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Question Solve for matrix XX in (A−5B)X=C(A-5B)X=C given A=[−5012],B=[−1203],C=[−2−6−3−5]A=\left[\begin{array}{cc}-5 & 0 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right], C=\left[\begin{array}{cc}-2 & -6 \\ -3 & -5\end{array}\right].

Studdy Solution

STEP 1

Assumptions
1. Matrix AA is given by A=[−5012]A=\left[\begin{array}{cc}-5 & 0 \\ 1 & 2\end{array}\right]
2. Matrix BB is given by B=[−1203]B=\left[\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right]
3. Matrix CC is given by C=[−2−6−3−5]C=\left[\begin{array}{cc}-2 & -6 \\ -3 & -5\end{array}\right]
4. We need to solve for matrix XX in the equation (A−5B)X=C(A-5B)X=C

STEP 2

First, we need to calculate the matrix (A−5B)(A-5B) by subtracting 55 times matrix BB from matrix AA.
(A−5B)=A−5B (A-5B) = A - 5B

STEP 3

Multiply matrix BB by 55.
5B=5[−1203]=[−510015] 5B = 5\left[\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right] = \left[\begin{array}{cc}-5 & 10 \\ 0 & 15\end{array}\right]

STEP 4

Subtract the matrix 5B5B from matrix AA.
(A−5B)=[−5012]−[−510015] (A-5B) = \left[\begin{array}{cc}-5 & 0 \\ 1 & 2\end{array}\right] - \left[\begin{array}{cc}-5 & 10 \\ 0 & 15\end{array}\right]

STEP 5

Perform the subtraction element-wise.
(A−5B)=[(−5)−(−5)0−101−02−15] (A-5B) = \left[\begin{array}{cc}(-5) - (-5) & 0 - 10 \\ 1 - 0 & 2 - 15\end{array}\right]

STEP 6

Calculate the resulting matrix.
(A−5B)=[0−101−13] (A-5B) = \left[\begin{array}{cc}0 & -10 \\ 1 & -13\end{array}\right]

STEP 7

Now we have the equation with the calculated matrix.
[0−101−13]X=C \left[\begin{array}{cc}0 & -10 \\ 1 & -13\end{array}\right]X = C

STEP 8

To solve for XX, we need to find the inverse of the matrix (A−5B)(A-5B), if it exists.
X=(A−5B)−1C X = (A-5B)^{-1}C

STEP 9

Calculate the determinant of (A−5B)(A-5B).
det(A−5B)=det([0−101−13])=(0)(−13)−(−10)(1) \text{det}(A-5B) = \text{det}\left(\left[\begin{array}{cc}0 & -10 \\ 1 & -13\end{array}\right]\right) = (0)(-13) - (-10)(1)

STEP 10

Find the value of the determinant.
det(A−5B)=0−(−10)=10 \text{det}(A-5B) = 0 - (-10) = 10

STEP 11

Since the determinant is not zero, the inverse of (A−5B)(A-5B) exists. Now we can calculate the inverse.
(A−5B)−1=1det(A−5B)[−1310−10] (A-5B)^{-1} = \frac{1}{\text{det}(A-5B)} \left[\begin{array}{cc}-13 & 10 \\ -1 & 0\end{array}\right]

STEP 12

Multiply the adjugate of (A−5B)(A-5B) by 1det(A−5B)\frac{1}{\text{det}(A-5B)}.
(A−5B)−1=110[−1310−10] (A-5B)^{-1} = \frac{1}{10} \left[\begin{array}{cc}-13 & 10 \\ -1 & 0\end{array}\right]

STEP 13

Simplify the inverse matrix.
(A−5B)−1=[−1.31−0.10] (A-5B)^{-1} = \left[\begin{array}{cc}-1.3 & 1 \\ -0.1 & 0\end{array}\right]

STEP 14

Now we can solve for XX by multiplying (A−5B)−1(A-5B)^{-1} with matrix CC.
X=(A−5B)−1C X = (A-5B)^{-1}C

STEP 15

Perform the matrix multiplication.
X=[−1.31−0.10][−2−6−3−5] X = \left[\begin{array}{cc}-1.3 & 1 \\ -0.1 & 0\end{array}\right]\left[\begin{array}{cc}-2 & -6 \\ -3 & -5\end{array}\right]

STEP 16

Multiply the matrices element-wise.
X=[(−1.3)(−2)+(1)(−3)(−1.3)(−6)+(1)(−5)(−0.1)(−2)+(0)(−3)(−0.1)(−6)+(0)(−5)] X = \left[\begin{array}{cc}(-1.3)(-2) + (1)(-3) & (-1.3)(-6) + (1)(-5) \\ (-0.1)(-2) + (0)(-3) & (-0.1)(-6) + (0)(-5)\end{array}\right]

STEP 17

Calculate the entries of matrix XX.
X=[2.6−37.8−50.2+00.6+0] X = \left[\begin{array}{cc}2.6 - 3 & 7.8 - 5 \\ 0.2 + 0 & 0.6 + 0\end{array}\right]

STEP 18

Simplify the entries of matrix XX.
X=[−0.42.80.20.6] X = \left[\begin{array}{cc}-0.4 & 2.8 \\ 0.2 & 0.6\end{array}\right]
The solution for the matrix XX is:
X=[−0.42.80.20.6] X = \left[\begin{array}{cc}-0.4 & 2.8 \\ 0.2 & 0.6\end{array}\right]

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