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  • Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 17 sub question (ii) math

Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 17 sub question (ii) math 

Answers (1)

Answer

: Hence, verify the distribution of matrix multiplication over matrix addition A(B+C)=A B +A C

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right], B=\left[\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right], C=\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \\

 

Consider,  

A(B+C) \\ \quad=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left(\left[\begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right]\right) \\\\\\ =\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left(\left[\begin{array}{ll} 0+1 & 1-1 \\ 1+0 & 1+1 \end{array}\right]\right) \\ =\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 1 & 2 \end{array}\right] \\

A(B+C)=\left[\begin{array}{cc} 2(1)+(-1)(1) & 2(0)+(-1)(2) \\ 1(1)+1(1) & 1(0)+1(2) \\ -1(1)+2(1) & -1(0)+2(2) \end{array}\right]

A(B+C)=\left[\begin{array}{cc} 2-1 & 0-2 \\ 1+1 & 0+2 \\ -1+2 & 0+4 \end{array}\right]\\\\ A(B+C)=\left[\begin{array}{cc} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{array}\right]

Now consider

A B+A C=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right] \\\\ \\ =\left[\begin{array}{cc} 2(0)+(-1)(1) & 2(1)+(-1)(1) \\ 1(0)+1(1) & 1(1)+1(1) \\ -1(0)+2(1) & -1(1)+2(1) \end{array}\right]+\left[\begin{array}{cc} 2(1)+(-1)(0) & 2(-1)+(-1)(1) \\ 1(1)+1(0) & 1(-1)+1(1) \\ -1(1)+2(0) & (-1)(-1)+2(1) \end{array}\right]

=\left[\begin{array}{cc} 0-1 & 2-1 \\ 0+1 & 1+1 \\ 0+2 & -1+2 \end{array}\right]+\left[\begin{array}{cc} 2+0 & -2-1 \\ 1+0 & -1+1 \\ -1+0 & 1+2 \end{array}\right] \\\\ A B+A C=\left[\begin{array}{cc} -1 & 1 \\ 1 & 2 \\ 2 & 1 \end{array}\right]+\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ -1 & 3 \end{array}\right]

A B+A C=\left[\begin{array}{cc} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{array}\right] .... (ii)

From equation i & ii

A(B+C)=A B+A C

 

 

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