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Bisectors of Angle between Two Lines

Solve the fol...

Question

Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
4x + 6y + 3 = 0

(ii) 5x + 2y = 3
3x + 2y = 5

(iii) 3x + 4y − 5 = 0
x − y + 3 = 0

(iv) 3x + y = 19
3x − y = 23

(v) 3x + 7y = 4
x + 2y = −1

(vi) 3x + y = 23
5x + 3y = 12

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Solution

$(i) The given system of equations can be written in matrix form as folllows: [\begin{matrix} 5 & 7 \\ 4 & 6 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - 2 \\ - 3 \end{matrix}] AX=B Here, A = [\begin{matrix} 5 & 7 \\ 4 & 6 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} - 2 \\ - 3 \end{matrix}] Now, |A| = [\begin{matrix} 5 & 7 \\ 4 & 6 \end{matrix}] = 30 - 28 = 2 \neq 0 The given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A = [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (6) = 6, C_{12} = {(- 1)}^{1 + 2} (4) = - 4 C_{21} = {(- 1)}^{2 + 1} (7) = - 7, C_{22} = {(- 1)}^{2 + 2} (5) = 5 adj A = {[\begin{matrix} 6 & - 4 \\ - 7 & 5 \end{matrix}]}^{T} = [\begin{matrix} 6 & - 7 \\ - 4 & 5 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A \Rightarrow A^{- 1} = \frac{1}{2} [\begin{matrix} 6 & - 7 \\ - 4 & 5 \end{matrix}] X = A^{- 1} B = \frac{1}{2} [\begin{matrix} 6 & - 7 \\ - 4 & 5 \end{matrix}] [\begin{matrix} - 2 \\ - 3 \end{matrix}] = \frac{1}{2} [\begin{matrix} - 12 + 21 \\ 8 - 15 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{9}{2} \\ \frac{- 7}{2} \end{matrix}] ∴ x = \frac{9}{2} and y = \frac{- 7}{2}$

$(ii) The given system of equations can be written in matrix form as follows: [\begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 3 \\ 5 \end{matrix}] AX = B Here, A = [\begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 3 \\ 5 \end{matrix}] Now, |A| = [\begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix}] = 10 - 6 = 4 \neq 0 The given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (2) = 2, C_{12} = {(- 1)}^{1 + 2} (3) = - 3 C_{21} = {(- 1)}^{2 + 1} (2) = - 2, C_{22} = {(- 1)}^{2 + 2} (5) = 5 ∴ adj A = {[\begin{matrix} 2 & - 3 \\ - 2 & 5 \end{matrix}]}^{T} = [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] X = A^{- 1} B = \frac{1}{4} [\begin{matrix} 2 & - 2 \\ - 3 & 5 \end{matrix}] [\begin{matrix} 3 \\ 5 \end{matrix}] = \frac{1}{4} [\begin{matrix} 6 - 10 \\ - 9 + 25 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{- 4}{4} \\ \frac{16}{4} \end{matrix}] ∴ x = - 1 and y = 4$

$(iii) The given system of equations can be written in matrix form as follows: [\begin{matrix} 3 & 4 \\ 1 & - 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 5 \\ - 3 \end{matrix}] AX =B Here, A = [\begin{matrix} 3 & 4 \\ 1 & - 1 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 5 \\ - 3 \end{matrix}] Now, |A| = [\begin{matrix} 3 & 4 \\ 1 & - 1 \end{matrix}] = - 3 - 4 = - 7 \neq 0 So, the given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (- 1) = - 1, C_{12} = {(- 1)}^{1 + 2} (1) = - 1 C_{21} = {(- 1)}^{2 + 1} (4) = - 4, C_{22} = {(- 1)}^{2 + 2} (3) = 3 adj A = {[\begin{matrix} - 1 & - 1 \\ - 4 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 1 & - 4 \\ - 1 & 3 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{- 7} [\begin{matrix} - 1 & - 4 \\ - 1 & 3 \end{matrix}] X = A^{- 1} B = \frac{1}{- 7} [\begin{matrix} - 1 & - 4 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 5 \\ - 3 \end{matrix}] = \frac{1}{- 7} [\begin{matrix} - 5 + 12 \\ - 5 - 9 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{7}{- 7} \\ \frac{- 14}{- 7} \end{matrix}] ∴ x = - 1 and y = 2$

$(iv) The given system of equations can be written in matrix form as follows: [\begin{matrix} 3 & 1 \\ 3 & - 1 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 19 \\ 23 \end{matrix}] AX=B Here, A = [\begin{matrix} 3 & 1 \\ 3 & - 1 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 19 \\ 23 \end{matrix}] Now, |A| = [\begin{matrix} 3 & 1 \\ 3 & - 1 \end{matrix}] = - 3 - 3 = - 6 \neq 0 So, the given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (- 1) = - 1, C_{12} = {(- 1)}^{1 + 2} (3) = - 3 C_{21} = {(- 1)}^{2 + 1} (1) = - 1, C_{22} = {(- 1)}^{2 + 2} (3) = 3 adj A = {[\begin{matrix} - 1 & - 3 \\ - 1 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 1 & - 1 \\ - 3 & 3 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{- 6} [\begin{matrix} - 1 & - 1 \\ - 3 & 3 \end{matrix}] X = A^{- 1} B = \frac{1}{- 6} [\begin{matrix} - 1 & - 1 \\ - 3 & 3 \end{matrix}] [\begin{matrix} 19 \\ 23 \end{matrix}] = \frac{1}{- 6} [\begin{matrix} - 19 - 23 \\ - 57 + 69 \end{matrix}] = [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{- 42}{- 6} \\ \frac{12}{- 6} \end{matrix}] ∴ x = 7 and y = - 2$

$(v) The given system of equations can be written in matrix form as follows: [\begin{matrix} 3 & 7 \\ 1 & 2 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 4 \\ - 1 \end{matrix}] AX = B Here, A = [\begin{matrix} 3 & 7 \\ 1 & 2 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 4 \\ - 1 \end{matrix}] Now, |A| = [\begin{matrix} 3 & 7 \\ 1 & 2 \end{matrix}] = 6 - 7 = - 1 \neq 0 So, the given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (2) = 2, C_{12} = {(- 1)}^{1 + 2} (1) = - 1 C_{21} = {(- 1)}^{2 + 1} (7) = - 7, C_{22} = {(- 1)}^{2 + 2} (3) = 3 adj A = {[\begin{matrix} 2 & - 1 \\ - 7 & 3 \end{matrix}]}^{T} = [\begin{matrix} 2 & - 7 \\ - 1 & 3 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{- 1} [\begin{matrix} 2 & - 7 \\ - 1 & 3 \end{matrix}] X = A^{- 1} B = \frac{1}{- 1} [\begin{matrix} 2 & - 7 \\ - 1 & 3 \end{matrix}] [\begin{matrix} 4 \\ - 1 \end{matrix}] = \frac{1}{- 1} [\begin{matrix} 8 + 7 \\ - 4 - 3 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{15}{- 1} \\ \frac{- 7}{- 1} \end{matrix}] ∴ x = - 15 and y = 7$

$(vi) The given system of equations can be written in matrix form as follows: [\begin{matrix} 3 & 1 \\ 5 & 3 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 7 \\ 12 \end{matrix}] AX =B H ere, A = [\begin{matrix} 3 & 1 \\ 5 & 3 \end{matrix}], X = [\begin{matrix} x \\ y \end{matrix}] and B = [\begin{matrix} 7 \\ 12 \end{matrix}] Now, |A| = [\begin{matrix} 3 & 1 \\ 5 & 3 \end{matrix}] = 9 - 5 = 4 \neq 0 So, the given system has a unique solution given by X = A^{- 1} B . {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A = [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} (3) = 3, C_{12} = {(- 1)}^{1 + 2} (5) = - 5 C_{21} = {(- 1)}^{2 + 1} (1) = - 1, C_{22} = {(- 1)}^{2 + 2} (3) = 3 adj A = {[\begin{matrix} 3 & - 5 \\ - 1 & 3 \end{matrix}]}^{T} = [\begin{matrix} 3 & - 1 \\ - 5 & 3 \end{matrix}] A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{4} [\begin{matrix} 3 & - 1 \\ - 5 & 3 \end{matrix}] X = A^{- 1} B = \frac{1}{4} [\begin{matrix} 3 & - 1 \\ - 5 & 3 \end{matrix}] [\begin{matrix} 7 \\ 12 \end{matrix}] = \frac{1}{4} [\begin{matrix} 21 - 12 \\ - 35 + 36 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} \frac{9}{4} \\ \frac{1}{4} \end{matrix}] ∴ x = \frac{9}{4} and y = \frac{1}{4}$

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Similar questions

Q. Evaluate:
(i)

(4 x + 2 y)^{2}

(5 x - y)^{2}

(3 x + 7 y) (3 x - 7 y)

Q. Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)

\frac{2}{x} - \frac{3}{y} + \frac{3}{z} = 10 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 10 \frac{3}{x} - \frac{1}{y} + \frac{2}{z} = 13

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii)

\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1, \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2; x, y, z \neq 0

(xiv) x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12

Q. By equating coefficients of variables, solve the following equations.
(i)

3 x - 4 y = 7; 5 x + 2 y = 3

5 x + 7 y = 17; 3 x - 2 y = 4

x - 2 y = - 10; 3 x - 5 y = - 12

4 x + y = 34; x + 4 y = 16

(i) (2 y + 3) (y - 2) - (5 y + 3) (y - 2)

(i i) (3 x - 4 y) (x + y) - (5 x + 7 y) (x - y)

[3 x - 2 5 x + 6 y - (2 z - y)] + 3 (z - 2 y)

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