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  • Find the value of x if [1 x 1] [1 3 2 ] =0

Find the value of x if
\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=0

Answers (1)

The given matrix equation is,

\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=0

We need to determine the value of x.

Let us compute L.H.S: \left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]

\begin{array}{l} \text { Let, } A=\left[\begin{array}{lll} 1 & \text { X } & 1 \end{array}\right] \\ \text { B }=\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right] \text { and } \\ C=\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right] \end{array}

Multiplication of any two matrices is only possible when the number of columns in A is equal to the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

First, let us compute

\text { A. } B=\left[\begin{array}{lll} 1 & \text { x } & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=D(\text { say })

Multiply 1st row of matrix A by matching members of 1st column of matrix B, then sum them up.
\\(1, x, 1).(1, 2, 15) = (1 $ \times $ 1) + (x $ \times $ 2) + (1 $ \times $ 15) \\$ \Rightarrow $ (1, x, 1).(1, 2, 15) = 1 + 2x + 15 \\$ \Rightarrow $ (1, x, 1).(1, 2, 15) = 2x + 16

\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=[(2 x+16)

Multiply 1st row of matrix A by matching members of 2nd column of matrix B, then sum them up.

\\(1, x, 1).(3, 5, 3) = (1 $ \times $ 3) + (x $ \times $ 5) + (1 $ \times $ 3) \\$ \Rightarrow $ (1, x, 1).(3, 5, 3) = 3 + 5x + 3 \\$ \Rightarrow $ (1, x, 1).(3, 5, 3) = 5x + 6

\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]=[(2 x+16) \quad(5 x+6) \quad]

Multiply 1st row of matrix A by matching members of 3rd column of matrix B, then sum them up.

\\(1, x, 1).(2, 1, 2) = (1 $ \times $ 2) + (x $ \times $ 1) + (1 $ \times $ 2) \\$ \Rightarrow $ (1, x, 1).(2, 1, 2) = 2 + x + 2 \\$ \Rightarrow $ (1, x, 1).(2, 1, 2) = x + 4

\left[\begin{array}{lll}1 & x & 1\end{array}\right]\left[\begin{array}{ccc}1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2\end{array}\right]=[(2 x+16) \quad(5 x+6) \quad(x+4)]$

So,

D=[(2 x+16) \quad(5 x+6) \quad(x+4)]

Now compute

\text { D. } C=[(2 x+16) \quad(5 x+6) \quad(x+4)]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right] $$

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

\\(2x + 16, 5x + 6, x + 4).(1, 2, x) = ((2x + 16) $ \times $ 1) + ((5x + 6) $ \times $ 2) + ((x + 4) $ \times $ x) \\$ \Rightarrow $ (2x + 16, 5x + 6, x + 4).(1, 2, x) = (2x + 16) + (10x + 12) + (x\textsuperscript{2} + 4x) \\$ \Rightarrow $ (2x + 16, 5x + 6, x + 4).(1, 2, x) = x\textsuperscript{2} + 2x + 10x + 4x + 16 + 12 \\$ \Rightarrow $ (2x + 16, 5x + 6, x + 4).(1, 2, x) = x\textsuperscript{2} + 16x + 28

\begin{aligned} &[(2 x+16) \quad(5 x+6) \quad(x+4)]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=\left[x^{2}+16 x+28\right]\\ &\text { So, we get, }\\ &\left[\begin{array}{lll} 1 & x & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ x \end{array}\right]=\left[x^{2}+16 x+28\right] \end{aligned}

Now, put L.H.S = R.H.S

[x\textsuperscript{2} + 16x + 28] = [0]

This means,

 \\x\textsuperscript{2} + 16x + 28 = 0 \\$ \Rightarrow $ x\textsuperscript{2} + 14x + 2x + 28 = 0 \\$ \Rightarrow $ x(x + 14) + 2(x + 14) = 0 \\$ \Rightarrow $ (x + 2)(x + 14) = 0 \\$ \Rightarrow $ (x + 2) = 0 or (x + 14) = 0 \\$ \Rightarrow $ x = -2 or x = -14

Thus, x = -2, -14.

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