Q1304: If $A=[\begin{bmatrix}1&-2&0\\ 2&-1&-1\\ 0&-2&1\end{bmatrix}...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1304

COMPETENCY BASED

UNDERSTAND

5 Marks 2024 AISSCE(Board Exam) LA

If $A=[\begin{bmatrix}1&-2&0\\ 2&-1&-1\\ 0&-2&1\end{bmatrix}],$ find $A^{-1}$ and use it to solve the following system of equations: $x-2y=10$, $2x-y-z=8$, $-2y+z=7$

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Detailed Solution

Step 1: Find the determinant of A

First, we need to find the determinant of matrix $A$ to check if it is invertible. $$|A| = \begin{vmatrix}1&-2&0\\ 2&-1&-1\\ 0&-2&1\end{vmatrix} = 1\begin{vmatrix}-1&-1\\ -2&1\end{vmatrix} - (-2)\begin{vmatrix}2&-1\\ 0&1\end{vmatrix} + 0\begin{vmatrix}2&-1\\ 0&-2\end{vmatrix}$$ $$|A| = 1((-1)(1) - (-1)(-2)) + 2((2)(1) - (-1)(0)) + 0$$ $$|A| = 1(-1 - 2) + 2(2 - 0) = 1(-3) + 2(2) = -3 + 4 = 1$$ Since $|A| = 1 \neq 0$, the matrix $A$ is invertible.

Step 2: Find the cofactors of A

Now, we find the cofactors of each element of matrix $A$: $C_{11} = \begin{vmatrix}-1&-1\\ -2&1\end{vmatrix} = -1 - 2 = -3$ $C_{12} = -\begin{vmatrix}2&-1\\ 0&1\end{vmatrix} = -(2 - 0) = -2$ $C_{13} = \begin{vmatrix}2&-1\\ 0&-2\end{vmatrix} = -4 - 0 = -4$ $C_{21} = -\begin{vmatrix}-2&0\\ -2&1\end{vmatrix} = -(-2 - 0) = 2$ $C_{22} = \begin{vmatrix}1&0\\ 0&1\end{vmatrix} = 1 - 0 = 1$ $C_{23} = -\begin{vmatrix}1&-2\\ 0&-2\end{vmatrix} = -(-2 - 0) = 2$ $C_{31} = \begin{vmatrix}-2&0\\ -1&-1\end{vmatrix} = 2 - 0 = 2$ $C_{32} = -\begin{vmatrix}1&0\\ 2&-1\end{vmatrix} = -(-1 - 0) = 1$ $C_{33} = \begin{vmatrix}1&-2\\ 2&-1\end{vmatrix} = -1 - (-4) = 3$

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

The cofactor matrix is: $C = \begin{bmatrix}-3&-2&-4\\ 2&1&2\\ 2&1&3\end{bmatrix}$ The adjoint of $A$ is the transpose of the cofactor matrix: $adj(A) = C^T = \begin{bmatrix}-3&2&2\\ -2&1&1\\ -4&2&3\end{bmatrix}$

Step 4: Find the inverse of A

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

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: $A = \begin{bmatrix}1&-2&0\\ 2&-1&-1\\ 0&-2&1\end{bmatrix}$, $X = \begin{bmatrix}x\\ y\\ z\end{bmatrix}$, and $B = \begin{bmatrix}10\\ 8\\ 7\end{bmatrix}$

Step 6: Solve for X using the inverse of A

To solve for $X$, we use the formula $X = A^{-1}B$: $X = \begin{bmatrix}-3&2&2\\ -2&1&1\\ -4&2&3\end{bmatrix} \begin{bmatrix}10\\ 8\\ 7\end{bmatrix} = \begin{bmatrix}(-3)(10) + (2)(8) + (2)(7)\\ (-2)(10) + (1)(8) + (1)(7)\\ (-4)(10) + (2)(8) + (3)(7)\end{bmatrix} = \begin{bmatrix}-30 + 16 + 14\\ -20 + 8 + 7\\ -40 + 16 + 21\end{bmatrix} = \begin{bmatrix}0\\ -5\\ -3\end{bmatrix}$ Therefore, $x = 0$, $y = -5$, and $z = -3$.

Final Answer: x = 0, y = -5, z = -3

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Pedagogical Audit

Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concept of matrix inverses and how to apply them to solve a system of linear 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 equations. This involves knowing the steps for calculating determinants, cofactors, adjoints, and matrix multiplication.

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 matrix inverses to solve a real-world problem (system of linear equations), demonstrating a higher-order thinking skill beyond simple recall.