Given A=[2−620],B=[−3240] and C=[4002].A = \begin{bmatrix}[r] 2 & -6 \ 2 & 0 \end{bmatrix}, B = \begin{bmatrix}[r] -3 & 2 \ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix}[r] 4 & 0 \ 0 & 2 \end{bmatrix}.A=[22−60],B=[−3420] and C=[4002].
Find the matrix X such that A + 2X = 2B + C.
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Putting values of A, B and C in A + 2X = 2B + C,
⇒[2−620]+2X=2[−3240]+[4002]⇒2X=[−6480]+[4002]−[2−620]⇒2X=[−6+4−24+0−(−6)8+0−20+2−0]⇒X=12[−41062]⇒X=[−2531]\Rightarrow \begin{bmatrix}[r] 2 & -6 \ 2 & 0 \end{bmatrix} + 2X = 2\begin{bmatrix}[r] -3 & 2 \ 4 & 0 \end{bmatrix} + \begin{bmatrix}[r] 4 & 0 \ 0 & 2 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] -6 & 4 \ 8 & 0 \end{bmatrix} + \begin{bmatrix}[r] 4 & 0 \ 0 & 2 \end{bmatrix} - \begin{bmatrix}[r] 2 & -6 \ 2 & 0 \end{bmatrix} \\[1em] \Rightarrow 2X = \begin{bmatrix}[r] -6 + 4 - 2 & 4 + 0 - (-6) \ 8 + 0 - 2 & 0 + 2 - 0 \end{bmatrix} \\[1em] \Rightarrow X = \dfrac{1}{2}\begin{bmatrix}[r] -4 & 10 \ 6 & 2 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -2 & 5 \ 3 & 1 \end{bmatrix}⇒[22−60]+2X=2[−3420]+[4002]⇒2X=[−6840]+[4002]−[22−60]⇒2X=[−6+4−28+0−24+0−(−6)0+2−0]⇒X=21[−46102]⇒X=[−2351]
Hence, matrix X = [−2531]\begin{bmatrix}[r] -2 & 5 \ 3 & 1 \end{bmatrix}[−2351].
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Solve the matrix equation [2150]−3X=[−7426]\begin{bmatrix} 2 & 1 \ 5 & 0 \end{bmatrix} - 3X = \begin{bmatrix}[r] -7 & 4 \ 2 & 6 \end{bmatrix}[2510]−3X=[−7246]
If [14−23]+2M=3[320−3],\begin{bmatrix}[r] 1 & 4 \ -2 & 3 \end{bmatrix} + 2\text{M} = 3\begin{bmatrix}[r] 3 & 2 \ 0 & -3 \end{bmatrix}, \\[1em][1−243]+2M=3[302−3], find the matrix M.
Find X and Y if
X + Y=[7025] and X - Y=[3003].\text{X + Y} = \begin{bmatrix} 7 & 0 \ 2 & 5 \end{bmatrix} \text{ and X - Y} = \begin{bmatrix} 3 & 0 \ 0 & 3 \end{bmatrix}.X + Y=[7205] and X - Y=[3003].
If 2[x79y−5]+[6−745]=[1072215],2\begin{bmatrix}[r] x & 7 \ 9 & y - 5 \end{bmatrix} + \begin{bmatrix}[r] 6 & -7 \ 4 & 5 \end{bmatrix} = \begin{bmatrix}[r] 10 & 7 \ 22 & 15 \end{bmatrix},2[x97y−5]+[64−75]=[1022715], find the values of x and y.
