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  • Find the matrix A such that [2 -1 1 0 -3 4 ]A = [-1 -8 -10]

Find the matrix A such that \left[\begin{array}{rr} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right] A=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]

 

Answers (1)

We are given that,

\left[\begin{array}{rr} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right] A=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]

As A is multiplied with a matrix of order 3×2 and gives a resultant matrix of order 3×3

For matrix multiplication to be possible A must have 2 rows and as resultant matrix is of 3rd order A must have 3 columns

∴ A is matrix of order 2×3

Let A = \left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & c \\ \mathrm{~d} & \mathrm{e} & \mathrm{f} \end{array}\right] where a, b, c, d, e and f are unknown variables.

\begin{aligned} &\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & \mathrm{b} & c \\ \mathrm{~d} & \mathrm{e} & \mathrm{f} \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \end{aligned}

∴ According to the rule of matrix multiplication we have-

\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right]

By equating the elements of 2 equal matrices, as both the matrices are equal to each other, we get-

a = 1 ; b = -2 and c = -5

also, we have,

\\2a - d = -1 \Rightarrow d = 2a + 1 = 2 + 1 = 3 \\ \therefore d = 3 \\2b - e = -8 \Rightarrow e = 2b + 8 = -4 + 8 = 4 \\ \therefore e = 4 \\Similarly, f = 2c + 10 = 0

\\ \therefore A = \begin{bmatrix} 1 &-2 &-5 \\3 &4 &0 \end{bmatrix}

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