Q1398: If $A=\begin{bmatrix}1&2&0\\ -2&-1&-2\\ 0&-1&1\end{bmatrix}$...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1398

COMPETENCY BASED

UNDERSTAND

5 Marks 2025 AISSCE(Board Exam) LA

If $A=\begin{bmatrix}1&2&0\\ -2&-1&-2\\ 0&-1&1\end{bmatrix}$, then find $A^{-1}$. Hence, solve the system of linear equations: $x-2y=10$, $2x-y-z=8$, $-2y+z=7$.

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Detailed Solution

Step 1: Find the determinant of A

We first find the determinant of matrix $A$ to check if it is invertible. $$|A| = \begin{vmatrix}1&2&0\\ -2&-1&-2\\ 0&-1&1\end{vmatrix} = 1\begin{vmatrix}-1&-2\\ -1&1\end{vmatrix} - 2\begin{vmatrix}-2&-2\\ 0&1\end{vmatrix} + 0\begin{vmatrix}-2&-1\\ 0&-1\end{vmatrix}$$ $$|A| = 1((-1)(1) - (-2)(-1)) - 2((-2)(1) - (-2)(0)) + 0$$ $$|A| = 1(-1 - 2) - 2(-2 - 0) = 1(-3) - 2(-2) = -3 + 4 = 1$$ 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 $A$: $C_{11} = \begin{vmatrix}-1&-2\\ -1&1\end{vmatrix} = -1 - 2 = -3$ $C_{12} = -\begin{vmatrix}-2&-2\\ 0&1\end{vmatrix} = -(-2 - 0) = 2$ $C_{13} = \begin{vmatrix}-2&-1\\ 0&-1\end{vmatrix} = 2 - 0 = 2$ $C_{21} = -\begin{vmatrix}2&0\\ -1&1\end{vmatrix} = -(2 - 0) = -2$ $C_{22} = \begin{vmatrix}1&0\\ 0&1\end{vmatrix} = 1 - 0 = 1$ $C_{23} = -\begin{vmatrix}1&2\\ 0&-1\end{vmatrix} = -(-1 - 0) = 1$ $C_{31} = \begin{vmatrix}2&0\\ -1&-2\end{vmatrix} = -4 - 0 = -4$ $C_{32} = -\begin{vmatrix}1&0\\ -2&-2\end{vmatrix} = -(-2 - 0) = 2$ $C_{33} = \begin{vmatrix}1&2\\ -2&-1\end{vmatrix} = -1 - (-4) = -1 + 4 = 3$

Step 3: Form the adjugate matrix

The adjugate matrix is the transpose of the cofactor matrix: $$adj(A) = \begin{bmatrix}C_{11}&C_{12}&C_{13}\\ C_{21}&C_{22}&C_{23}\\ C_{31}&C_{32}&C_{33}\end{bmatrix}^T = \begin{bmatrix}-3&2&2\\ -2&1&1\\ -4&2&3\end{bmatrix}^T = \begin{bmatrix}-3&-2&-4\\ 2&1&2\\ 2&1&3\end{bmatrix}$$

Step 4: Find the inverse of A

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

Step 5: Express the system of linear equations in matrix form

The given system of linear equations is: $x - 2y = 10$ $2x - y - z = 8$ $-2y + z = 7$ We can rewrite this system 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 the system using the inverse

To solve the system, we need to find $X = A^{-1}B$. However, the matrix $A$ in the system of equations is different from the matrix $A$ given in the question. Let's call the matrix in the system of equations $A_{sys}$. $A_{sys} = \begin{bmatrix}1& -2& 0\\ 2& -1& -1\\ 0& -2& 1\end{bmatrix}$ Notice that $A_{sys}^T = \begin{bmatrix}1& 2& 0\\ -2& -1& -2\\ 0& -1& 1\end{bmatrix} = A$ Therefore, $A_{sys} = A^T$. We know that $(A^T)^{-1} = (A^{-1})^T$. So, we need to find the transpose of $A^{-1}$: $(A^{-1})^T = \begin{bmatrix}-3&2&2\\ -2&1&1\\ -4&2&3\end{bmatrix}$ Now, we can find $X = (A^{-1})^T B$: $X = \begin{bmatrix}-3&2&2\\ -2&1&1\\ -4&2&3\end{bmatrix} \begin{bmatrix}10\\ 8\\ 7\end{bmatrix} = \begin{bmatrix}-3(10) + 2(8) + 2(7)\\ -2(10) + 1(8) + 1(7)\\ -4(10) + 2(8) + 3(7)\end{bmatrix} = \begin{bmatrix}-30 + 16 + 14\\ -20 + 8 + 7\\ -40 + 16 + 21\end{bmatrix} = \begin{bmatrix}0\\ -5\\ -3\end{bmatrix}$ Thus, $x = 0$, $y = -5$, 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 inverse and its application in solving a system of linear equations. They must also understand how to calculate the inverse of a matrix and apply it to solve the given system.

Knowledge Dimension: PROCEDURAL

Justification: The question requires the student to apply a series of well-defined steps to find the inverse of a matrix and then use it to solve a system of linear equations. This involves knowing the procedures for calculating determinants, cofactors, adjugate, 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 concepts of matrices and determinants to solve a real-world problem, which is a system of linear equations. This requires more than just rote memorization of formulas; it requires understanding and application of the concepts.