If A = [2x01]and B =[43601],\begin{bmatrix}[r] 2 & x \ 0 & 1 \end{bmatrix} \text{and B } = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix},[20x1]and B =[40361], find the value of x, given that A2 = B.
14 Likes
Given,
A2 = B
⇒[2x01][2x01]=[43601]⇒[2×2+x×02×x+x×10×2+1×00×x+1×1]=[43601]⇒[4+02x+x0+00+1]=[43601]⇒[43x01]=[43601]\Rightarrow \begin{bmatrix}[r] 2 & x \ 0 & 1 \end{bmatrix} \begin{bmatrix}[r] 2 & x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times 2 + x \times 0 & 2 \times x + x \times 1 \ 0 \times 2 + 1 \times 0 & 0 \times x + 1 \times 1 \end{bmatrix} = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 + 0 & 2x + x \ 0 + 0 & 0 + 1 \end{bmatrix} = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 3x \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 4 & 36 \ 0 & 1 \end{bmatrix} \\[1em]⇒[20x1][20x1]=[40361]⇒[2×2+x×00×2+1×02×x+x×10×x+1×1]=[40361]⇒[4+00+02x+x0+1]=[40361]⇒[403x1]=[40361]
By definition of equality of matrices we get,
⇒ 3x = 36∴ x = 12.
Hence, the value of x = 12.
Answered By
6 Likes
If [3425]=[abcd][1001],\begin{bmatrix}[r] 3 & 4 \ 2 & 5 \end{bmatrix} = \begin{bmatrix}[r] a & b \ c & d \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix},[3245]=[acbd][1001], write down the values of a, b, c and d.
Find the value of x given that A2 = B where,
A=[21201] and B=[4x01].A = \begin{bmatrix}[r] 2 & 12 \ 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 4 & x \ 0 & 1 \end{bmatrix}.A=[20121] and B=[40x1].
If A = [3x01] and B=[9160−y],\begin{bmatrix}[r] 3 & x \ 0 & 1 \end{bmatrix} \text{ and B} = \begin{bmatrix}[r] 9 & 16 \ 0 & -y \end{bmatrix},[30x1] and B=[9016−y], find x and y when A2 = B.
Find x, y if [−2031][−12x]+3[−21]=2[y3].\begin{bmatrix}[r] -2 & 0 \ 3 & 1 \end{bmatrix} \begin{bmatrix}[r] -1 \ 2x \end{bmatrix} + 3\begin{bmatrix}[r] -2 \ 1 \end{bmatrix} = 2\begin{bmatrix}[r] y \ 3 \end{bmatrix}.[−2301][−12x]+3[−21]=2[y3].
