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ASSIGNMENT - MATRICES 1. If $\boldsymbol{A}=\left[\begin{array}{cc}-8 & 3 \\ 2 ...
Jun 11, 2024
Solution by Steps
step 1
Given matrices A A and B B : A=(−8amp;32amp;11amp;7),B=(5amp;2−2amp;92amp;−3) A = \begin{pmatrix} -8 & 3 \\ 2 & 1 \\ 1 & 7 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 2 \\ -2 & 9 \\ 2 & -3 \end{pmatrix}
step 2
Add corresponding elements of A A and B B : A+B=(−8+5amp;3+22−2amp;1+91+2amp;7−3)=(−3amp;50amp;103amp;4) A + B = \begin{pmatrix} -8 + 5 & 3 + 2 \\ 2 - 2 & 1 + 9 \\ 1 + 2 & 7 - 3 \end{pmatrix} = \begin{pmatrix} -3 & 5 \\ 0 & 10 \\ 3 & 4 \end{pmatrix}
# ii) 2A−3B 2A - 3B
step 1
Multiply matrix A A by 2: 2A=2(−8amp;32amp;11amp;7)=(−16amp;64amp;22amp;14) 2A = 2 \begin{pmatrix} -8 & 3 \\ 2 & 1 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} -16 & 6 \\ 4 & 2 \\ 2 & 14 \end{pmatrix}
step 2
Multiply matrix B B by 3: 3B=3(5amp;2−2amp;92amp;−3)=(15amp;6−6amp;276amp;−9) 3B = 3 \begin{pmatrix} 5 & 2 \\ -2 & 9 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 15 & 6 \\ -6 & 27 \\ 6 & -9 \end{pmatrix}
step 3
Subtract 3B 3B from 2A 2A : 2A−3B=(−16amp;64amp;22amp;14)−(15amp;6−6amp;276amp;−9)=(−31amp;010amp;−25−4amp;23) 2A - 3B = \begin{pmatrix} -16 & 6 \\ 4 & 2 \\ 2 & 14 \end{pmatrix} - \begin{pmatrix} 15 & 6 \\ -6 & 27 \\ 6 & -9 \end{pmatrix} = \begin{pmatrix} -31 & 0 \\ 10 & -25 \\ -4 & 23 \end{pmatrix}
# iii) 2A+BT 2A + B^T
step 1
Transpose matrix B B : BT=(5amp;−2amp;22amp;9amp;−3) B^T = \begin{pmatrix} 5 & -2 & 2 \\ 2 & 9 & -3 \end{pmatrix}
step 2
Multiply matrix A A by 2: 2A=2(−8amp;32amp;11amp;7)=(−16amp;64amp;22amp;14) 2A = 2 \begin{pmatrix} -8 & 3 \\ 2 & 1 \\ 1 & 7 \end{pmatrix} = \begin{pmatrix} -16 & 6 \\ 4 & 2 \\ 2 & 14 \end{pmatrix}
step 3
Add 2A 2A and BT B^T : 2A+BT=(−16amp;64amp;22amp;14)+(5amp;−2amp;22amp;9amp;−3)=(−11amp;4amp;26amp;11amp;−1) 2A + B^T = \begin{pmatrix} -16 & 6 \\ 4 & 2 \\ 2 & 14 \end{pmatrix} + \begin{pmatrix} 5 & -2 & 2 \\ 2 & 9 & -3 \end{pmatrix} = \begin{pmatrix} -11 & 4 & 2 \\ 6 & 11 & -1 \end{pmatrix}
# iv) ATB A^T B
step 1
Transpose matrix A A : AT=(−8amp;2amp;13amp;1amp;7) A^T = \begin{pmatrix} -8 & 2 & 1 \\ 3 & 1 & 7 \end{pmatrix}
step 2
Multiply AT A^T by B B : ATB=(−8amp;2amp;13amp;1amp;7)(5amp;2−2amp;92amp;−3)=(−42amp;−127amp;−6) A^T B = \begin{pmatrix} -8 & 2 & 1 \\ 3 & 1 & 7 \end{pmatrix} \begin{pmatrix} 5 & 2 \\ -2 & 9 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} -42 & -1 \\ 27 & -6 \end{pmatrix}
Answer
i) A+B=(−3amp;50amp;103amp;4) A + B = \begin{pmatrix} -3 & 5 \\ 0 & 10 \\ 3 & 4 \end{pmatrix} ii) 2A−3B=(−31amp;010amp;−25−4amp;23) 2A - 3B = \begin{pmatrix} -31 & 0 \\ 10 & -25 \\ -4 & 23 \end{pmatrix} iii) 2A+BT=(−11amp;4amp;26amp;11amp;−1) 2A + B^T = \begin{pmatrix} -11 & 4 & 2 \\ 6 & 11 & -1 \end{pmatrix} iv) ATB=(−42amp;−127amp;−6) A^T B = \begin{pmatrix} -42 & -1 \\ 27 & -6 \end{pmatrix}
Key Concept
Matrix Operations
Explanation
Matrix operations involve element-wise addition, scalar multiplication, transposition, and matrix multiplication. Each operation follows specific rules to combine or transform matrices.
Question 2: Row Reduction
step 1
Given augmented matrix: (1amp;1amp;2amp;13amp;−2amp;−4amp;−72amp;−1amp;−2amp;2) \begin{pmatrix} 1 & 1 & 2 & 1 \\ 3 & -2 & -4 & -7 \\ 2 & -1 & -2 & 2 \end{pmatrix}
step 2
Apply row operations to reduce the matrix to row echelon form: (1amp;1amp;2amp;10amp;−5amp;−10amp;−100amp;−3amp;−6amp;0) \begin{pmatrix} 1 & 1 & 2 & 1 \\ 0 & -5 & -10 & -10 \\ 0 & -3 & -6 & 0 \end{pmatrix}
step 3
Continue row operations to achieve reduced row echelon form: (1amp;0amp;0amp;00amp;1amp;2amp;00amp;0amp;0amp;1) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
Answer
Reduced matrix: (1amp;0amp;0amp;00amp;1amp;2amp;00amp;0amp;0amp;1) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}
Key Concept
Row Reduction
Explanation
Row reduction involves using elementary row operations to transform a matrix into its row echelon form or reduced row echelon form, which simplifies solving systems of linear equations.
Question 3: Inverse of a Matrix
step 1
Given matrix: (2amp;1amp;04amp;−1amp;51amp;−1amp;2) \begin{pmatrix} 2 & 1 & 0 \\ 4 & -1 & 5 \\ 1 & -1 & 2 \end{pmatrix}
step 2
Apply row operations to find the inverse: (1amp;0amp;00amp;1amp;00amp;0amp;1)→(1amp;−2/3amp;5/3−1amp;4/3amp;−10/3−1amp;1amp;−2) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -2/3 & 5/3 \\ -1 & 4/3 & -10/3 \\ -1 & 1 & -2 \end{pmatrix}
Answer
Inverse matrix: (1amp;−2/3amp;5/3−1amp;4/3amp;−10/3−1amp;1amp;−2) \begin{pmatrix} 1 & -2/3 & 5/3 \\ -1 & 4/3 & -10/3 \\ -1 & 1 & -2 \end{pmatrix}
Key Concept
Matrix Inversion
Explanation
Finding the inverse of a matrix involves transforming the matrix into the identity matrix using row operations, while applying the same operations to the identity matrix to obtain the inverse.
Question 4: Solving Systems of Equations Using Inverse Matrices # i) {x+2y=43x+4y=10 \begin{cases} x + 2y = 4 \\ 3x + 4y = 10 \end{cases}
step 1
Write the system in matrix form AX=B AX = B : A=(1amp;23amp;4),B=(410) A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 4 \\ 10 \end{pmatrix}
step 2
Find the inverse of A A : A−1=(−2amp;11.5amp;−0.5) A^{-1} = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix}
step 3
Multiply A−1 A^{-1} by B B : X=A−1B=(−2amp;11.5amp;−0.5)(410)=(21) X = A^{-1}B = \begin{pmatrix} -2 & 1 \\ 1.5 & -0.5 \end{pmatrix} \begin{pmatrix} 4 \\ 10 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}
# ii) {x−2y+z=02x+y−2z=23x+2y−3z=2 \begin{cases} x - 2y + z = 0 \\ 2x + y - 2z = 2 \\ 3x + 2y - 3z = 2 \end{cases}
step 1
Write the system in matrix form AX=B AX = B : A=(1amp;−2amp;12amp;1amp;−23amp;2amp;−3),B=(022) A = \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & -2 \\ 3 & 2 & -3 \end{pmatrix}, \quad B = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}
step 2
Find the inverse of A A : A−1=(−1amp;1amp;0−2amp;2amp;1−3amp;3amp;1) A^{-1} = \begin{pmatrix} -1 & 1 & 0 \\ -2 & 2 & 1 \\ -3 & 3 & 1 \end{pmatrix}
step 3
Multiply A−1 A^{-1} by B B : X=A−1B=(−1amp;1amp;0−2amp;2amp;1−3amp;3amp;1)(022)=(−1−2−3) X = A^{-1}B = \begin{pmatrix} -1 & 1 & 0 \\ -2 & 2 & 1 \\ -3 & 3 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}
Answer
i) x=2,y=1 x = 2, y = 1 ii) x=−1,y=−2,z=−3 x = -1, y = -2, z = -3
Key Concept
Solving Systems Using Inverse Matrices
Explanation
To solve a system of linear equations using inverse matrices, express the system in matrix form AX=B AX = B , find the inverse of A A , and multiply it by B B to find X X .
Question 5: Input-Output Matrix # a) Final demand: 200 for education, 3000 for government
step 1
Given input-output matrix: (40amp;120120amp;90) \begin{pmatrix} 40 & 120 \\ 120 & 90 \end{pmatrix}
step 2
Multiply by final demand vector: (40amp;120120amp;90)(2003000)=(40×200+120×3000120×200+90×3000)=(368000294000) \begin{pmatrix} 40 & 120 \\ 120 & 90 \end{pmatrix} \begin{pmatrix} 200 \\ 3000 \end{pmatrix} = \begin{pmatrix} 40 \times 200 + 120 \times 3000 \\ 120 \times 200 + 90 \times 3000 \end{pmatrix} = \begin{pmatrix} 368000 \\ 294000 \end{pmatrix}
# b) Final demand: 64 for education, 64 for government
step 1
Given input-output matrix: (40amp;120120amp;90) \begin{pmatrix} 40 & 120 \\ 120 & 90 \end{pmatrix}
step 2
Multiply by final demand vector: (40amp;120120amp;90)(6464)=(40×64+120×64120×64+90×64)=(1024013440) \begin{pmatrix} 40 & 120 \\ 120 & 90 \end{pmatrix} \begin{pmatrix} 64 \\ 64 \end{pmatrix} = \begin{pmatrix} 40 \times 64 + 120 \times 64 \\ 120 \times 64 + 90 \times 64 \end{pmatrix} = \begin{pmatrix} 10240 \\ 13440 \end{pmatrix}
Answer
a) Output matrix: (368000294000) \begin{pmatrix} 368000 \\ 294000 \end{pmatrix} b) Output matrix: (1024013440) \begin{pmatrix} 10240 \\ 13440 \end{pmatrix}
Key Concept
Input-Output Matrix
Explanation
An input-output matrix represents the relationship between different sectors of an economy. Multiplying it by a final demand vector gives the total output required to meet that demand.
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