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Mathematics

Find matrix X such that [5−7−23]\begin{bmatrix} 5 & -7 \ -2 & 3 \end{bmatrix} X = [−16−672]\begin{bmatrix} -16 & -6 \ 7 & 2 \end{bmatrix}.

Matrices

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Answer

Let Y = [5−7−23]\begin{bmatrix} 5 & -7 \ -2 & 3 \end{bmatrix}.

Then, YX = [−16−672]\begin{bmatrix} -16 & -6 \ 7 & 2 \end{bmatrix}

Since YX exists, we have:

Number of rows of X = Number of columns in Y = 2

Number of columns of X = Number of columns in YX = 2

Order of X is 2 × 2.

Let X = [abcd]\begin{bmatrix} a & b \ c & d \end{bmatrix}.

Then,

⇒[5−7−23]×[abcd]=[−16−672]⇒[5a−7c5b−7d−2a+3c−2b+3d]=[−16−672]\Rightarrow \begin{bmatrix} 5 & -7 \ -2 & 3 \end{bmatrix} \times \begin{bmatrix} a & b \ c & d \end{bmatrix}= \begin{bmatrix} -16 & -6 \ 7 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5a - 7c & 5b - 7d \ -2a + 3c & -2b + 3d \end{bmatrix} = \begin{bmatrix} -16 & -6 \ 7 & 2 \end{bmatrix} \\[1em]

Solving for a and c :

∴ -2a + 3c = 7

⇒ 2a = 3c - 7

⇒ a = 3c−72\dfrac{3c - 7}{2} …(1)

∴ 5a - 7c = -16 …….(2)

Substituting value of a from equation (1) in (2), we get :

⇒5(3c−72)−7c=−16⇒(15c−352)−7c=−16⇒15c−352−7c×22=−16⇒15c−35−14c2=−16⇒c−352=−16⇒c−35=−16×2⇒c−35=−32⇒c=−32+35⇒c=3.\Rightarrow 5\Big(\dfrac{3c - 7}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \Big(\dfrac{15c - 35}{2}\Big) - 7c = -16 \\[1em] \Rightarrow \dfrac{15c - 35}{2} - \dfrac{7c \times 2}{2} = -16 \\[1em] \Rightarrow \dfrac{15c - 35 - 14c}{2} = -16 \\[1em] \Rightarrow \dfrac{c - 35}{2} = -16 \\[1em] \Rightarrow c - 35 = -16 \times 2 \\[1em] \Rightarrow c - 35 = -32 \\[1em] \Rightarrow c = -32 + 35 \\[1em] \Rightarrow c = 3.

Substituting value of c in equation (1), we get:

⇒ a = 3(3)−72\dfrac{3(3) - 7}{2}

⇒ a = 9−72\dfrac{9 - 7}{2}

⇒ a = 22\dfrac{2}{2}

⇒ a = 1.

Solving for b and d:

∴ -2b + 3d = 2

⇒ 2b = 3d - 2

⇒ b = 3d−22\dfrac{3d - 2}{2} ….(3)

∴ 5b - 7d = -6 ……(4)

Substituting value of b from equation (3) in (4), we get:

⇒5(3d−22)−7d=−6⇒(15d−102)−7d=−6⇒15d−102−7d×22=−6⇒15d−10−14d2=−6⇒d−102=−6⇒d−10=−6×2⇒d−10=−12⇒d=−12+10⇒d=−2.\Rightarrow 5\Big(\dfrac{3d - 2}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \Big(\dfrac{15d - 10}{2}\Big) - 7d = -6 \\[1em] \Rightarrow \dfrac{15d - 10}{2} - \dfrac{7d \times 2}{2} = -6 \\[1em] \Rightarrow \dfrac{15d - 10 - 14d}{2} = -6 \\[1em] \Rightarrow \dfrac{d - 10}{2} = -6 \\[1em] \Rightarrow d - 10 = -6 \times 2 \\[1em] \Rightarrow d - 10 = -12 \\[1em] \Rightarrow d = -12 + 10 \\[1em] \Rightarrow d = -2.

Substituting value of d in equation (3), we get:

⇒ b = 3(−2)−22\dfrac{3(-2) - 2}{2}

⇒ b = −6−22\dfrac{-6 - 2}{2}

⇒ b = −82\dfrac{-8}{2}

⇒ b = -4.

∴X=[1−43−2]\therefore X = \begin{bmatrix} 1 & -4 \ 3 & -2 \end{bmatrix}.

Hence, X = [1−43−2]\begin{bmatrix} 1 & -4 \ 3 & -2 \end{bmatrix}.

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