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Find A, if \left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right] \mathrm{A}=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right].

Answers (1)

We have the matrix,

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right] \mathrm{A}=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]

We need to find the matrix A.

Let us check what the order of the given matrices is.

We know that order of a matrix is the number of rows and columns in a matrix.

If a given matrix has M rows and N columns, the order of matrix is M × N.

Order of  \left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]= X($say)

Number of rows = 3

$ \Rightarrow $ M = 3

Number of column = 1

$ \Rightarrow $ N = 1

Then, order of matrix X =M $ \times $ N

$ \Rightarrow $ Order of matrix X = 3 $ \times $ 1

Order of \left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]=\mathrm{Y}(\text { say })

Number of rows = 3

$ \Rightarrow $ M = 3

Number of columns = 3

$ \Rightarrow $  N = 3

Then, order of matrix Y = M $ \times $ N

$ \Rightarrow $  Order of matrix Y = 3 $ \times $ 3

We must note that, when a matrix of order 1 $ \times $ 3  is multiplied to the matrix X, only then matrix Y is produced.

Let matrix A be of order 1 $ \times $ 3, and can be represented as

A=\left[\begin{array}{lll}a & b & c\end{array}\right]$ Then, we have $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c}\end{array}\right]=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]$ Take $\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]\left[\begin{array}{lll}a & b & c\end{array}\right]$

In order to carry out the multiplication of two matrices, A and B, the number of columns in A must equal 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.

So, we get,

\text { X. } A=\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]

Multiply 1st row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.

(4)(a) = 4a

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & \\ & \end{bmatrix}

Multiply 1st row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.

(4)(b) = 4b

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b \\ & \end{bmatrix}

Multiply 1st row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.

(4)(c) = 4c

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ & \end{bmatrix}

Multiply 2nd row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.

(1)(a) = a

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& \end{bmatrix}

Multiply 2nd row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.

(1)(b) = b

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b\end{bmatrix}

Multiply 2nd row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.

(1)(c) = c

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\end{bmatrix}

Multiply 3rd row of matrix X by matching member of 1st column of matrix A, then finally end by summing it up.

(3)(a) = 3a

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & & \end{bmatrix}

Multiply 3rd row of matrix X by matching member of 2nd column of matrix A, then finally end by summing it up.

(3)(b) = 3b

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & 3b & \end{bmatrix}

Multiply 3rd row of matrix X by matching member of 3rd column of matrix A, then finally end by summing it up.

(3)(c) = 3c

\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \text { a } & \text { b } & c \end{array}\right]= \begin{bmatrix} 4a & 4b & 4c\\ a& b & c\\ 3a & 3b &3c \end{bmatrix}

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

\begin{array}{l} \Rightarrow\left[\begin{array}{l} 4 \\ 1 \\ 3 \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & \mathrm{c} \end{array}\right]=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ccc} 4 \mathrm{a} & 4 \mathrm{~b} & 4 \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \mathrm{c} \\ 3 \mathrm{a} & 3 \mathrm{~b} & 3 \mathrm{c} \end{array}\right]=\left[\begin{array}{lll} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right] \end{array}

Since, the matrices have the same order, we can say,

\\4a = -4 $ \ldots $ (i) \\4b = 8 $ \ldots $ (ii) \\4c = 4 $ \ldots $ (iii) \\a = -1 $ \ldots $ (iv) \\b = 2 $ \ldots $ (v) \\c = 1 $ \ldots $ (vi) \\3a = -3 $ \ldots $ (vii) \\3b = 6 $ \ldots $ (viii) \\3c = 3 $ \ldots $ (ix)

From equation (i), we can determine the value of a,

4a = -4

$ \Rightarrow $ a = -1

From equation (ii), we can determine the value of b,

4b = 8

$ \Rightarrow $ b = 2

From equation (iii), we can determine the value of c,

4c = 4

$ \Rightarrow $ c = 1

And it will satisfy other equations (iv), (v), (vi), (vii), (viii) and (ix) too.

Thus, the matrix A is

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

Posted by

infoexpert22

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