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If A = [1−12−1], B =[x14−1]\begin{bmatrix}[r] 1 & -1 \ 2 & -1 \end{bmatrix}, \text{ B =} \begin{bmatrix}[r] x & 1 \ 4 & -1 \end{bmatrix} and A2 + B2 = (A + B)2, find the value of x.

State, whether A2 + B2 and (A + B)2 are always equal or not.

Matrices

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Answer

Given,

A2 + B2 = (A + B)2

⇒[1−12−1]2+[x14−1]2=([1−12−1]+[x14−1])2⇒[1−12−1][1−12−1]+[x14−1][x14−1]=([1+x06−2])2⇒[1×1+(−1)×21×−1+(−1)×(−1)2×1+(−1)×22×−1+(−1)×(−1)]+[x×x+1×4x×1+1×(−1)4×x+(−1)×44×1+(−1)×(−1)]=[(1+x)2+0×6(1+x)×0+0×−26(1+x)+(−2)×66×0+(−2)×(−2)]⇒[1−2−1+12−2−2+1]+[x2+4x−14x−44+1]=[(1+x)206+6x−124]⇒[−100−1]+[x2+4x−14x−45]=[(1+x)206x−64]⇒[x2+4−1x−14x−45−1]=[(1+x)206x−64]⇒[x2+3x−14x−44]=[(1+x)206x−64]\Rightarrow \begin{bmatrix}[r] 1 & -1 \ 2 & -1 \end{bmatrix}^2 + \begin{bmatrix}[r] x & 1 \ 4 & -1 \end{bmatrix}^2 = \Big(\begin{bmatrix}[r] 1 & -1 \ 2 & -1 \end{bmatrix} + \begin{bmatrix}[r] x & 1 \ 4 & -1 \end{bmatrix}\Big)^2 \\[1em] \Rightarrow \begin{bmatrix}[r] 1 & -1 \ 2 & -1 \end{bmatrix}\begin{bmatrix}[r] 1 & -1 \ 2 & -1 \end{bmatrix} + \begin{bmatrix}[r] x & 1 \ 4 & -1 \end{bmatrix}\begin{bmatrix}[r] x & 1 \ 4 & -1 \end{bmatrix} \\[1em] = \Big(\begin{bmatrix}[r] 1 + x & 0 \ 6 & -2 \end{bmatrix}\Big)^2 \\[1em] \Rightarrow \begin{bmatrix}[r] 1 \times 1 + (-1) \times 2 & 1 \times -1 + (-1) \times(-1) \ 2 \times 1 + (-1) \times 2 & 2 \times -1 + (-1) \times (-1) \end{bmatrix} + \begin{bmatrix}[r] x \times x + 1 \times 4 & x \times 1 + 1 \times(-1) \ 4 \times x + (-1) \times 4 & 4 \times 1 + (-1) \times (-1) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] (1 + x)^2 + 0 \times 6 & (1 + x) \times 0 + 0 \times -2 \ 6(1 + x) + (-2) \times 6 & 6 \times 0 + (-2) \times(-2) \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 1 - 2 & -1 + 1 \ 2 - 2 & -2 + 1 \end{bmatrix} + \begin{bmatrix}[r] x^2 + 4 & x - 1 \ 4x - 4 & 4 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] (1 + x)^2 & 0 \ 6 + 6x - 12 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -1 & 0 \ 0 & -1 \end{bmatrix} + \begin{bmatrix}[r] x^2 + 4 & x - 1 \ 4x - 4 & 5 \end{bmatrix} = \begin{bmatrix}[r] (1 + x)^2 & 0 \ 6x - 6 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x^2 + 4 - 1 & x - 1 \ 4x - 4 & 5 - 1 \end{bmatrix} = \begin{bmatrix}[r] (1 + x)^2 & 0 \ 6x - 6 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x^2 + 3 & x - 1 \ 4x - 4 & 4 \end{bmatrix} = \begin{bmatrix}[r] (1 + x)^2 & 0 \ 6x - 6 & 4 \end{bmatrix}

⇒ x - 1 = 0

⇒ x = 1.

A2 + B2 and (A + B)2 are not always equal. As,

By formula,

(A + B)2 = A2 + B2 + 2AB.

Hence, x = 1.

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