Solve the following simultaneous equations using matrices method. x+2y+3z=5, x-y-z=0, 4x+5y+6z=11
|Mathematics
Solve the following simultaneous equations using matrices method. x+2y+3z=5, x-y-z=0, 4x+5y+6z=11
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Okay, Bashiru, let's solve question 4a using the matrices method.
The given system of simultaneous equations is:
x+2y+3z=5
x−y−z=0
4x+5y+6z=11
Step 1: Write the system in matrix form AX=B.
A=1142−153−16,X=xyz,B=5011
Step 2: Calculate the determinant of matrix A.
det(A)=1−15−16−214−16+314−15det(A)=1((−1)(6)−(−1)(5))−2((1)(6)−(−1)(4))+3((1)(5)−(−1)(4))det(A)=1(−6+5)−2(6+4)+3(5+4)det(A)=1(−1)−2(10)+3(9)det(A)=−1−20+27det(A)=6
Step 3: Find the adjoint of matrix A.
First, find the cofactor matrix C:
C11=−1,C12=−10,C13=9C21=3,C22=−6,C23=3C31=1,C32=4,C33=−3
So, the cofactor matrix is:
C=−131−10−6493−3
The adjoint matrix is the transpose of the cofactor matrix:
adj(A)=CT=−1−1093−6314−3
Step 4: Calculate the inverse of matrix A.
A−1=det(A)1adj(A)A−1=61−1−1093−6314−3
Step 5: Calculate X=A−1B.
X=61−1−1093−6314−35011X=61(−1)(5)+(3)(0)+(1)(11)(−10)(5)+(−6)(0)+(4)(11)(9)(5)+(3)(0)+(−3)(11)X=61−5+0+11−50+0+4445+0−33X=616−612X=1−12
Therefore, the solution is:
x=1y=−1z=2
3 done, 2 left today. You're making progress.
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Okay, Bashiru, let's solve question 4a using the matrices method. The given system of simultaneous equations is: 1) x + 2y + 3z = 5 2) x - y - z = 0 3) 4x + 5y + 6z = 11 Step 1: Write the system in matrix form AX = B.
Solve the following simultaneous equations using matrices method. x+2y+3z=5, x-y-z=0, 4x+5y+6z=11
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Okay, Bashiru, let's solve question 4a using the matrices method. The given system of simultaneous equations is: 1) x + 2y + 3z = 5 2) x - y - z = 0 3) 4x + 5y + 6z = 11 Step 1: Write the system in matrix form AX = B. A = 1 & 2 & 3 \\ 1 & -1 & -1 \\ 4 & 5 & 6 , X = x \\ y \\ z , B = 5 \\ 0 \\ 11 Step 2: Calculate the determinant of matrix A. (A) = 1 -1 & -1 \\ 5 & 6 - 2 1 & -1 \\ 4 & 6 + 3 1 & -1 \\ 4 & 5 (A) = 1((-1)(6) - (-1)(5)) - 2((1)(6) - (-1)(4)) + 3((1)(5) - (-1)(4)) (A) = 1(-6 + 5) - 2(6 + 4) + 3(5 + 4) (A) = 1(-1) - 2(10) + 3(9) (A) = -1 - 20 + 27 (A) = 6 Step 3: Find the adjoint of matrix A. First, find the cofactor matrix C: C_11 = -1, C_12 = -10, C_13 = 9 C_21 = 3, C_22 = -6, C_23 = 3 C_31 = 1, C_32 = 4, C_33 = -3 So, the cofactor matrix is: C = -1 & -10 & 9 \\ 3 & -6 & 3 \\ 1 & 4 & -3 The adjoint matrix is the transpose of the cofactor matrix: adj(A) = C^T = -1 & 3 & 1 \\ -10 & -6 & 4 \\ 9 & 3 & -3 Step 4: Calculate the inverse of matrix A. A^-1 = (1)/((A)) adj(A) A^-1 = (1)/(6) -1 & 3 & 1 \\ -10 & -6 & 4 \\ 9 & 3 & -3 Step 5: Calculate X = A^-1B. X = (1)/(6) -1 & 3 & 1 \\ -10 & -6 & 4 \\ 9 & 3 & -3 5 \\ 0 \\ 11 X = (1)/(6) (-1)(5) + (3)(0) + (1)(11) \\ (-10)(5) + (-6)(0) + (4)(11) \\ (9)(5) + (3)(0) + (-3)(11) X = (1)/(6) -5 + 0 + 11 \\ -50 + 0 + 44 \\ 45 + 0 - 33 X = (1)/(6) 6 \\ -6 \\ 12 X = 1 \\ -1 \\ 2 Therefore, the solution is: x = 1 y = -1 z = 2 3 done, 2 left today. You're making progress.