Problem 182 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]\). Show that \(A(B+C)=A B+A C\).

Step-by-Step Solution

Answer

Therefore, it is shown that \(A(B+C)=AB+AC\), confirming the distributive property of matrix multiplication over addition.

Step 1: Calculate B+C

The first step is to add matrices B and C. We add corresponding elements in the matrices: \(B+C=\left[\begin{array}{ll}2 & 1 \ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \ 7 & 4\end{array}\right]=\left[\begin{array}{ll}2+5 & 1+1 \ 4+7 & 2+4\end{array}\right]=\left[\begin{array}{ll}7 & 2 \ 11 & 6\end{array}\right]\)

Step 2: Calculate A(B+C)

Now, we will multiply matrix \(A\) with \(B+C\): \(A(B+C)=\left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]\left[\begin{array}{ll}7 & 2 \ 11 & 6\end{array}\right]=\left[\begin{array}{ll}1*7+2*11 & 1*2+2*6 \ 3*7+4*11 & 3*2+4*6\end{array}\right]=\left[\begin{array}{ll}29 & 14 \ 71 & 30\end{array}\right]\)

Step 3: Calculate AB and AC

Now, calculate the matrices AB and AC individually: \(AB=\left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]\left[\begin{array}{ll}2 & 1 \ 4 & 2\end{array}\right]=\left[\begin{array}{ll}1*2+2*4 & 1*1+2*2 \ 3*2+4*4 & 3*1+4*2\end{array}\right]=\left[\begin{array}{ll}10 & 5 \ 22 & 11\end{array}\right]\). \(AC=\left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]\left[\begin{array}{ll}5 & 1 \ 7 & 4\end{array}\right]=\left[\begin{array}{ll}1*5+2*7 & 1*1+2*4 \ 3*5+4*7 & 3*1+4*4\end{array}\right]=\left[\begin{array}{ll}19 & 9 \ 43 & 19\end{array}\right]\)

Step 4: Calculate AB + AC

The final step is to add the two matrices AB and AC: \(AB+AC=\left[\begin{array}{ll}10 & 5 \ 22 & 11\end{array}\right]+\left[\begin{array}{ll}19 & 9 \ 43 & 19\end{array}\right]=\left[\begin{array}{ll}10+19 & 5+9 \ 22+43 & 11+19\end{array}\right]=\left[\begin{array}{ll}29 & 14 \ 71 & 30\end{array}\right]\)

Key Concepts

Matrix Addition

Matrix addition is a straightforward process that involves adding two matrices of the same dimensions by summing their corresponding elements. Consider two matrices, \(B\) and \(C\), each of size 2x2:

• \(B = \left[\begin{array}{ll}2 & 1 \ 4 & 2\end{array}\right]\)
• \(C = \left[\begin{array}{ll}5 & 1 \ 7 & 4\end{array}\right]\)

To add \(B\) and \(C\), you simply add each corresponding element:\[B+C=\left[\begin{array}{ll}2 & 1 \ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \ 7 & 4\end{array}\right]=\left[\begin{array}{ll}2+5 & 1+1 \ 4+7 & 2+4\end{array}\right]=\left[\begin{array}{ll}7 & 2 \ 11 & 6\end{array}\right]\]This operation combines each element of the matrices in a similar way to how you add ordinary numbers. It's crucial that both matrices are of the same size to perform matrix addition.

Matrix Multiplication

Matrix multiplication is a bit more complex than matrix addition and requires the number of columns in the first matrix to match the number of rows in the second matrix. If \(A\) is a 2x2 matrix like:\[A = \left[\begin{array}{ll}1 & 2 \ 3 & 4\end{array}\right]\]It can be multiplied with another 2x2 matrix like \(B+C\):\[B+C = \left[\begin{array}{ll}7 & 2 \ 11 & 6\end{array}\right]\]To find \(A(B+C)\), we multiply and sum across the rows of \(A\) and the columns of \(B+C\):

• First row, first column: \(1 \times 7 + 2 \times 11 = 29\)
• First row, second column: \(1 \times 2 + 2 \times 6 = 14\)
• Second row, first column: \(3 \times 7 + 4 \times 11 = 71\)
• Second row, second column: \(3 \times 2 + 4 \times 6 = 30\)

So, the product \(A(B+C)\) results in:\[\left[\begin{array}{ll}29 & 14 \ 71 & 30\end{array}\right]\]This systematic approach of multiplying rows by columns is essential in matrix algebra.

Distributive Property of Matrices

The distributive property is a fundamental concept in mathematics and applies to matrices as well. It states that for matrices \(A\), \(B\), and \(C\) of appropriate dimensions:\[A(B+C) = AB + AC\]To demonstrate this, we need to compute each product separately and then show they are equal:

• \(AB = \left[\begin{array}{ll}10 & 5 \ 22 & 11\end{array}\right]\)
• \(AC = \left[\begin{array}{ll}19 & 9 \ 43 & 19\end{array}\right]\)

Adding \(AB\) and \(AC\) gives:\[AB + AC = \left[\begin{array}{ll}10+19 & 5+9 \ 22+43 & 11+19\end{array}\right] = \left[\begin{array}{ll}29 & 14 \ 71 & 30\end{array}\right]\]This result matches \(A(B+C)\). This validates that the distributive property holds for matrices, showing that operations involving matrices are consistent with ordinary numeric algebra.