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  • Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question13

Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 13

Answers (1)

Answer:

Hence, prove A B=B A=0_{3 \times 3}

Hint: matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: A=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right], B=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]

Consider,

A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{lll}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right] \\\\ \\ A B=\left[\begin{array}{ccc}0\left(a^{2}\right)+c(a b)+(-b)(a c) & (0)(a b)+(c)\left(b^{2}\right)+(-b)(b c) & (0)(a c)+c(b c)+(-b)\left(c^{2}\right) \\ (-c)\left(a^{2}\right)+(0)(a b)+a(a c) & (-c)(a b)+(0)\left(b^{2}\right)+a(b c) & (-c)(a c)+0(b c)+a\left(c^{2}\right) \\ b\left(a^{2}\right)+(-a)(a b)+0(a c) & (b)(a b)+(-a)\left(b^{2}\right)+0(b c) & (b)(a c)+(-a)(b c)+o\left(c^{2}\right)\end{array}\right]\) \\\\ \\A B=\left[\begin{array}{ccc}0+a b c-a b c & 0+b^{2} c-b^{2} c & 0+b c^{2}-b c^{2} \\ -a^{2} c+0+a^{2} c & -a b c+0+a b c & -a c^{2}+0+a c^{2} \\ a^{2} b-a^{2} b+0 & a b^{2}-a b^{2}+0 & a b c-a b c+0\end{array}\right]

A B=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] . . . . (i)

Again consider,

 

B A=\left[\begin{array}{ccc}a^{2} & a b & a c \\ a b & b^{2} & b c \\ a c & b c & c^{2}\end{array}\right]\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right] \\\\ \\ B A=\left[\begin{array}{ccc}\left(a^{2}\right)(0)+(a b)(-c)+(a c)(b) & a^{2}(c)+a b(0)+a c(-a) & a^{2}(-b)+a b(a)+a c(0) \\ a b(0)+b^{2}(-c)+b c(b) & a b(c)+b^{2}(0)+b c(-a) & a b(-b)+b^{2}(a)+b c(0) \\ a c(0)+b c(-c)+c^{2}(b) & a c(c)+b c(0)+c^{2}(-a) & a c(-b)+b c(a)+c^{2}(0)\end{array}\right] \\\\\\ B A=\left[\begin{array}{ccc}0-a b c+a b c & a^{2} c+0-a^{2} c & -a^{2} b+a^{2} b+0 \\ 0-b^{2} c+b^{2} c & a b c+0-a b c & -a b^{2}+a b^{2}+0 \\ 0-b c^{2}+b c^{2} & a c^{2}+0-a c^{2} & -a b c+a b c+0\end{array}\right] \\\\\\ B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \quad \ldots(i i)

From equation (i) & (ii)

A B=B A=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \\\\ A B=B A=0_{3 \times 3}

Hence, proved

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