Q1285: Solve the following system of equations, using matrices: $\f...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1285

COMPETENCY BASED

APPLY

5 Marks 2024 AISSCE(Board Exam) LA

Solve the following system of equations, using matrices: $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$ $\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$ , $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$ where x, y, $z\ne0$

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AI Tutor Explanation

Detailed Solution

Step 1: Define new variables

Let $p = \frac{1}{x}$, $q = \frac{1}{y}$, and $r = \frac{1}{z}$. The given system of equations can be rewritten as: $2p + 3q + 10r = 4$ $4p - 6q + 5r = 1$ $6p + 9q - 20r = 2$

Step 2: Express the system in matrix form

The system can be expressed in matrix form as $AX = B$, where: $A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}$, $X = \begin{bmatrix} p \\ q \\ r \end{bmatrix}$, and $B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$

Step 3: Calculate the determinant of matrix A

$\det(A) = 2 \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3 \begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} + 10 \begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix}$ $\det(A) = 2((-6)(-20) - (5)(9)) - 3((4)(-20) - (5)(6)) + 10((4)(9) - (-6)(6))$ $\det(A) = 2(120 - 45) - 3(-80 - 30) + 10(36 + 36)$ $\det(A) = 2(75) - 3(-110) + 10(72)$ $\det(A) = 150 + 330 + 720 = 1200$

Step 4: Calculate the adjoint of matrix A

First, find the matrix of cofactors: $C_{11} = (-6)(-20) - (5)(9) = 120 - 45 = 75$ $C_{12} = -(4(-20) - 5(6)) = -(-80 - 30) = 110$ $C_{13} = (4)(9) - (-6)(6) = 36 + 36 = 72$ $C_{21} = -(3(-20) - 10(9)) = -(-60 - 90) = 150$ $C_{22} = (2)(-20) - (10)(6) = -40 - 60 = -100$ $C_{23} = -(2(9) - 3(6)) = -(18 - 18) = 0$ $C_{31} = (3)(5) - (-6)(10) = 15 + 60 = 75$ $C_{32} = -(2(5) - 10(4)) = -(10 - 40) = 30$ $C_{33} = (2)(-6) - (3)(4) = -12 - 12 = -24$ The matrix of cofactors is: $\begin{bmatrix} 75 & 110 & 72 \\ 150 & -100 & 0 \\ 75 & 30 & -24 \end{bmatrix}$ The adjoint of A is the transpose of the cofactor matrix: $adj(A) = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

Step 5: Calculate the inverse of matrix A

$A^{-1} = \frac{1}{\det(A)} adj(A) = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$

Step 6: Solve for X

$X = A^{-1}B = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$ $X = \frac{1}{1200} \begin{bmatrix} (75)(4) + (150)(1) + (75)(2) \\ (110)(4) + (-100)(1) + (30)(2) \\ (72)(4) + (0)(1) + (-24)(2) \end{bmatrix}$ $X = \frac{1}{1200} \begin{bmatrix} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{bmatrix}$ $X = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac{600}{1200} \\ \frac{400}{1200} \\ \frac{240}{1200} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{5} \end{bmatrix}$ So, $p = \frac{1}{2}$, $q = \frac{1}{3}$, and $r = \frac{1}{5}$.

Step 7: Find x, y, and z

Since $p = \frac{1}{x}$, $q = \frac{1}{y}$, and $r = \frac{1}{z}$, we have: $x = \frac{1}{p} = \frac{1}{\frac{1}{2}} = 2$ $y = \frac{1}{q} = \frac{1}{\frac{1}{3}} = 3$ $z = \frac{1}{r} = \frac{1}{\frac{1}{5}} = 5$

Final Answer: x = 2, y = 3, z = 5

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the knowledge of matrices and determinants to solve a system of linear equations.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to follow a specific procedure (using matrices) to solve the system of equations. This involves knowing the steps for finding the inverse of a matrix and using it to solve the equations.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply matrix algebra to solve a system of linear equations, which is a key competency in the curriculum.