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  • Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 52

Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 52

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Answer: Hence proved,

A^{2}+A=A(A+I)

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}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]

To prove:

A^{2}+A=A(A+I)

I is identity matrix

Taking LHS

\\\\ A^{2}+A=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}1+0-0 & 0+0-3 & -3+0-3 \\ 2+2+0 & 0+1+3 & -6+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right] \\\\\\=\left[\begin{array}{ccc}1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]

=\left[\begin{array}{ccc}1+1 & -3+0 & -6-3 \\ 4+2 & 4+1 & 0+3 \\ 2+0 & 2+1 & 4+1\end{array}\right]=\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right]

Now, taking RHS

\\\\ A(A+I)=\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left(\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]+\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\right)\\\\\\ =\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1+1 & 0+0 & -3+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1\end{array}\right] =\left[\begin{array}{ccc}1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2\end{array}\right]\\\\\\ =\left[\begin{array}{ccc}2+0+0 & 0+0-3 & -3+0-6 \\ 4+2+0 & 0+2+3 & -6+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2\end{array}\right]\\\\ =\left[\begin{array}{ccc}2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5\end{array}\right]

Therefore, LHS=RHS

Hence proved,A^{2}+A=A(A+I)

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