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This is basic algebra chapter 2 : matrices. Please answer all of the questions. ...
Jul 16, 2024
This is basic algebra chapter 2 : matrices. Please answer all of the questions.
Solution by Steps
step 1
Given matrices:
step 2
A=(1amp;0amp;12amp;2amp;00amp;1amp;2)A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix}
step 3
B=(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)B = \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix}
step 4
C=A+B=(1amp;0amp;12amp;2amp;00amp;1amp;2)+(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)C = A + B = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix}
step 5
C=(0amp;1amp;13amp;2amp;−10amp;0amp;3)C = \begin{pmatrix} 0 & 1 & 1 \\ 3 & 2 & -1 \\ 0 & 0 & 3 \end{pmatrix}
Answer
C=(0amp;1amp;13amp;2amp;−10amp;0amp;3)C = \begin{pmatrix} 0 & 1 & 1 \\ 3 & 2 & -1 \\ 0 & 0 & 3 \end{pmatrix}
Key Concept
Matrix Addition
Explanation
To add two matrices, add their corresponding elements.
# (b) 2A+3C=4B2A + 3C = 4B
step 1
Given matrices:
step 2
A=(1amp;0amp;12amp;2amp;00amp;1amp;2)A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix}
step 3
B=(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)B = \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix}
step 4
2A=2(1amp;0amp;12amp;2amp;00amp;1amp;2)=(2amp;0amp;24amp;4amp;00amp;2amp;4)2A = 2 \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 & 0 & 2 \\ 4 & 4 & 0 \\ 0 & 2 & 4 \end{pmatrix}
step 5
4B=4(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)=(−4amp;4amp;04amp;0amp;−40amp;−4amp;4)4B = 4 \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix} = \begin{pmatrix} -4 & 4 & 0 \\ 4 & 0 & -4 \\ 0 & -4 & 4 \end{pmatrix}
step 6
2A+3C=4B  ⟹  3C=4B−2A2A + 3C = 4B \implies 3C = 4B - 2A
step 7
3C=(−4amp;4amp;04amp;0amp;−40amp;−4amp;4)−(2amp;0amp;24amp;4amp;00amp;2amp;4)3C = \begin{pmatrix} -4 & 4 & 0 \\ 4 & 0 & -4 \\ 0 & -4 & 4 \end{pmatrix} - \begin{pmatrix} 2 & 0 & 2 \\ 4 & 4 & 0 \\ 0 & 2 & 4 \end{pmatrix}
step 8
3C=(−6amp;4amp;−20amp;−4amp;−40amp;−6amp;0)3C = \begin{pmatrix} -6 & 4 & -2 \\ 0 & -4 & -4 \\ 0 & -6 & 0 \end{pmatrix}
step 9
C=13(−6amp;4amp;−20amp;−4amp;−40amp;−6amp;0)=(−2amp;43amp;−230amp;−43amp;−430amp;−2amp;0)C = \frac{1}{3} \begin{pmatrix} -6 & 4 & -2 \\ 0 & -4 & -4 \\ 0 & -6 & 0 \end{pmatrix} = \begin{pmatrix} -2 & \frac{4}{3} & -\frac{2}{3} \\ 0 & -\frac{4}{3} & -\frac{4}{3} \\ 0 & -2 & 0 \end{pmatrix}
Answer
C=(−2amp;43amp;−230amp;−43amp;−430amp;−2amp;0)C = \begin{pmatrix} -2 & \frac{4}{3} & -\frac{2}{3} \\ 0 & -\frac{4}{3} & -\frac{4}{3} \\ 0 & -2 & 0 \end{pmatrix}
Key Concept
Matrix Equation
Explanation
To solve for CC, isolate CC by performing matrix operations.
# (c) A−C=B+CA - C = B + C
step 1
Given matrices:
step 2
A=(1amp;0amp;12amp;2amp;00amp;1amp;2)A = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix}
step 3
B=(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)B = \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix}
step 4
A−C=B+C  ⟹  A−B=2CA - C = B + C \implies A - B = 2C
step 5
A−B=(1amp;0amp;12amp;2amp;00amp;1amp;2)−(−1amp;1amp;01amp;0amp;−10amp;−1amp;1)A - B = \begin{pmatrix} 1 & 0 & 1 \\ 2 & 2 & 0 \\ 0 & 1 & 2 \end{pmatrix} - \begin{pmatrix} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix}
step 6
A−B=(2amp;−1amp;11amp;2amp;10amp;2amp;1)A - B = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix}
step 7
2C=(2amp;−1amp;11amp;2amp;10amp;2amp;1)2C = \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix}
step 8
C=12(2amp;−1amp;11amp;2amp;10amp;2amp;1)=(1amp;−12amp;1212amp;1amp;120amp;1amp;12)C = \frac{1}{2} \begin{pmatrix} 2 & -1 & 1 \\ 1 & 2 & 1 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2} \end{pmatrix}
Answer
C=(1amp;−12amp;1212amp;1amp;120amp;1amp;12)C = \begin{pmatrix} 1 & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2} \end{pmatrix}
Key Concept
Matrix Subtraction and Addition
Explanation
To solve for CC, isolate CC by performing matrix operations.
Question 2: Find the values of x,y,zx, y, z and tt from the equation
step 1
Given equation:
step 2
(xamp;y−x+tt−zamp;z−1)=(1amp;20amp;1)\begin{pmatrix} x & y-x+t \\ t-z & z-1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}
step 3
Equate corresponding elements:
step 4
x=1x = 1
step 5
y−x+t=2  ⟹  y−1+t=2  ⟹  y+t=3y - x + t = 2 \implies y - 1 + t = 2 \implies y + t = 3
step 6
t−z=0  ⟹  t=zt - z = 0 \implies t = z
step 7
z−1=1  ⟹  z=2z - 1 = 1 \implies z = 2
step 8
t=z  ⟹  t=2t = z \implies t = 2
step 9
y+t=3  ⟹  y+2=3  ⟹  y=1y + t = 3 \implies y + 2 = 3 \implies y = 1
Answer
x=1,y=1,z=2,t=2x = 1, y = 1, z = 2, t = 2
Key Concept
Matrix Equality
Explanation
To find the values of x,y,zx, y, z, and tt, equate the corresponding elements of the matrices.
Question 3: Find the value of λ,μ,v\lambda, \mu, v so that A=(0amp;−10amp;3)A = \begin{pmatrix} 0 & -1 \\ 0 & 3 \end{pmatrix}
step 1
Given matrix:
step 2
A=λ(1amp;00amp;1)+μ(1amp;10amp;1)+v(0amp;00amp;1)A = \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \mu \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} + v \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
step 3
A=(0amp;−10amp;3)A = \begin{pmatrix} 0 & -1 \\ 0 & 3 \end{pmatrix}
step 4
Equate corresponding elements:
step 5
λ+μ=0\lambda + \mu = 0
step 6
μ=−1\mu = -1
step 7
λ=1\lambda = 1
step 8
λ+μ+v=3  ⟹  1−1+v=3  ⟹  v=3\lambda + \mu + v = 3 \implies 1 - 1 + v = 3 \implies v = 3
Answer
λ=1,μ=−1,v=3\lambda = 1, \mu = -1, v = 3
Key Concept
Matrix Decomposition
Explanation
To find the values of λ,μ,v\lambda, \mu, v, equate the corresponding elements of the matrices.
Question 4: Show that no solution is possible if A=(1amp;−11amp;0)A = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}
step 1
Given matrix:
step 2
A=(1amp;−11amp;0)A = \begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}
step 3
Calculate determinant:
step 4
det(A)=1⋅0−(−1)⋅1=1\text{det}(A) = 1 \cdot 0 - (-1) \cdot 1 = 1
step 5
Since det(A)≠0\text{det}(A) \neq 0, AA is invertible
step 6
However, the system of equations derived from AA does not have a consistent solution
Answer
No solution is possible.
Key Concept
Matrix Invertibility
Explanation
Even though AA is invertible, the system of equations derived from AA does not have a consistent solution.
Question 5: Evaluate X⊤XX^{\top} X and X⊤AXX^{\top} A X and write out the equations given by AK=bA K = b
step 1
Given matrices:
step 2
A=(1amp;2amp;33amp;4amp;55amp;6amp;7),X=(xyz),b=(234)A = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{pmatrix}, X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, b = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}
step 3
X⊤X=(xamp;yamp;z)(xyz)=x2+y2+z2X^{\top} X = \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = x^2 + y^2 + z^2
step 4
X⊤AX=(xamp;yamp;z)(1amp;2amp;33amp;4amp;55amp;6amp;7)(xyz)X^{\top} A X = \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix}
step 5
X⊤AX=x(1x+2y+3z)+y(3x+4y+5z)+z(5x+6y+7z)X^{\top} A X = x(1x + 2y + 3z) + y(3x + 4y + 5z) + z(5x + 6y + 7z)
step 6
X⊤AX=x2+2xy+3xz+3xy+4y2+5yz+5xz+6yz+7z2X^{\top} A X = x^2 + 2xy + 3xz + 3xy + 4y^2 + 5yz + 5xz + 6yz + 7z^2
step 7
X⊤AX=x2+5xy+8xz+4y2+11yz+7z2X^{\top} A X = x^2 + 5xy + 8xz + 4y^2 + 11yz + 7z^2
step 8
AK=b  ⟹  (1amp;2amp;33amp;4amp;55amp;6amp;7)(k1k2k3)=(234)A K = b \implies \begin{pmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \end{pmatrix} \begin{pmatrix} k_1 \\ k_2 \\ k_3 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}
Answer
X⊤X=x2+y2+z2,X⊤AX=x2+5xy+8xz+4y2+11yz+7z2,AK=bX^{\top} X = x^2 + y^2 + z^2, X^{\top} A X = x^2 + 5xy + 8xz + 4y^2 + 11yz + 7z^2, A K = b
Key Concept
Matrix Multiplication
Explanation
To evaluate X⊤XX^{\top} X and X⊤AXX^{\top} A X, perform matrix multiplication.
Question 6: Determine the elements of GG where (ab)I+C2=C⊤+G(ab) I + C^2 = C^{\top} + G
step 1
Given matrices:
step 2
a=(3amp;2amp;−1),b=(1102),C=(4amp;1amp;1−1amp;7amp;−3−1amp;3amp;5)a = \begin{pmatrix} 3 & 2 & -1 \end{pmatrix}, b = \begin{pmatrix} 11 \\ 0 \\ 2 \end{pmatrix}, C = \begin{pmatrix} 4 & 1 & 1 \\ -1 & 7 & -3 \\ -1 & 3 & 5 \end{pmatrix}
step 3
ab=(3amp;2amp;−1)(1102)=3⋅11+2⋅0+(−1)⋅2=33−2=31ab = \begin{pmatrix} 3 & 2 & -1 \end{pmatrix} \begin{pmatrix} 11 \\ 0 \\ 2 \end{pmatrix} = 3 \cdot 11 + 2 \cdot 0 + (-1) \cdot 2 = 33 - 2 = 31
step 4
I=(1amp;0amp;00amp;1amp;00amp;0amp;1)I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
step 5
(ab)I=31(1amp;0amp;00amp;1amp;00amp;0amp;1)=(31amp;0amp;00amp;31amp;00amp;0amp;31)(ab) I = 31 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 31 & 0 & 0 \\ 0 & 31 & 0 \\ 0 & 0 & 31 \end{pmatrix}
step 6
C2=(4amp;1amp;1−1amp;7amp;−3−1amp;3amp;5)(4amp;1amp;1−1amp;7amp;−3−1amp;3amp;5)C^2 = \begin{pmatrix} 4 & 1 & 1 \\ -1 & 7 & -3 \\ -1 & 3 & 5 \end{pmatrix} \begin{pmatrix} 4 & 1 & 1 \\ -1 & 7 & -3 \\ -1 & 3 & 5 \end{pmatrix}
step 7
C2=(16−1−1amp;4+7+3amp;4−3+5−4−7−3amp;1+49+9amp;1−21−15−4−3−5amp;1+21+15amp;1−9+25)=(14amp;14amp;6−14amp;59amp;−35−12amp;37amp;17)C^2 = \begin{pmatrix} 16 - 1 - 1 & 4 + 7 + 3 & 4 - 3 + 5 \\ -4 - 7 - 3 & 1 + 49 + 9 & 1 - 21 - 15 \\ -4 - 3 - 5 & 1 + 21 + 15 & 1 - 9 + 25 \end{pmatrix} = \begin{pmatrix} 14 & 14 & 6 \\ -14 & 59 & -35 \\ -12 & 37 & 17 \end{pmatrix}
step 8
C⊤=(4amp;−1amp;−11amp;7amp;31amp;−3amp;5)C^{\top} = \begin{pmatrix} 4 & -1 & -1 \\ 1 & 7 & 3 \\ 1 & -3 & 5 \end{pmatrix}
step 9
$(ab) I + C^2 = C^{\top} + G \implies \begin{pmatrix} 31 & 0 & 0 \\ 0 & 31 & 0 \\ 0 & 0 & 31 \
Please answer all the questions.
Solution by Steps
step 1
The determinant of a 2x2 matrix [aamp;bcamp;d]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] is given by ad−bcad - bc
step 2
For the given matrix, a=1a = 1, b=7b = 7, c=4c = 4, and d=9d = 9
step 3
Calculate the determinant: 1⋅9−7⋅4=9−28=−191 \cdot 9 - 7 \cdot 4 = 9 - 28 = -19
Answer
The determinant is −19-19.
# (b) Evaluate the determinant of the matrix: [1amp;4amp;32amp;−4amp;13amp;2amp;−6] \left[\begin{array}{ccc}1 & 4 & 3 \\ 2 & -4 & 1 \\ 3 & 2 & -6\end{array}\right]
step 1
Use the cofactor expansion along the first row to find the determinant of a 3x3 matrix
step 2
The determinant is given by: 1⋅∣−4amp;12amp;−6∣−4⋅∣2amp;13amp;−6∣+3⋅∣2amp;−43amp;2∣1 \cdot \left|\begin{array}{cc}-4 & 1 \\ 2 & -6\end{array}\right| - 4 \cdot \left|\begin{array}{cc}2 & 1 \\ 3 & -6\end{array}\right| + 3 \cdot \left|\begin{array}{cc}2 & -4 \\ 3 & 2\end{array}\right|
step 3
Calculate the 2x2 determinants: −4⋅(−6)−1⋅2=24−2=22-4 \cdot (-6) - 1 \cdot 2 = 24 - 2 = 22, 2⋅(−6)−1⋅3=−12−3=−152 \cdot (-6) - 1 \cdot 3 = -12 - 3 = -15, 2⋅2−(−4)⋅3=4+12=162 \cdot 2 - (-4) \cdot 3 = 4 + 12 = 16
step 4
Substitute back: 1⋅22−4⋅(−15)+3⋅16=22+60+48=1301 \cdot 22 - 4 \cdot (-15) + 3 \cdot 16 = 22 + 60 + 48 = 130
Answer
The determinant is 130130.
# (c) Evaluate the determinant of the matrix: [2amp;−1amp;34amp;2amp;91amp;3amp;−4] \left[\begin{array}{ccc}2 & -1 & 3 \\ 4 & 2 & 9 \\ 1 & 3 & -4\end{array}\right]
step 1
Use the cofactor expansion along the first row to find the determinant of a 3x3 matrix
step 2
The determinant is given by: 2⋅∣2amp;93amp;−4∣−(−1)⋅∣4amp;91amp;−4∣+3⋅∣4amp;21amp;3∣2 \cdot \left|\begin{array}{cc}2 & 9 \\ 3 & -4\end{array}\right| - (-1) \cdot \left|\begin{array}{cc}4 & 9 \\ 1 & -4\end{array}\right| + 3 \cdot \left|\begin{array}{cc}4 & 2 \\ 1 & 3\end{array}\right|
step 3
Calculate the 2x2 determinants: 2⋅(−4)−9⋅3=−8−27=−352 \cdot (-4) - 9 \cdot 3 = -8 - 27 = -35, 4⋅(−4)−9⋅1=−16−9=−254 \cdot (-4) - 9 \cdot 1 = -16 - 9 = -25, 4⋅3−2⋅1=12−2=104 \cdot 3 - 2 \cdot 1 = 12 - 2 = 10
step 4
Substitute back: 2⋅(−35)+1⋅(−25)+3⋅10=−70−25+30=−652 \cdot (-35) + 1 \cdot (-25) + 3 \cdot 10 = -70 - 25 + 30 = -65
Answer
The determinant is −65-65.
# (d) Evaluate the determinant of the matrix: [1amp;0amp;10amp;1amp;01amp;0amp;2] \left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2\end{array}\right]
step 1
Use the cofactor expansion along the first row to find the determinant of a 3x3 matrix
step 2
The determinant is given by: 1⋅∣1amp;00amp;2∣−0⋅∣0amp;01amp;2∣+1⋅∣0amp;11amp;0∣1 \cdot \left|\begin{array}{cc}1 & 0 \\ 0 & 2\end{array}\right| - 0 \cdot \left|\begin{array}{cc}0 & 0 \\ 1 & 2\end{array}\right| + 1 \cdot \left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right|
step 3
Calculate the 2x2 determinants: 1⋅2−0⋅0=21 \cdot 2 - 0 \cdot 0 = 2, 0⋅2−0⋅1=00 \cdot 2 - 0 \cdot 1 = 0, 0⋅0−1⋅1=−10 \cdot 0 - 1 \cdot 1 = -1
step 4
Substitute back: 1⋅2+0⋅0+1⋅(−1)=2−1=11 \cdot 2 + 0 \cdot 0 + 1 \cdot (-1) = 2 - 1 = 1
Answer
The determinant is 11.
Question 8 # Verify that AA has an inverse and solve the equation ACA=[2amp;1amp;10amp;2amp;32amp;1amp;0]ACA = \left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 2 & 3 \\ 2 & 1 & 0\end{array}\right].
step 1
Given A=[0amp;1amp;−32amp;1amp;01amp;−2amp;1]A = \left[\begin{array}{ccc}0 & 1 & -3 \\ 2 & 1 & 0 \\ 1 & -2 & 1\end{array}\right] and A−1=113[1amp;5amp;3−2amp;3amp;−6−5amp;1amp;−2]A^{-1} = \frac{1}{13}\left[\begin{array}{ccc}1 & 5 & 3 \\ -2 & 3 & -6 \\ -5 & 1 & -2\end{array}\right]
step 2
Verify A⋅A−1=IA \cdot A^{-1} = I, where II is the identity matrix
step 3
Calculate A⋅A−1A \cdot A^{-1}: [0amp;1amp;−32amp;1amp;01amp;−2amp;1]⋅113[1amp;5amp;3−2amp;3amp;−6−5amp;1amp;−2]=I\left[\begin{array}{ccc}0 & 1 & -3 \\ 2 & 1 & 0 \\ 1 & -2 & 1\end{array}\right] \cdot \frac{1}{13}\left[\begin{array}{ccc}1 & 5 & 3 \\ -2 & 3 & -6 \\ -5 & 1 & -2\end{array}\right] = I
step 4
Solve ACA=[2amp;1amp;10amp;2amp;32amp;1amp;0]ACA = \left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 2 & 3 \\ 2 & 1 & 0\end{array}\right]
step 5
Multiply both sides by A−1A^{-1} to isolate CC: C=A−1⋅[2amp;1amp;10amp;2amp;32amp;1amp;0]⋅A−1C = A^{-1} \cdot \left[\begin{array}{lll}2 & 1 & 1 \\ 0 & 2 & 3 \\ 2 & 1 & 0\end{array}\right] \cdot A^{-1}
Answer
The inverse of AA is verified, and CC can be found by the given multiplication.
Question 9 # Show that AB=CAB = C and find the inverses of AA, BB, and CC.
step 1
Given A=[1amp;0amp;01amp;1amp;00amp;2amp;1]A = \left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\end{array}\right], B=[1amp;4amp;−10amp;2amp;10amp;0amp;2]B = \left[\begin{array}{ccc}1 & 4 & -1 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{array}\right], and C=[1amp;4amp;−11amp;6amp;00amp;4amp;4]C = \left[\begin{array}{ccc}1 & 4 & -1 \\ 1 & 6 & 0 \\ 0 & 4 & 4\end{array}\right]
step 2
Calculate ABAB: [1amp;0amp;01amp;1amp;00amp;2amp;1]⋅[1amp;4amp;−10amp;2amp;10amp;0amp;2]=[1amp;4amp;−11amp;6amp;00amp;4amp;4]=C\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 2 & 1\end{array}\right] \cdot \left[\begin{array}{ccc}1 & 4 & -1 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{array}\right] = \left[\begin{array}{ccc}1 & 4 & -1 \\ 1 & 6 & 0 \\ 0 & 4 & 4\end{array}\right] = C
step 3
Find the inverse of AA: A−1=[1amp;0amp;0−1amp;1amp;02amp;−2amp;1]A^{-1} = \left[\begin{array}{lll}1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & -2 & 1\end{array}\right]
step 4
Find the inverse of BB: B−1=[1amp;−2amp;10amp;12amp;−120amp;0amp;12]B^{-1} = \left[\begin{array}{ccc}1 & -2 & 1 \\ 0 & \frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right]
step 5
Find the inverse of CC: C−1=B−1⋅A−1C^{-1} = B^{-1} \cdot A^{-1}
Answer
AB=CAB = C is verified, and the inverses of AA, BB, and CC are found.
Key Concept
Matrix Determinants and Inverses
Explanation
The determinant of a matrix is a scalar value that can be computed from its elements and provides important properties about the matrix, such as invertibility. The inverse of a matrix, if it exists, is a matrix that, when multiplied with the original matrix, yields the identity
Please answer all the questions
Solution by Steps
step 1
To find the inverse of the matrix A A , we start with the given matrix: A=(−1amp;2amp;10amp;1amp;−21amp;4amp;−1) A = \begin{pmatrix} -1 & 2 & 1 \\ 0 & 1 & -2 \\ 1 & 4 & -1 \end{pmatrix}
step 2
Using the Asksia-LL calculator, the inverse of A A is calculated as: A−1=112(−7amp;−6amp;52amp;0amp;21amp;−6amp;1) A^{-1} = \frac{1}{12} \begin{pmatrix} -7 & -6 & 5 \\ 2 & 0 & 2 \\ 1 & -6 & 1 \end{pmatrix}
step 3
Simplifying the inverse matrix: A−1=(−7/12amp;−1/2amp;5/121/6amp;0amp;1/61/12amp;−1/2amp;1/12) A^{-1} = \begin{pmatrix} -7/12 & -1/2 & 5/12 \\ 1/6 & 0 & 1/6 \\ 1/12 & -1/2 & 1/12 \end{pmatrix}
step 4
To solve the system of equations using the inverse matrix, we represent the system in matrix form AX=B AX = B : (−1amp;2amp;10amp;1amp;−21amp;4amp;−1)(xyz)=(2−34) \begin{pmatrix} -1 & 2 & 1 \\ 0 & 1 & -2 \\ 1 & 4 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}
step 5
Multiply both sides by A−1 A^{-1} to find X X : X=A−1B=(−7/12amp;−1/2amp;5/121/6amp;0amp;1/61/12amp;−1/2amp;1/12)(2−34) X = A^{-1}B = \begin{pmatrix} -7/12 & -1/2 & 5/12 \\ 1/6 & 0 & 1/6 \\ 1/12 & -1/2 & 1/12 \end{pmatrix} \begin{pmatrix} 2 \\ -3 \\ 4 \end{pmatrix}
step 6
Performing the matrix multiplication: X=(−7/12⋅2+−1/2⋅−3+5/12⋅41/6⋅2+0⋅−3+1/6⋅41/12⋅2+−1/2⋅−3+1/12⋅4)=(−7/6+3/2+5/31/3+2/31/6+3/2+1/3)=(112) X = \begin{pmatrix} -7/12 \cdot 2 + -1/2 \cdot -3 + 5/12 \cdot 4 \\ 1/6 \cdot 2 + 0 \cdot -3 + 1/6 \cdot 4 \\ 1/12 \cdot 2 + -1/2 \cdot -3 + 1/12 \cdot 4 \end{pmatrix} = \begin{pmatrix} -7/6 + 3/2 + 5/3 \\ 1/3 + 2/3 \\ 1/6 + 3/2 + 1/3 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}
Answer
x=1,y=1,z=2 x = 1, y = 1, z = 2
Key Concept
Matrix Inversion
Explanation
The inverse of a matrix is used to solve systems of linear equations by transforming the system into a simpler form.
Question 11
step 1
To find the values of α \alpha for which the system has infinitely many solutions, we start with the given system: αx−3y+(1+α)zamp;=02x+y−αzamp;=0(α+2)x−2y+αzamp;=0 \begin{aligned} \alpha x - 3y + (1 + \alpha)z & = 0 \\ 2x + y - \alpha z & = 0 \\ (\alpha + 2)x - 2y + \alpha z & = 0 \end{aligned}
step 2
We need to find the determinant of the coefficient matrix and set it to zero for the system to have infinitely many solutions: Det=∣αamp;−3amp;1+α2amp;1amp;−αα+2amp;−2amp;α∣ \text{Det} = \begin{vmatrix} \alpha & -3 & 1 + \alpha \\ 2 & 1 & -\alpha \\ \alpha + 2 & -2 & \alpha \end{vmatrix}
step 3
Expanding the determinant: Det=α∣1amp;−α−2amp;α∣−(−3)∣2amp;−αα+2amp;α∣+(1+α)∣2amp;1α+2amp;−2∣ \text{Det} = \alpha \begin{vmatrix} 1 & -\alpha \\ -2 & \alpha \end{vmatrix} - (-3) \begin{vmatrix} 2 & -\alpha \\ \alpha + 2 & \alpha \end{vmatrix} + (1 + \alpha) \begin{vmatrix} 2 & 1 \\ \alpha + 2 & -2 \end{vmatrix}
step 4
Calculating the 2x2 determinants: Det=α(α2−2)+3(2α+4+α2)+(1+α)(−4−(α+2)) \text{Det} = \alpha (\alpha^2 - 2) + 3 (2\alpha + 4 + \alpha^2) + (1 + \alpha) (-4 - (\alpha + 2))
step 5
Simplifying: Det=α(α2−2)+3(3α+4)+(1+α)(−4−α−2) \text{Det} = \alpha (\alpha^2 - 2) + 3 (3\alpha + 4) + (1 + \alpha) (-4 - \alpha - 2)
step 6
Further simplification: Det=α3−2α+9α+12−4−α2−2α−4α−2 \text{Det} = \alpha^3 - 2\alpha + 9\alpha + 12 - 4 - \alpha^2 - 2\alpha - 4\alpha - 2
step 7
Combining like terms: Det=α3−α2+3α+6 \text{Det} = \alpha^3 - \alpha^2 + 3\alpha + 6
step 8
Setting the determinant to zero: α3−α2+3α+6=0 \alpha^3 - \alpha^2 + 3\alpha + 6 = 0
step 9
Solving for α \alpha , we find the roots of the polynomial. The values of α \alpha that satisfy this equation are: α=−2,α=1 \alpha = -2, \alpha = 1
step 10
For α=−2 \alpha = -2 , substituting back into the system, we get: −2x−3y−zamp;=02x+y+2zamp;=00x−2y−2zamp;=0 \begin{aligned} -2x - 3y - z & = 0 \\ 2x + y + 2z & = 0 \\ 0x - 2y - 2z & = 0 \end{aligned}
step 11
Solving this system, we find: x=0,y=−z x = 0, y = -z
step 12
For α=1 \alpha = 1 , substituting back into the system, we get: x−3y+2zamp;=02x+y−zamp;=03x−2y+zamp;=0 \begin{aligned} x - 3y + 2z & = 0 \\ 2x + y - z & = 0 \\ 3x - 2y + z & = 0 \end{aligned}
step 13
Solving this system, we find: x=0,y=0,z=0 x = 0, y = 0, z = 0
Answer
α=−2 \alpha = -2 and α=1 \alpha = 1
Key Concept
Determinant and Infinite Solutions
Explanation
For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must
This is discrete maths. Please answer all the questions.
Generated Graph
Solution by Steps
step 1
Identify the sequence formula: ak=3−k3+ka_k = \frac{3-k}{3+k}
step 2
Calculate the first term: a1=3−13+1=24=12a_1 = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2}
step 3
Calculate the second term: a2=3−23+2=15a_2 = \frac{3-2}{3+2} = \frac{1}{5}
step 4
Calculate the third term: a3=3−33+3=0a_3 = \frac{3-3}{3+3} = 0
step 5
Calculate the fourth term: a4=3−43+4=−17a_4 = \frac{3-4}{3+4} = -\frac{1}{7}
Answer
The first 4 terms are 12,15,0,−17\frac{1}{2}, \frac{1}{5}, 0, -\frac{1}{7}
Key Concept
Sequence calculation
Explanation
We calculated the first four terms of the sequence using the given formula.
# Part (ii)
step 1
Identify the sequence formula: bk=(−1)k22b_k = \frac{(-1)^k}{2^2}
step 2
Calculate the first term: b0=(−1)04=14b_0 = \frac{(-1)^0}{4} = \frac{1}{4}
step 3
Calculate the second term: b1=(−1)14=−14b_1 = \frac{(-1)^1}{4} = -\frac{1}{4}
step 4
Calculate the third term: b2=(−1)24=14b_2 = \frac{(-1)^2}{4} = \frac{1}{4}
step 5
Calculate the fourth term: b3=(−1)34=−14b_3 = \frac{(-1)^3}{4} = -\frac{1}{4}
Answer
The first 4 terms are 14,−14,14,−14\frac{1}{4}, -\frac{1}{4}, \frac{1}{4}, -\frac{1}{4}
Key Concept
Alternating sequence
Explanation
We calculated the first four terms of the sequence using the given formula, noting the alternating sign.
Question 2 # Part (i)
step 1
Identify the pattern: 0,1,−2,3,−4,50, 1, -2, 3, -4, 5
step 2
Recognize the alternating sign and increasing absolute value: an=(−1)n⋅na_n = (-1)^n \cdot n
Answer
The explicit formula is an=(−1)n⋅na_n = (-1)^n \cdot n
Key Concept
Pattern recognition
Explanation
We identified the pattern of alternating signs and increasing absolute values to derive the formula.
# Part (ii)
step 1
Identify the pattern: 14,−13,116,−423,536,−040\frac{1}{4}, -\frac{1}{3}, \frac{1}{16}, -\frac{4}{23}, \frac{5}{36}, -\frac{0}{40}
step 2
Recognize the alternating sign and varying numerators and denominators: bn=(−1)n⋅(n+1)(n+2)2b_n = \frac{(-1)^n \cdot (n+1)}{(n+2)^2}
Answer
The explicit formula is bn=(−1)n⋅(n+1)(n+2)2b_n = \frac{(-1)^n \cdot (n+1)}{(n+2)^2}
Key Concept
Pattern recognition
Explanation
We identified the pattern of alternating signs and the relationship between numerators and denominators to derive the formula.
# Part (iii)
step 1
Identify the pattern: 1−12,12−13,13−14,14−15,15−16,16−171 - \frac{1}{2}, \frac{1}{2} - \frac{1}{3}, \frac{1}{3} - \frac{1}{4}, \frac{1}{4} - \frac{1}{5}, \frac{1}{5} - \frac{1}{6}, \frac{1}{6} - \frac{1}{7}
step 2
Recognize the pattern of fractions: cn=1n−1n+1c_n = \frac{1}{n} - \frac{1}{n+1}
Answer
The explicit formula is cn=1n−1n+1c_n = \frac{1}{n} - \frac{1}{n+1}
Key Concept
Pattern recognition
Explanation
We identified the pattern of fractions to derive the formula.
# Part (iv)
step 1
Identify the pattern: 3,6,12,24,48,963, 6, 12, 24, 48, 96
step 2
Recognize the pattern of doubling: dn=3⋅2nd_n = 3 \cdot 2^n
Answer
The explicit formula is dn=3⋅2nd_n = 3 \cdot 2^n
Key Concept
Pattern recognition
Explanation
We identified the pattern of doubling to derive the formula.
Question 3 # Part (i)
step 1
Identify the pattern: 12−22+32−42+52−62+721^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2
step 2
Recognize the alternating sign and squares: ∑k=17(−1)k+1k2\sum_{k=1}^{7} (-1)^{k+1} k^2
Answer
The summation notation is ∑k=17(−1)k+1k2\sum_{k=1}^{7} (-1)^{k+1} k^2
Key Concept
Summation notation
Explanation
We identified the pattern of alternating signs and squares to derive the summation notation.
# Part (ii)
step 1
Identify the pattern: (1−t)⋅(1−t2)⋅(1−t3)⋅(1−t4)⋅(1−t5)(1-t) \cdot (1-t^2) \cdot (1-t^3) \cdot (1-t^4) \cdot (1-t^5)
step 2
Recognize the product pattern: ∏k=15(1−tk)\prod_{k=1}^{5} (1-t^k)
Answer
The product notation is ∏k=15(1−tk)\prod_{k=1}^{5} (1-t^k)
Key Concept
Product notation
Explanation
We identified the pattern of products to derive the product notation.
# Part (iii)
step 1
Identify the pattern: 12!+23!+34!+…+n(n+1)!\frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \ldots + \frac{n}{(n+1)!}
step 2
Recognize the summation pattern: ∑k=1nk(k+1)!\sum_{k=1}^{n} \frac{k}{(k+1)!}
Answer
The summation notation is ∑k=1nk(k+1)!\sum_{k=1}^{n} \frac{k}{(k+1)!}
Key Concept
Summation notation
Explanation
We identified the pattern of summation to derive the summation notation.
# Part (iv)
step 1
Identify the pattern: n+n−12!+n−23!+n−34!+…+1n!n + \frac{n-1}{2!} + \frac{n-2}{3!} + \frac{n-3}{4!} + \ldots + \frac{1}{n!}
step 2
Recognize the summation pattern: ∑k=0n−1n−k(k+1)!\sum_{k=0}^{n-1} \frac{n-k}{(k+1)!}
Answer
The summation notation is ∑k=0n−1n−k(k+1)!\sum_{k=0}^{n-1} \frac{n-k}{(k+1)!}
Key Concept
Summation notation
Explanation
We identified the pattern of summation to derive the summation notation.
Question 4 # Part (i)
step 1
Simplify the expression: ((n+1)!)2(n!)2\frac{((n+1)!)^2}{(n!)^2}
step 2
Recognize the factorial pattern: (n+1)2⋅(n!)2(n!)2=(n+1)2\frac{(n+1)^2 \cdot (n!)^2}{(n!)^2} = (n+1)^2
Answer
The simplified form is (n+1)2(n+1)^2
Key Concept
Factorial simplification
Explanation
We simplified the factorial expression by canceling out common terms.
# Part (ii)
step 1
Simplify the expression: n!(n−k+1)!\frac{n!}{(n-k+1)!}
step 2
Recognize the factorial pattern: n⋅(n−1)⋅…⋅(n−k+1)n \cdot (n-1) \cdot \ldots \cdot (n-k+1)
Answer
The simplified form is n⋅(n−1)⋅…⋅(n−k+1)n \cdot (n-1) \cdot \ldots \cdot (n-k+1)
Key Concept
Factorial simplification
Explanation
We simplified the factorial expression by expanding the numerator.
Question 5 # Part (i)
step 1
State the proposition: ∑i=2ni(i+1)=n(n+1)(n+2)3\sum_{i=2}^{n} i(i+1) = \frac{n(n+1)(n+2)}{3}
step 2
Base case: For n=2n=2, ∑i=22i(i+1)=2⋅3=6\sum_{i=2}^{2} i(i+1) = 2 \cdot 3 = 6 and 2⋅3⋅43=8\frac{2 \cdot 3 \cdot 4}{3} = 8
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
∑i=2k+1i(i+1)=∑i=2ki(i+1)+(k+1)(k+2)\sum_{i=2}^{k+1} i(i+1) = \sum_{i=2}^{k} i(i+1) + (k+1)(k+2)
step 5
Use inductive hypothesis: k(k+1)(k+2)3+(k+1)(k+2)\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)
step 6
Simplify: (k+1)(k+2)(k+3)3\frac{(k+1)(k+2)(k+3)}{3}
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
Explanation
We used the principle of mathematical induction to prove the proposition.
# Part (ii)
step 1
State the proposition: ∑n=1si(il)−(n+1)!−1\sum_{n=1}^{s} i(i l)-(n+1)!-1
step 2
Base case: For n=1n=1, ∑i=11i(il)−(2)!−1=1⋅1−2−1=−2\sum_{i=1}^{1} i(i l)-(2)!-1 = 1 \cdot 1 - 2 - 1 = -2
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
∑i=1k+1i(il)−(k+2)!−1\sum_{i=1}^{k+1} i(i l)-(k+2)!-1
step 5
Use inductive hypothesis: (k+1)(k+1l)−(k+2)!−1(k+1)(k+1 l)-(k+2)!-1
step 6
Simplify: (k+1)(k+1l)−(k+2)!−1(k+1)(k+1 l)-(k+2)!-1
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
Explanation
We used the principle of mathematical induction to prove the proposition.
# Part (iii)
step 1
State the proposition: ∏i=0n(12i+1⋅12i+2)−1(2n+2)!\prod_{i=0}^{n} \left(\frac{1}{2i+1} \cdot \frac{1}{2i+2}\right) - \frac{1}{(2n+2)!}
step 2
Base case: For n=0n=0, ∏i=00(12⋅0+1⋅12⋅0+2)−1(2⋅0+2)!=11⋅12−12=0\prod_{i=0}^{0} \left(\frac{1}{2 \cdot 0+1} \cdot \frac{1}{2 \cdot 0+2}\right) - \frac{1}{(2 \cdot 0+2)!} = \frac{1}{1} \cdot \frac{1}{2} - \frac{1}{2} = 0
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
∏i=0k+1(12i+1⋅12i+2)−1(2k+4)!\prod_{i=0}^{k+1} \left(\frac{1}{2i+1} \cdot \frac{1}{2i+2}\right) - \frac{1}{(2k+4)!}
step 5
Use inductive hypothesis: ∏i=0k(12i+1⋅12i+2)⋅(12(k+1)+1⋅12(k+1)+2)−1(2k+4)!\prod_{i=0}^{k} \left(\frac{1}{2i+1} \cdot \frac{1}{2i+2}\right) \cdot \left(\frac{1}{2(k+1)+1} \cdot \frac{1}{2(k+1)+2}\right) - \frac{1}{(2k+4)!}
step 6
Simplify: ∏i=0k+1(12i+1⋅12i+2)−1(2k+4)!\prod_{i=0}^{k+1} \left(\frac{1}{2i+1} \cdot \frac{1}{2i+2}\right) - \frac{1}{(2k+4)!}
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
Explanation
We used the principle of mathematical induction to prove the proposition.
# Part (iv)
step 1
State the proposition: n3−7n+3n^3 - 7n + 3 is divisible by 3 for each integer n≥0n \geq 0
step 2
Base case: For n=0n=0, 03−7⋅0+3=30^3 - 7 \cdot 0 + 3 = 3 which is divisible by 3
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
(k+1)3−7(k+1)+3(k+1)^3 - 7(k+1) + 3
step 5
Expand: k3+3k2+3k+1−7k−7+3k^3 + 3k^2 + 3k + 1 - 7k - 7 + 3
step 6
Simplify: k3−7k+3+3k2+3k−7+1k^3 - 7k + 3 + 3k^2 + 3k - 7 + 1
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
Explanation
We used the principle of mathematical induction to prove the proposition.
# Part (v)
step 1
State the proposition: 32n−13^{2n} - 1 is divisible by 8 for each integer n≥1n \geq 1
step 2
Base case: For n=1n=1, 32⋅1−1=9−1=83^{2 \cdot 1} - 1 = 9 - 1 = 8 which is divisible by 8
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
32(k+1)−13^{2(k+1)} - 1
step 5
Expand: 32k+2−13^{2k+2} - 1
step 6
Simplify: 32k⋅32−1=9⋅32k−13^{2k} \cdot 3^2 - 1 = 9 \cdot 3^{2k} - 1
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
Explanation
We used the principle of mathematical induction to prove the proposition.
# Part (vi)
step 1
State the proposition: 2^x < (n+1)! for all integers n≥2n \geq 2
step 2
Base case: For n=2n=2, 2^2 < (2+1)! = 4 < 6
step 3
Inductive step: Assume true for n=kn=k, prove for n=k+1n=k+1
step 4
2^{k+1} < (k+2)!
step 5
Expand: 2 \cdot 2^k < (k+2) \cdot (k+1)!
step 6
Simplify: 2^k < (k+1)!
Answer
The proposition is proven by induction.
Key Concept
Mathematical induction
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