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Problem 111

Let \((\mathrm{r} \times \mathrm{s})\) denote a matrix with shape $(\mathrm{r} \times \mathrm{s})$. Find the shape of the following products if the product is defined (i) \((2 \times 3)(3 \times 4)\) (iii) \((1 \times 2)(3 \times 1)\) (v) \((3 \times 4)(3 \times 4)\) (ii) \((4 \times 1)(1 \times 2)\) (iv) \((5 \times 2)(2 \times 3)\) (vi) \((2 \times 2)(2 \times 4)\)

Short Answer

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(i) \((2 \times 4)\) (iii) Not defined (v) Not defined (ii) \((4 \times 2)\) (iv) \((5 \times 3)\) (vi) \((2 \times 4)\)
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Step 1: (i) \((2 \times 3)(3 \times 4)\)

In this case, matrix A has 2 rows and 3 columns, and matrix B has 3 rows and 4 columns. Since the number of columns in A (3) is equal to the number of rows in B (3), we can multiply these matrices. The resulting matrix will have 2 rows (from A) and 4 columns (from B), so the shape is \((2\times 4)\).

Step 2: (iii) \((1 \times 2)(3 \times 1)\)

Matrix A has 1 row and 2 columns, while matrix B has 3 rows and 1 column. The number of columns in A (2) is not equal to the number of rows in B (3), so the product is not defined.

Step 3: (v) \((3 \times 4)(3 \times 4)\)

Both matrices A and B have 3 rows and 4 columns. The number of columns in A (4) is not equal to the number of rows in B (3), so the product is not defined.

Step 4: (ii) \((4 \times 1)(1 \times 2)\)

Matrix A has 4 rows and 1 column, while matrix B has 1 row and 2 columns. Since the number of columns in A (1) is equal to the number of rows in B (1), we can multiply these matrices. The resulting matrix will have 4 rows (from A) and 2 columns (from B), so the shape is \((4\times 2)\).

Step 5: (iv) \((5 \times 2)(2 \times 3)\)

Matrix A has 5 rows and 2 columns, while matrix B has 2 rows and 3 columns. Since the number of columns in A (2) is equal to the number of rows in B (2), we can multiply these matrices. The resulting matrix will have 5 rows (from A) and 3 columns (from B), so the shape is \((5\times 3)\).

Step 6: (vi) \((2 \times 2)(2 \times 4)\)

Matrix A has 2 rows and 2 columns, while matrix B has 2 rows and 4 columns. Since the number of columns in A (2) is equal to the number of rows in B (2), we can multiply these matrices. The resulting matrix will have 2 rows (from A) and 4 columns (from B), so the shape is \((2\times 4)\).

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Most popular questions from this chapter

Chapter 4

Find (i) \(A^{2}\) (ii) \(A^{3}\) (iii) \(A^{4}\) when $A=\begin{array}{rl}1 & 2 \mid \\ & \mid-1\end{array}$

Chapter 4

Show that in matrix arithmetic we can have the following: (a) \(\quad \mathrm{AB} \neq \mathrm{BA}\). (b) \(\quad \mathrm{A} \neq 0, \mathrm{~B} \neq 0\) and yet, \(\mathrm{AB}=0\). (c) \(\quad A \neq 0\) and \(A^{2}=0\). (d) \(\quad \mathrm{A} \neq 0, \mathrm{~A}^{2} \neq 0\) and \(\mathrm{A}^{3}=0\). (e) \(\mathrm{A}^{2}=\mathrm{A}\) with \(\mathrm{A} \neq 0\) and $\mathrm{A} \neq \mathrm{I}$. (f) \(\quad A^{2}=\) I with \(A \neq I\) and \(A \neq-I\).

Chapter 4

a) If \(\mathrm{A}=\left(\mathrm{a}_{\mathrm{ij}}\right)\) is a $\mathrm{p} \times \mathrm{q}\( matrix and \)\mathrm{B}=\left(\mathrm{b}_{i j}\right)$ is a \(\mathrm{q} \times \mathrm{r}\) matrix prove \(A B\) is the \(p \times r\) matrix \(\left(c_{i j}\right)\) where $c_{i j}={ }^{q} \sum_{k=1} a_{i k} b_{k j}, \quad i=1,2, \ldots, p$ \(\mathrm{J}=1,2, \ldots, \mathrm{r}\) b) If $\mathrm{A}=\begin{array}{ccccc} & 2 & 1 & 1 \mid & \mid 2 & 1 \mid \\\ & \mid-1 & 2 & 3 \mid & \text { and } B=1-1 & 1 \mid, \text { find } A B \text { . } \\ & \mid 1 & 0 & 1 \mid & \mid 2 & -1 \mid\end{array}$

Chapter 4

If $\mathrm{A}=\begin{array}{cccccc}2 & -2 & 4 \mid & \text { and } & \mathrm{B}=10 & 1 & -3\end{array} \mid$ \(\mid-1 \quad 1\) \(1 \mid\) 113 find \(2 \mathrm{~A}+\mathrm{B}\).

Chapter 4

Prove \((\mathrm{AB}) \mathrm{C}=\mathrm{A}(\mathrm{BC})\) where $\mathrm{A}=\mid 5 \quad \begin{array}{ccc}12 & 2 & 3 \mid \\ -3 & 4 \mid,\end{array}$ $\begin{array}{rlrr}\mathrm{B}= & 12 & -1 & 1 & 0 \\ 10 & 2 & 2 & 21 \\ 13 & 0 & -1 & 31\end{array}$ and $\quad C=\begin{array}{ccc}11 & 0 & 2 \mid \\ 12 & -3 & 0 \mid \\ 12 & 1 & 0 \mid\end{array}$

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