The number of matrices A = <math><mrow><mrow><mo>(</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mi>a</mi></mtd><mtd columnalign="center"><mi>b</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>c</mi></mtd><mtd columnalign="center"><mi>d</mi></mtd></mtr></mtable></mrow><mo>)</mo></mrow><mo>,</mo></mrow></math> where a, b, c, d <math><mrow><mo>∈</mo></mrow></math> <math><mrow><mrow><mo>{</mo><mrow><mo>−</mo><mn>1</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>0</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>1</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>2</mn><mo>,</mo><mtext> </mtext><mtext> </mtext><mn>3</mn><mo>,</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>,</mo><mn>1</mn><mn>0</mn></mrow><mo>}</mo></mrow><mo>,</mo></mrow></math> such that A = A1, is……..

Ask & Answer Question

Maths Class 12th Matrices Maths NCERT Exemplar Solutions Class 12th Chapter Three

The number of matrices A = $(\begin{matrix} a & b \\ c & d \end{matrix}),$ where a, b, c, d $\in$ ${- 1, 0, 1, 2, 3, . . . . . . ., 10},$ such that A = A1, is……..

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

A = $(\begin{matrix} a & b \\ c & d \end{matrix})$

$A^{2} = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} a^{2} + b c & a b + b d \\ a c + d c & a c + d^{2} \end{matrix})$

a2 + bc = bc + d2 = 1 ………. (i)

and b (a + d) = c (a + d) = 0 ……… (ii)

Case 1

b = c = 0

then possible ordered pair of

(a, d) $\equiv$ (1, 1) (-1, -1) (-1, 1) (1, -1) total 4 possible case

Case 2

a = -d

then (a, d) $\equiv$ (-1, 1) (1, -1)

then bc = 0

now if b = 0

then possible choice for {-1, 0, 1, 2, …….10} = 12

Similarly if c = 0 then possible choice for $b \in {- 1, 0, 1, 2, . . . . . . 10}$ is = 12

but (0, 0) counted twice

$∴$ bc = 0 in (12 + 12 – 1) = 23 ways

$∴$ total number of ways = 2 × 23 = 46

$∴$ total number of required matrices = 46 + 4 = 50

A = <math><mrow><mrow><mo> (</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mi>a</mi></mtd><mtd columnalign="center"><mi>b</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>c</mi></mtd><mtd columnalign="center"><mi>d</mi></mtd></mtr></mtable></mrow><mo>)</mo></mrow></mrow></math><math><mrow><msup><mrow><mi>A</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>=</mo><mrow><mo> (</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mi>a</mi></mtd><mtd columnalign="center"><mi>b</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>c</mi></mtd><mtd columnalign="center"><mi>d</mi></mtd></mtr></mtable></mrow><mo>)</mo></mrow><mrow><mo> (</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><mi>a</mi></mtd><mtd columnalign="center"><mi>b</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>c</mi></mtd><mtd columnalign="center"><mi>d</mi></mtd></mtr></mtable></mrow><mo>)</mo></mrow><mo>=</mo><mrow><mo> (</mo><mrow><mtable><mtr columnalign="center"><mtd columnalign="center"><msup><mrow><mi>a</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><mi>b</mi><mi>c</mi></mtd><mtd columnalign="center"><mi>a</mi><mi>b</mi><mo>+</mo><mi>b</mi><mi>d</mi></mtd></mtr><mtr columnalign="center"><mtd columnalign="center"><mi>a</mi><mi>c</mi><mo>+</mo><mi>d</mi><mi>c</mi></mtd><mtd columnalign="center"><mi>a</mi><mi>c</mi><mo>+</mo><msup><mrow><mi>d</mi></mrow><mrow><mn>2</mn></mrow></msup></mtd></mtr></mtable></mrow><mo>)</mo></mrow></mrow></math>a2 + bc = bc + d2 = 1 ………. (i)and b (a + d) = c (a + d) = 0 ……… (ii)Case 1b = c = 0then possible ordered pair of (a, d) <math><mrow><mo>≡</mo></mrow></math>  (1, 1) (-1, -1) (-1, 1) (1, -1) total 4 possible caseCase 2a = -dthen (a, d) <math><mrow><mo>≡</mo></mrow></math>  (-1, 1) (1, -1)then bc = 0now if b = 0then possible choice for {-1, 0, 1, 2, …….10} = 12Similarly if c = 0 then possible choice for <math><mrow><mi>b</mi><mo>∈</mo><mrow><mo> {</mo><mrow><mo>−</mo><mn>1</mn><mo>, </mo><mn>0</mn><mo>, </mo><mn>1</mn><mo>, </mo><mn>2</mn><mo>, </mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mn>1</mn><mn>0</mn></mrow><mo>}</mo></mrow></mrow></math> is = 12but (0, 0) counted twice<math><mrow><mo>∴</mo></mrow></math> bc = 0 in (12 + 12 – 1) = 23 ways<math><mrow><mo>∴</mo></mrow></math> total number of ways = 2 × 23 = 46<math><mrow><mo>∴</mo></mrow></math> total number of required matrices = 46 + 4 = 50

0 Upvotes 0 Downvotes

The number of matrices A = $(\begin{matrix} a & b \\ c & d \end{matrix}),$ where a, b, c, d $\in$ ${- 1, 0, 1, 2, 3, . . . . . . ., 10},$ such that A = A1, is……..

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

The number of matrices A = (abcd), where a, b, c, d ∈ {−1, 0, 1, 2, 3,.......,10}, such that A = A1, is……..

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

The number of matrices A = $(\begin{matrix} a & b \\ c & d \end{matrix}),$ where a, b, c, d $\in$ ${- 1, 0, 1, 2, 3, . . . . . . ., 10},$ such that A = A1, is……..