Class 12 Matrices · Case Study4 Marksmatrix_applications
Class 12 Maths · Matrices · Case Study #31
Amit, Biraj and Chirag create the square matrices A=\(\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\), B=\(\begin{bmatrix}4 & 0\\1 & 5\end{bmatrix}\) and C=\(\begin{bmatrix}2 & 0\\1 & -2\end{bmatrix}\) respectively.
Part (1) 1 Marks
Find A+(B+C).
Answer: \(\begin{bmatrix}7 & 2\\1 & 6\end{bmatrix}\)
- First B+C=[[6,0],[2,3]]. Adding A entry by entry gives [[7,2],[1,6]].
- Therefore A+(B+C)=\(\begin{bmatrix}7 & 2\\1 & 6\end{bmatrix}\).
Part (2) 1 Marks
Evaluate (Aᵀ)ᵀ.
Answer: \(\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\)
- Aᵀ=[[1,−1],[2,3]]. Transposing once more returns [[1,2],[−1,3]].
- Thus (Aᵀ)ᵀ=A=\(\begin{bmatrix}1 & 2\\-1 & 3\end{bmatrix}\).
Part (3) 2 Marks
Find AC−BC.
Answer: \(\begin{bmatrix}-4 & -4\\-6 & 4\end{bmatrix}\)
- Compute AC by row-column multiplication: [[4,−4],[1,−6]].
- Similarly BC=[[8,0],[7,−10]]; subtract to get [[4−8,−4−0],[1−7,−6−(−10)]].
- Therefore AC−BC=\(\begin{bmatrix}-4 & -4\\-6 & 4\end{bmatrix}\).
Part (4) 2 Marks
Alternatively, find (a+b)B for a=4,b=−2.
Answer: \(\begin{bmatrix}8 & 0\\2 & 10\end{bmatrix}\)
- The scalar sum is a+b=4−2=2.
- Multiply each entry of B by 2: 2[[4,0],[1,5]]=[[8,0],[2,10]].
- Thus (a+b)B=\(\begin{bmatrix}8 & 0\\2 & 10\end{bmatrix}\).