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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 20

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 20

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Answer: Hence proved,A^{3}=p I+q A+r A^{2}

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}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]  and I is the identity matrix of order 3

So,

I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]

Consider, LHS side

A^{2}=A A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q &r \end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc}0+0+0 & 0+0+0 & 0+1+0 \\ 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2}\end{array}\right] \\\\\\ A^{2}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\

Now, A^3 = A^2 A

A^{3}=\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{ccc} 0+0+p & 0+0+q & 0+0+r \\ 0+0+p r & p+0+q r & 0+q+r^{2} \\ 0+0+p q+p r^{2} & p r+0+q^{2}+q r^{2} & 0+p+q r+q r+r^{3} \end{array}\right]\\\\\\\ A^{3}=\left[\begin{array}{ccc} p & q & r \\ p r & p+q r & q+r^{2} \\ p q+p r^{2} & p r+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right] ....(i)

Now consider RHS  p I+q A+r A^{2}

Then put the values in the equation, we get.

p I+q A+r A^{2}

=p\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]+q\left[\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{array}\right]+r\left[\begin{array}{ccc} 0 & 0 & 1 \\ p & q & r \\ p r & p+q r & q+r^{2} \end{array}\right]\\\\ \\ =\left[\begin{array}{ccc} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{array}\right]+\left[\begin{array}{ccc} 0 & q & 0 \\ 0 & 0 & q \\ q p & q^{2} & q r \end{array}\right]+\left[\begin{array}{ccc} 0 & 0 & r \\ r p & r q & r^{2} \\ p r^{2} & r p+q r^{2} & r q+r^{3} \end{array}\right]\\ \\\\\\ =\left[\begin{array}{ccc} p+0+0 & 0+q+0 & 0+0+r \\ 0+0+r p & p+0+q r & 0+q+r^{2} \\ 0+p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\ \\\\ =\left[\begin{array}{ccc} p & q & r \\ r p & p+q r & q+r^{2} \\ p q+p r^{2} & r p+q^{2}+q r^{2} & p+2 q r+r^{3} \end{array}\right]\\-----ii

From equation i & equation ii

A^{3}=p I+q A+r A^{2}

Hence, proved.

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