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  • If A =[1 0 -1] , then verify that A^2 + A = A(A + I), where I is 3 × 3 unit matrix.

If A = \begin{bmatrix} 1 &0 &-1 \\2 & 1 & 3\\0 &1 &1 \end{bmatrix}, then verify that A^2 + A = A(A + I), where I is 3 × 3 unit matrix.

 

Answers (1)

We are given the following matrix A, such that

A = \begin{bmatrix} 1 &0 &-1 \\2 & 1 & 3\\0 &1 &1 \end{bmatrix}.

We need to verify  A^2 + A = A(A + I)

Take L.H.S: A\textsuperscript{2} + A.

Solve for A\textsuperscript{2}.

A\textsuperscript{2} = A.A

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

Multiply 1st row of matrix A by matching members of 1st column of matrix A, then finally sum them up.

\\(1, 0, -1)(1, 2, 0) = (1 \times 1) + (0 \times 2) + (-1 \times 0) \\ \Rightarrow (1, 0, -1)(1, 2, 0) = 1 + 0 + 0 \\ \Rightarrow (1, 0, -1)(1, 2, 0) = 1

\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]= \begin{bmatrix} 1 & & \\ & & \\ & & \end{bmatrix}

Similarly, repeat steps to find other elements.

\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right] =\left[\begin{array}{ccc} 1 & (1 \times 0)+(0 \times 1)+(-1 \times 1) & (1 \times-1)+(0 \times 3)+(-1 \times 1) \\ (2 \times 1)+(1 \times 2)+(3 \times 0) & (2 \times 0)+(1 \times 1)+(3 \times 1) & (2 x-1)+(1 \times 3)+(3 \times 1) \\ (0 \times 1)+(1 \times 2)+(1 \times 0) & (0 \times 0)+(1 \times 1)+(1 \times 1) & (0 \times-1)+(1 \times 3)+(1 \times 1) \end{array}\right]

\begin{aligned} &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0+0-1 & -1+0-1 \\ 2+2+0 & 0+1+3 & -2+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right]\\ &\text { Now, add } A^{2} \text { and } A \text { , }\\ &A^{2}+A=\left[\begin{array}{ccc} 1 & -1 & -2 \\ 4 & 4 & 4 \\ 2 & 2 & 4 \end{array}\right]+\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\\ \end{aligned}

\begin{aligned} &\Rightarrow A^{2}+A=\left[\begin{array}{ccc} 1+1 & -1+0 & -2-1 \\ 4+2 & 4+1 & 4+3 \\ 2+0 & 2+1 & 4+1 \end{array}\right]\\ &\Rightarrow A^{2}+A=\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right] \end{aligned}

Take R.H.S: A(A + I)

First, let us solve for (A + I).

\begin{aligned} &A+1=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]+\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &\Rightarrow A+1=\left[\begin{array}{ccc} 1+1 & 0+0 & -1+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1 \end{array}\right]\\ &\Rightarrow A+I=\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right]\\ &\text { Multiply }(\mathrm{A}+1) \text { from } \mathrm{A} \text { . }\\ &A(A+I)=\left[\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & -1 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{array}\right]\\ &\Rightarrow A(A+1) \end{aligned}

\begin{array}{l} =\left[\begin{array}{ccc} (1 \times 2)+(0 \times 2)+(-1 \times 0) & (1 \times 0)+(0 \times 2)+(-1 \times 1) & (1 \times-1)+(0 \times 3)+(-1 \times 2) \\ (2 \times 2)+(1 \times 2)+(3 \times 0) & (2 \times 0)+(1 \times 2)+(3 \times 1) & (2 \times-1)+(1 \times 3)+(3 \times 2) \\ (0 \times 2)+(1 \times 2)+(1 \times 0) & (0 \times 0)+(1 \times 2)+(1 \times 1) & (0 \times-1)+(1 \times 3)+(1 \times 2) \end{array}\right] \\ \Rightarrow A(A+1)=\left[\begin{array}{ccc} 2+0+0 & 0+0-1 & -1+0-2 \\ 4+2+0 & 0+2+3 & -2+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2 \end{array}\right] \\ \Rightarrow A(A+1)=\left[\begin{array}{ccc} 2 & -1 & -3 \\ 6 & 5 & 7 \\ 2 & 3 & 5 \end{array}\right] \end{array}

Since, L.H.S = R.H.S.

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

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