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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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