Q1330: If $A=[\begin{bmatrix}2&1&-3\\ 3&2&1\\ 1&2&-1\end{bmatrix}],...

Class CBSE Class 12 Mathematics Matrices and Determinants Q #1330

COMPETENCY BASED

APPLY

5 Marks 2024 AISSCE(Board Exam) LA

If $A=[\begin{bmatrix}2&1&-3\\ 3&2&1\\ 1&2&-1\end{bmatrix}],$ find $A^{-1}$ and hence solve the following system of equations: $2x+y-3z=13$, $3x+2y+z=4$, $x+2y-z=8$

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AI Tutor Explanation

Detailed Solution

Step 1: Find the determinant of A

First, we need to find the determinant of matrix $A$. $$|A| = \begin{vmatrix}2&1&-3\\ 3&2&1\\ 1&2&-1\end{vmatrix} = 2(2(-1) - 1(2)) - 1(3(-1) - 1(1)) + (-3)(3(2) - 2(1))$$ $$|A| = 2(-2 - 2) - 1(-3 - 1) - 3(6 - 2) = 2(-4) - 1(-4) - 3(4) = -8 + 4 - 12 = -16$$ Since $|A| = -16 \neq 0$, the inverse $A^{-1}$ exists.

Step 2: Find the cofactors of A

Now, we find the cofactors of each element of matrix $A$: $C_{11} = \begin{vmatrix}2&1\\ 2&-1\end{vmatrix} = -2 - 2 = -4$ $C_{12} = -\begin{vmatrix}3&1\\ 1&-1\end{vmatrix} = -(-3 - 1) = 4$ $C_{13} = \begin{vmatrix}3&2\\ 1&2\end{vmatrix} = 6 - 2 = 4$ $C_{21} = -\begin{vmatrix}1&-3\\ 2&-1\end{vmatrix} = -(-1 - (-6)) = -(5) = -5$ $C_{22} = \begin{vmatrix}2&-3\\ 1&-1\end{vmatrix} = -2 - (-3) = 1$ $C_{23} = -\begin{vmatrix}2&1\\ 1&2\end{vmatrix} = -(4 - 1) = -3$ $C_{31} = \begin{vmatrix}1&-3\\ 2&1\end{vmatrix} = 1 - (-6) = 7$ $C_{32} = -\begin{vmatrix}2&-3\\ 3&1\end{vmatrix} = -(2 - (-9)) = -11$ $C_{33} = \begin{vmatrix}2&1\\ 3&2\end{vmatrix} = 4 - 3 = 1$

Step 3: Form the adjugate of A

The adjugate of $A$ is the transpose of the cofactor matrix: $$adj(A) = \begin{bmatrix} -4 & 4 & 4 \\ -5 & 1 & -3 \\ 7 & -11 & 1 \end{bmatrix}^T = \begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix}$$

Step 4: Calculate the inverse of A

The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} adj(A)$: $$A^{-1} = \frac{1}{-16} \begin{bmatrix} -4 & -5 & 7 \\ 4 & 1 & -11 \\ 4 & -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & \frac{5}{16} & -\frac{7}{16} \\ -\frac{1}{4} & -\frac{1}{16} & \frac{11}{16} \\ -\frac{1}{4} & \frac{3}{16} & -\frac{1}{16} \end{bmatrix}$$

Step 5: Solve the system of equations

The system of equations can be written in matrix form as $AX = B$, where $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix}$. To solve for $X$, we use $X = A^{-1}B$: $$X = \begin{bmatrix} \frac{1}{4} & \frac{5}{16} & -\frac{7}{16} \\ -\frac{1}{4} & -\frac{1}{16} & \frac{11}{16} \\ -\frac{1}{4} & \frac{3}{16} & -\frac{1}{16} \end{bmatrix} \begin{bmatrix} 13 \\ 4 \\ 8 \end{bmatrix} = \begin{bmatrix} \frac{13}{4} + \frac{20}{16} - \frac{56}{16} \\ -\frac{13}{4} - \frac{4}{16} + \frac{88}{16} \\ -\frac{13}{4} + \frac{12}{16} - \frac{8}{16} \end{bmatrix} = \begin{bmatrix} \frac{52 + 20 - 56}{16} \\ \frac{-52 - 4 + 88}{16} \\ \frac{-52 + 12 - 8}{16} \end{bmatrix} = \begin{bmatrix} \frac{16}{16} \\ \frac{32}{16} \\ \frac{-48}{16} \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}$$ Thus, $x = 1$, $y = 2$, and $z = -3$.

Final Answer: x = 1, y = 2, z = -3

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student needs to apply the knowledge of finding the inverse of a matrix and using it 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 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.