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  • If P =[x 0 0 0 y 0 0 0 z] , prove that PQ =[xa 0 0 ] =QP

 \text{If } P=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right], Q=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right] prove that P Q=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right]=Q P.

Answers (1)

We have the following given matrices P and Q, such that

P=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right], Q=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]

We have to prove that: 

\quad\\ P Q=\left[\begin{array}{ccc} x a & 0 & 0 \\ 0 & y b & 0 \\ 0 & 0 & z c \end{array}\right]=Q P

Proof: First, we shall compute PQ.

\mathrm{PQ}=\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]

 

For carrying out the multiplication of two matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

Order of P = 3 × 3

And order of Q = 3 × 3

Number of columns of matrix P = Number of rows of matrix Q = 3

So, P and Q can be multiplied.

So, multiply 1st row of matrix P by matching members of 1st column of matrix Q, then finally sum them up.

(x, 0, 0)(a, 0, 0) = (x × a) + (0 × 0) + (0 × 0)

⇒ (x, 0, 0)(a, 0, 0) = xa

\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right] = \begin{bmatrix} xa & & \\ & & \\ & & \end{bmatrix}

Multiply 1st row of matrix P by matching members of 2nd column of matrix Q, then finally sum them up

\\(x, 0, 0)(a, 0, 0) = (x $ \times $ a) + (0 $ \times $ 0) + (0 $ \times $ 0) \\$ \Rightarrow $ (x, 0, 0)(a, 0, 0) = xa

\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right] = \begin{bmatrix} xa & 0& \\ & & \\ & & \end{bmatrix}

Similarly, let us fill for other elements.

\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]=\left[\begin{array}{ccc} \text { Xa } & 0 & (\mathrm{x} \times 0)+(0 \times 0)+(0 \times \mathrm{c}) \\ (0 \times \mathrm{a})+(\mathrm{y} \times 0)+(0 \times 0) & (0 \times 0)+(\mathrm{y} \times \mathrm{b})+(0 \times 0) & (0 \times 0)+(\mathrm{y} \times 0)+(0 \times \mathrm{c}) \\ (0 \times \mathrm{a})+(0 \times 0)+(\mathrm{z} \times 0) & (0 \times 0)+(0 \times \mathrm{b})+(\mathrm{z} \times 0) & (0 \times 0)+(0 \times 0)+(\mathrm{z} \times \mathrm{c}) \end{array}\right]

\Rightarrow\left[\begin{array}{ccc}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\left[\begin{array}{ccc}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0+0+0 \\ 0+0+0 & 0+\mathrm{yb}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+\mathrm{z} \mathrm{c}\end{array}\right]$ $\Rightarrow\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{C}\end{array}\right]$ \\So,\\ $\mathrm{PQ}=\left[\begin{array}{ccc}\mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{C}\end{array}\right] \ldots$

Now, we shall compute QP.

\mathrm{QP}=\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]\\$ Multiply $1^{\text {st }}$ row of matrix $\mathrm{Q}$ by matching members of $1^{\text {st }}$ column of matrix $\mathrm{P}$, then finally sum them up.\\ $(\mathrm{a}, 0,0)(\mathrm{x}, 0,0)=(\mathrm{a} \times \mathrm{x})+(0 \times 0)+(0 \times 0)$ \\$\Rightarrow(a, 0,0)(x, 0,0)=x a+0+0$ \\$\Rightarrow(a, 0,0)(x, 0,0)=x a$ $\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]\left[\begin{array}{lll}\mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z}\end{array}\right]=[\mathrm{xa}

Similarly, let us fill the other elements.

\begin{aligned} &\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]\\ &=\left[\begin{array}{ccc} \text { xa } & (a \times 0)+(0 \times y)+(0 \times 0) & (a \times 0)+(0 \times 0)+(0 \times z) \\ (0 \times x)+(b \times 0)+(0 \times 0) & (0 \times 0)+(b \times y)+(0 \times 0) & (0 \times 0)+(b \times 0)+(0 \times z) \\ (0 \times x)+(0 \times 0)+(c \times 0) & (0 \times 0)+(0 \times y)+(c \times 0) & (0 \times 0)+(0 \times 0)+(c \times z) \end{array}\right]\\ \end{aligned}

\begin{aligned} &\Rightarrow\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{xa} & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+\mathrm{yb}+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+\mathrm{z} \mathrm{c} \end{array}\right]\\ &\Rightarrow\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c} \end{array}\right]\left[\begin{array}{lll} \mathrm{x} & 0 & 0 \\ 0 & \mathrm{y} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right]=\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \mathrm{c} \end{array}\right] \end{aligned}

\begin{aligned} &\text { So, }\\ &\begin{aligned} \mathrm{QP}=&\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{z} \end{array}\right] \\ \mathrm{Thus}, &PQ=\left[\begin{array}{ccc} \mathrm{xa} & 0 & 0 \\ 0 & \mathrm{yb} & 0 \\ 0 & 0 & \mathrm{zc} \end{array}\right]=\mathrm{QP} \end{aligned} \end{aligned}

 

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