Class CBSE Class 12 Mathematics Matrices and Determinants Q #1450
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
Let $A=\begin{bmatrix}1\\ 4\\ -2\end{bmatrix}$ and $C=\begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$ be two matrices. Then, find the matrix B if $AB=C$.
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Detailed Solution

Step 1: Define the unknown matrix B

Let $B = \begin{bmatrix} x & y & z \end{bmatrix}$. Since $A$ is a $3 \times 1$ matrix and $B$ is a $1 \times 3$ matrix, the product $AB$ will be a $3 \times 3$ matrix.

Step 2: Compute the matrix product AB

We have $$AB = \begin{bmatrix}1\\ 4\\ -2\end{bmatrix} \begin{bmatrix} x & y & z \end{bmatrix} = \begin{bmatrix} x & y & z \\ 4x & 4y & 4z \\ -2x & -2y & -2z \end{bmatrix}$$

Step 3: Set up the equation AB = C

We are given that $AB = C$, where $C = \begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$. Therefore, we have $$\begin{bmatrix} x & y & z \\ 4x & 4y & 4z \\ -2x & -2y & -2z \end{bmatrix} = \begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$$

Step 4: Solve for x, y, and z

By comparing the entries of the matrices, we get the following equations: $x = 3$ $y = 4$ $z = 2$ $4x = 12$ $4y = 16$ $4z = 8$ $-2x = -6$ $-2y = -8$ $-2z = -4$ The first three equations give us $x = 3$, $y = 4$, and $z = 2$. The other equations are consistent with these values.

Step 5: Write the matrix B

Therefore, $B = \begin{bmatrix} 3 & 4 & 2 \end{bmatrix}$.

Final Answer: $\begin{bmatrix} 3 & 4 & 2 \end{bmatrix}$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because the student needs to recall the definition of matrix multiplication and apply it to find the unknown matrix.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply the procedure of matrix multiplication and solving for unknowns in a matrix equation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of matrix multiplication as defined in the textbook.

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