant of a matrix and its adjugate: $\det(\text{adj } A) = (\det A)^{n-1}$. Since $A$ is a $3 \times 3$ matrix (indicated by the dimensions of $\text{adj } A$ and $X$), we have $n=3$. Thus, the property becomes $\det(\text{adj } A) = (\det A)^{3-1} = (\det A)^2$.
Step 1: Calculate $\det(\text{adj } A)$
Calculate the determinant of the given adjugate matrix:
$ \det(\text{adj } A) = \begin{vmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{vmatrix} $
Expanding along the first row:
$ \det(\text{adj } A) = 4 \begin{vmatrix} 0 & 5 \\ -2 & 3 \end{vmatrix} - 2 \begin{vmatrix} -5 & 5 \\ 1 & 3 \end{vmatrix} + 2 \begin{vmatrix} -5 & 0 \\ 1 & -2 \end{vmatrix} $
$ \det(\text{adj } A) = 4((0)(3) - (5)(-2)) - 2((-5)(3) - (5)(1)) + 2((-5)(-2) - (0)(1)) $
$ \det(\text{adj } A) = 4(0 + 10) - 2(-15 - 5) + 2(10 - 0) $
$ \det(\text{adj } A) = 4(10) - 2(-20) + 2(10) = 40 + 40 + 20 = 100 $
Step 2: Find $\det A$
Using the relation $(\det A)^2 = \det(\text{adj } A)$, we get:
$ (\det A)^2 = 100 $
$ \det A = \pm \sqrt{100} = \pm 10 $
This indicates that matrix $A$ is invertible, and a unique solution $X$ exists.
Calculate Solution Sum $x+y+z$
We use the relationship $(\det A) X = (\text{adj } A) B$. This allows us to find $X$ without explicitly finding $A$ or $A^{-1}$.
Step 3: Compute the product $(\text{adj } A) B$
$ (\text{adj } A) B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} $
Performing matrix multiplication:
$ (\text{adj } A) B = \begin{bmatrix} (4)(4) + (2)(0) + (2)(2) \\ (-5)(4) + (0)(0) + (5)(2) \\ (1)(4) + (-2)(0) + (3)(2) \end{bmatrix} = \begin{bmatrix} 16 + 0 + 4 \\ -20 + 0 + 10 \\ 4 + 0 + 6 \end{bmatrix} = \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} $
Step 4: Find the sum $x+y+z$
We have the equation $(\det A) X = (\text{adj } A) B$, which is:
$ (\det A) \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} $
To find the sum $x+y+z$, we can sum the components of the vector equation:
$ (\det A)(x) = 20 $
$ (\det A)(y) = -10 $
$ (\det A)(z) = 10 $
Summing these gives:
$ (\det A)x + (\det A)y + (\det A)z = 20 + (-10) + 10 $
$ (\det A)(x+y+z) = 20 $
Therefore, the sum $x+y+z$ is:
$ x+y+z = \frac{20}{\det A} $
Final Value $|x+y+z|$
Now substitute the possible values for $\det A$ ($\pm 10$) into the expression for the sum:
- Case 1: If $\det A = 10$, then $x+y+z = \frac{20}{10} = 2$.
- Case 2: If $\det A = -10$, then $x+y+z = \frac{20}{-10} = -2$.
In both possible cases, the absolute value of the sum is:
$ |x+y+z| = |2| = |-2| = 2 $
Similar Questions
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Important Questions from Algebra
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