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If \left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{A} , find A

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We are given the following matrix equation,

\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{A}

We need to determine the value of A.

Take L.H.S:  \left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]

\begin{aligned} &\begin{array}{l} \text { Let us solve } \\ \left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]=\mathrm{XY}(\text { say }) \\ , \text { where } \end{array}\\ &X=\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\\ &Y=\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right]\\ &\text { Then, }\\ &X Y=\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] \end{aligned}

Order of X = 1 × 3

Order of Y = 3 × 3

Then, the order of matrix Z (say) = 1 × 3 [Let Z = XY]

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

\\(2, 1, 3)(-1, -1, 0) = (2 $ \times $ -1) + (1 $ \times $ -1) + (3 $ \times $ 0) \\$ \Rightarrow $ (2, 1, 3)(-1, -1, 0) = -2 - 1 + 0 \\$ \Rightarrow $ (2, 1, 3)(-1, -1, 0) = -3

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

Multiply 1st row of matrix X by matching members of 2nd column of matrix Y, then finally sum them up.

\\(2, 1, 3)(0, 1, 1) = (2 $ \times $ 0) + (1 $ \times $ 1) + (3 $ \times $ 1) \\$ \Rightarrow $ (2, 1, 3)(0, 1, 1) = 0 + 1 + 3 \\$ \Rightarrow $ (2, 1, 3)(0, 1, 1) = 4

\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] = \begin{bmatrix} -3 & 4& \end{bmatrix}

Multiply 1st row of matrix X by matching members of 3rd column of matrix Y, then finally sum them up.

\\(2, 1, 3)(-1, 0, 1) = (2 $ \times $ -1) + (1 $ \times $ 0) + (3 $ \times $ 1) \\$ \Rightarrow $ (2, 1, 3)(-1, 0, 1) = -2 + 0 + 3 \\$ \Rightarrow $ (2, 1, 3)(-1, 0, 1) = 1

\left[\begin{array}{lll} 2 & 1 & 3 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right] = \begin{bmatrix} -3 & 4&1 \end{bmatrix}

So, we have,

\begin{array}{l} \mathrm{Z}=\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right] \\ \text { Now, multiplying } \mathrm{Z} \text { by }\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right]=\mathrm{Q}(\text { say }) \\ \mathrm{ZQ}=\left[\begin{array}{lll} -3 & 4 & 1 \end{array}\right]\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right] \end{array}

Order of Z = 1 × 3

Order of Q = 3 × 1

Then, order of the resulting matrix = 1 × 1

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

\\(-3, 4, 1)(1, 0, -1) = (-3 $ \times $ 1) + (4 $ \times $ 0) + (1 $ \times $ -1) \\$ \Rightarrow $ (-3, 4, 1)(1, 0, -1) = -3 + 0 - 1 \\$ \Rightarrow $ (-3, 4, 1)(1, 0, -1) = -4

\\ \left[\begin{array}{lll}-3 & 4 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=[-4]$ \\Now, since $\left[\begin{array}{lll}2 & 1 & 3\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 0 \\ -1\end{array}\right]=\mathrm{A}$ \\Thus, $A=[-4]$

 

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