I If A -1-202 130-21- find A-1- Using A-1- solve the system of linear equations x-2 y-10-2 x-y-3 z-8-2 y-z-7ii A -3-422 35101- find A-1 and hence solve the following system of equations-3 x-4 y-2 z-1-2 x-3 y-5 z-7- x-z-2iii A -1-202130-21 and B -72-6-21-3-42 5- find A B- Hence- solve the system of equations-x-2 y-10-2 x-y-3 z-8 and -2 y-z-7iv If A -120-2-1-20-11- find A-1- Using A-1- solve the system of linear equations x-2 y-10-2 x-y-z-8-2 y-z-7v Given A -22-4-42-42-15- B -1-10234012- find BA and use this to solve the system of equations y-2 z-7- x-y-3-2 x-3 y-4 z-17

(i) #160;Here, #160;A=1 202130 21 #160;A=1 #160;1+6+22 0+0 4 0 #160; #160; #160; #160; #160;=7+4+0 #160; #160; #160; #160; #160;=11 ...

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Byju's Answer

Standard XII

Mathematics

Solving a system of linear equation in two variables

i If A =1-202...

Question

(i) If $A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}]$ , find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii) $A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]$ , find A⁻¹ and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii) $A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]$ , find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If $A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]$ , find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given $A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]$ , find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

Open in App

Solution

$(i) Here, A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] |A| =1 (1 + 6) + 2 (2 - 0) + 0 (- 4 - 0) = 7 + 4 + 0 = 11 {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} |\begin{matrix} 1 & 3 \\ - 2 & 1 \end{matrix}| = 7, C_{12} = {(- 1)}^{1 + 2} |\begin{matrix} 2 & 3 \\ 0 & 1 \end{matrix}| = - 2, C_{13} = {(- 1)}^{1 + 3} |\begin{matrix} 2 & 1 \\ 0 & - 2 \end{matrix}| = - 4 C_{21} = {(- 1)}^{2 + 1} |\begin{matrix} - 2 & 0 \\ - 2 & 1 \end{matrix}| = 2, C_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}| = 1, C_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 1 & - 2 \\ 0 & - 2 \end{matrix}| = 2 C_{31} = {(- 1)}^{3 + 1} |\begin{matrix} - 2 & 0 \\ 1 & 3 \end{matrix}| = - 6, C_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 1 & 0 \\ 2 & 3 \end{matrix}| = - 3, C_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix}| = 5 ∴ adj A = {[\begin{matrix} 7 & - 2 & - 4 \\ 2 & 1 & 2 \\ - 6 & - 3 & 5 \end{matrix}]}^{T} = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] \Rightarrow A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] or, A X = B where, A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}] Now, ∴ X = A^{- 1} B \Rightarrow X = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}] \Rightarrow X = \frac{1}{11} [\begin{matrix} 70 + 16 - 42 \\ - 20 + 8 - 21 \\ - 40 + 16 + 35 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{11} [\begin{matrix} 44 \\ - 33 \\ 11 \end{matrix}] ∴ x = 4, y = - 3 and z = 1$

$(ii) Here, A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}] |A| =3 (3 - 0) + 4 (2 - 5) + 2 (0 - 3) = 9 - 12 - 6 = - 9 {Let C}_{i j} {be the cofactors of the elements a}_{i j} in A= [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} |\begin{matrix} 3 & 5 \\ 0 & 1 \end{matrix}| = 3, C_{12} = {(- 1)}^{1 + 2} |\begin{matrix} 2 & 5 \\ 1 & 1 \end{matrix}| = 3, C_{13} = {(- 1)}^{1 + 3} |\begin{matrix} 2 & 3 \\ 1 & 0 \end{matrix}| = - 3 C_{21} = {(- 1)}^{2 + 1} |\begin{matrix} - 4 & 2 \\ 0 & 1 \end{matrix}| = 4, C_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}| = 1, C_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 3 & - 4 \\ 1 & 0 \end{matrix}| = - 4 C_{31} = {(- 1)}^{3 + 1} |\begin{matrix} - 4 & 2 \\ 3 & 5 \end{matrix}| = - 26, C_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 3 & 2 \\ 2 & 5 \end{matrix}| = - 11, C_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 3 & - 4 \\ 2 & 3 \end{matrix}| = 17 adj A = {[\begin{matrix} 3 & 3 & - 3 \\ 4 & 1 & - 4 \\ - 26 & - 11 & 17 \end{matrix}]}^{T} = [\begin{matrix} 3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17 \end{matrix}] \Rightarrow A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{- 9} [\begin{matrix} 3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17 \end{matrix}] A X = B Here, A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} - 1 \\ 7 \\ 2 \end{matrix}] X = A^{- 1} B \Rightarrow X = \frac{1}{- 9} [\begin{matrix} 3 & 4 & - 26 \\ 3 & 1 & - 11 \\ - 3 & - 4 & 17 \end{matrix}] [\begin{matrix} - 1 \\ 7 \\ 2 \end{matrix}] \Rightarrow X = \frac{1}{- 9} [\begin{matrix} - 3 + 28 - 52 \\ - 3 + 7 - 22 \\ 3 - 28 + 34 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{- 9} [\begin{matrix} - 27 \\ - 18 \\ 9 \end{matrix}] ∴ x = 3, y = 2 and z = - 1$

$(iii) Here, A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] A B = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] \Rightarrow A B = [\begin{matrix} 7 + 4 + 0 & 2 - 2 + 0 & - 6 + 6 + 0 \\ 14 - 2 - 12 & 4 + 1 + 6 & - 12 - 3 + 15 \\ 0 + 4 - 4 & 0 - 2 + 2 & 0 + 6 + 5 \end{matrix}] = [\begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix}] \Rightarrow A B = 11 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A B = 11 I_{3} \Rightarrow \frac{1}{11} A B = I_{3} \Rightarrow (\frac{1}{11} B) A = I_{3} \Rightarrow A^{- 1} = \frac{1}{11} B \Rightarrow A^{- 1} = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] X = A^{- 1} B X = \frac{1}{11} [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}] [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{11} [\begin{matrix} 70 + 16 - 42 \\ - 20 + 9 - 21 \\ - 40 + 16 + 35 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{11} [\begin{matrix} 44 \\ - 33 \\ 11 \end{matrix}] ∴ x = 4, y = - 3 and z = 1$

$(iv) Here, A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}] |A| = 1 (- 1 - 2) + 2 (2) = - 3 + 4 = 1 Let C_{i j} be the cofactors of the elements a_{i j} in A = [a_{i j}] . Then, C_{11} = {(- 1)}^{1 + 1} |\begin{matrix} - 1 & - 2 \\ - 1 & 1 \end{matrix}| = - 3, C_{12} = {(- 1)}^{1 + 2} |\begin{matrix} - 2 & - 2 \\ 0 & 1 \end{matrix}| = 2, C_{13} = {(- 1)}^{1 + 3} |\begin{matrix} - 2 & - 1 \\ 0 & - 1 \end{matrix}| = 2 C_{21} = {(- 1)}^{2 + 1} |\begin{matrix} 2 & 0 \\ - 1 & 1 \end{matrix}| = - 2, C_{22} = {(- 1)}^{2 + 2} |\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}| = 1, C_{23} = {(- 1)}^{2 + 3} |\begin{matrix} 1 & 2 \\ 0 & - 1 \end{matrix}| = 1 C_{31} = {(- 1)}^{3 + 1} |\begin{matrix} 2 & 0 \\ - 1 & - 2 \end{matrix}| = - 4, C_{32} = {(- 1)}^{3 + 2} |\begin{matrix} 1 & 0 \\ - 2 & - 2 \end{matrix}| = 2, C_{33} = {(- 1)}^{3 + 3} |\begin{matrix} 1 & 2 \\ - 2 & - 1 \end{matrix}| = 3 ∴ adj A = {[\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}]}^{T} = [\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}] \Rightarrow A^{- 1} = \frac{1}{|A|} adj A = \frac{1}{1} [\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}] = [\begin{matrix} - 3 & - 2 & - 4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{matrix}] We know that, {(A^{T})}^{- 1} = {(A^{- 1})}^{T} . Here, C = A^{T} i . e ., C = [\begin{matrix} 1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1 \end{matrix}] ∴ C^{- 1} = [\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}] or, C X = B where, C = [\begin{matrix} 1 & - 2 & 0 \\ 2 & - 1 & - 1 \\ 0 & - 2 & 1 \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}] and B = [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}] Now, ∴ X = C^{- 1} B \Rightarrow X = [\begin{matrix} - 3 & 2 & 2 \\ - 2 & 1 & 1 \\ - 4 & 2 & 3 \end{matrix}] [\begin{matrix} 10 \\ 8 \\ 7 \end{matrix}] \Rightarrow X = [\begin{matrix} - 30 + 16 + 14 \\ - 20 + 8 + 7 \\ - 40 + 16 + 21 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 0 \\ - 5 \\ - 3 \end{matrix}] ∴ x = 0, y = - 5 and z = - 3 .$

$(v) Here, A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}] and B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}] B A = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}] [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}] \Rightarrow B A = [\begin{matrix} 2 + 4 + 0 & 2 - 2 + 0 & - 4 + 4 + 0 \\ 4 - 12 + 8 & 4 + 6 - 4 & - 8 - 12 + 20 \\ 0 - 4 + 4 & 0 + 2 - 2 & 0 - 4 + 10 \end{matrix}] = [\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix}] \Rightarrow B A = 6 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow B A = 6 I_{3} \Rightarrow B (\frac{1}{6} A) = I_{3} \Rightarrow B^{- 1} = \frac{1}{6} A \Rightarrow B^{- 1} = \frac{1}{6} [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}] Now, B X = C where, B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}], X = [\begin{matrix} x \\ y \\ z \end{matrix}] and C = [\begin{matrix} 3 \\ 17 \\ 7 \end{matrix}] ∴ X = B^{- 1} C \Rightarrow X = \frac{1}{6} [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}] [\begin{matrix} 3 \\ 17 \\ 7 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 6 + 34 - 28 \\ - 12 + 34 - 28 \\ 6 - 17 + 35 \end{matrix}] \Rightarrow [\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{6} [\begin{matrix} 12 \\ - 6 \\ 24 \end{matrix}] ∴ x = 2, y = - 1 and z = 4 .$

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