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  • If A = [3 5], B = [7 3], then find a non-zero matrix C such that AC = BC.

If A = [3\: \: 5], B = [7\: \: 3], then find a non-zero matrix C such that AC = BC.

Answers (1)

We have the given matrices A and B, such that

A = [3\: \: 5], B = [7\: \: 3]

We need to find matrix C, such that AC = BC.

Let C be a non-zero matrix of order 2 × 1, such that

C=\begin{bmatrix} X\\Y \end{bmatrix}

But order of C can be 2 × 1, 2 × 2, 2 × 3, 2 × 4, …

[ if we have to multiply two given 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.

∴, number of columns in matrix A = number of rows in matrix C = 2]

Take AC.

AC=\begin{bmatrix} 3 &5 \end{bmatrix}\begin{bmatrix} X\\Y \end{bmatrix}

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

\begin{aligned} &(3,5)(x, y)=(3 \times x)+(5 \times y)\\ &\Rightarrow(3,5)(x, y)=3 x+5 y\\ &\left[\begin{array}{ll} 3 & 5 \end{array}\right]\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \end{array}\right]=[3 \mathrm{x}+5 \mathrm{y}]\\ &\Rightarrow A C=[3 x+5 y]\\ &\text { Now, take BC. }\\ &\mathrm{BC}=\left[\begin{array}{ll} 7 & 3 \end{array}\right]\left[\begin{array}{l} \mathrm{X} \\ \mathrm{y} \end{array}\right] \end{aligned}

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

\begin{array}{l} (7,3)(x, y)=(7 \times x)+(3 \times y) \\ \Rightarrow(7,3)(x, y)=7 x+3 y \\ {[7 \quad 3]\left[\begin{array}{l} x \\ y \end{array}\right]=[7 x+3 y]} \\ \Rightarrow B C=[7 x+3 y] \end{array}

 

And,

AC = BC

 \\$ \Rightarrow $ [3x + 5y] = [7x + 3y] \\$ \Rightarrow $ 3x + 5y = 7x + 3y \\$ \Rightarrow $ 7x - 3x = 5y - 3y \\$ \Rightarrow $ 4x = 2y \\$ \Rightarrow $ y = 2x

Then, we have,

$$ C=\left[\begin{array}{l} x \\ 2 x \end{array}\right] $$
since, $\mathrm{C}$$ is of orders, $2 \times 1,2 \times 2,2 \times 3, \ldots$
C=\left[\begin{array}{c}x \\ 2 x\end{array}\right]=\left[\begin{array}{cc}x & x \\ 2 x & 2 x\end{array}\right]=\left[\begin{array}{ccc}x & x & x \\ 2 x & 2 x & 2 x\end{array}\right]=\cdots$
In general,
 $$ C=\left[\begin{array}{l} \mathrm{k} \\ 2 \mathrm{k} \end{array}\right]=\left[\begin{array}{ll} \mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{k} & \mathrm{k} & \mathrm{k} \\ 2 \mathrm{k} & 2 \mathrm{k} & 2 \mathrm{k} \end{array}\right]=\cdots $$
Where, k can be any real number.

Posted by

infoexpert22

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