Q1349: If $A=[\begin{bmatrix}1&2&-3\\ 2&0&-3\\ 1&2&0\end{bmatrix}],...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1349

COMPETENCY BASED

REMEMBER

5 Marks 2024 AISSCE(Board Exam) LA

If $A=[\begin{bmatrix}1&2&-3\\ 2&0&-3\\ 1&2&0\end{bmatrix}],$ then find $A^{-1}$ and hence solve the following system of equations: $x+2y-3z=1$, $2x-3z=2$, $x+2y=3$

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Detailed Solution

Step 1: Find the determinant of A

First, we need to find the determinant of matrix $A$. $$|A| = \begin{vmatrix}1&2&-3\\ 2&0&-3\\ 1&2&0\end{vmatrix} = 1(0 - (-6)) - 2(0 - (-3)) + (-3)(4 - 0) = 6 - 6 - 12 = -12$$ Since $|A| \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Find the cofactors of A

Now, we find the cofactors of each element of matrix $A$: $C_{11} = (0 - (-6)) = 6$ $C_{12} = -(0 - (-3)) = -3$ $C_{13} = (4 - 0) = 4$ $C_{21} = -(0 - (-6)) = -6$ $C_{22} = (0 - (-3)) = 3$ $C_{23} = -(2 - (-3)) = -5$ $C_{31} = (-6 - 0) = -6$ $C_{32} = -(-3 - (-6)) = -3$ $C_{33} = (0 - 4) = -4$

Step 3: Form the cofactor matrix and find the adjoint of A

The cofactor matrix is: $$C = \begin{bmatrix}6&-3&4\\ -6&3&-5\\ -6&-3&-4\end{bmatrix}$$ The adjoint of $A$ is the transpose of the cofactor matrix: $$adj(A) = C^T = \begin{bmatrix}6&-6&-6\\ -3&3&-3\\ 4&-5&-4\end{bmatrix}$$

Step 4: Calculate the inverse of A

The inverse of $A$ is given by: $$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-12} \begin{bmatrix}6&-6&-6\\ -3&3&-3\\ 4&-5&-4\end{bmatrix} = \begin{bmatrix}-1/2&1/2&1/2\\ 1/4&-1/4&1/4\\ -1/3&5/12&1/3\end{bmatrix}$$

Step 5: Express the system of equations in matrix form

The given system of equations can be written in matrix form as $AX = B$, where: $$A = \begin{bmatrix}1&2&-3\\ 2&0&-3\\ 1&2&0\end{bmatrix}, X = \begin{bmatrix}x\\ y\\ z\end{bmatrix}, B = \begin{bmatrix}1\\ 2\\ 3\end{bmatrix}$$

Step 6: Solve for X using the inverse of A

To solve for $X$, we use the formula $X = A^{-1}B$: $$X = \begin{bmatrix}-1/2&1/2&1/2\\ 1/4&-1/4&1/4\\ -1/3&5/12&1/3\end{bmatrix} \begin{bmatrix}1\\ 2\\ 3\end{bmatrix} = \begin{bmatrix}(-1/2) + (1) + (3/2)\\ (1/4) - (1/2) + (3/4)\\ (-1/3) + (5/6) + (1)\end{bmatrix} = \begin{bmatrix}2\\ 1/2\\ 5/6\end{bmatrix}$$ Therefore, $x = 2$, $y = 1/2$, and $z = 5/6$.

Final Answer: x = 2, y = 1/2, z = 5/6

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the formula for finding the inverse of a matrix and apply it to solve the system of equations. The question requires recalling the steps involved in finding the inverse and then using it to solve the equations.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to follow a specific procedure to find the inverse of a matrix and then use that inverse to solve a system of linear equations. This involves a series of steps and algorithms.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concept of matrices and determinants to solve a system of linear equations, which is a practical application of the topic.